Business Mathematics and Statistics 9788173819759

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Business Mathematics and Statistics (Algebra, Geometry and Business Statistics)

Ram Krishna Ghosh Suranjan Saha

BUSINESS MATHEMATICS AND STATISTICS [For BCom First Year General & Honours Students]

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Written according to the BCom, First Year New Syllabus ofCalcutta University. ALro covers the syllabi of Vidyasagar, North Bengal and Burdwan Universities. Useful to the students ofMBA, CA, ICWAI and other professional courses as well.

BUSINESS MATHEMATICS AND STATISTICS [For BCom First Year General and Honours Students and Other Professional Examinations.]

Ram Krishna Ghosh Formerly, Head, Department of Mathematics, St. Xavier's College, Kolkata Sometime Senior Lecrurer in Mathematics, University of Cape Coast, Ghana Sometime Pan-time Lecturer in Mathematics, University of Kalyani, Kalyani, WB Formerly, President, Institute of Modern Management, Kolkata and Director, Academy of Civil Services, Kolkata

Suranjan Saha Formerly, Head, Department of Mathematics & Statistics, Heramba Chandra College [City College, South Kolkata] Formerly, Pan-time Lecturer in Mathematics & Statistics, Sivanath Sastri College and Shri Shikshayatan College, Kolkata Sometime Part-rime Honorary Lecturer in Business Mathematics & Statistics, ICWAI

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BUSINESS MATHEMATICS AND STATISTICS • Ram Krishna Ghosh, Suranjan Saha

© Copyright reserved by the Authors Publication, Distribution, and Promotion Rights reserved by the Publisher All rights reserved. No part of the text in general, and the figures, diagrams, page layout, and cover design in particular, may be reproduced or transmitted in any form or by any means-electronic, mechanical, photocopying, recording, or by any information storage and retrieval system-without the prior written permission of the Publisher First Published: 1979 Thirteenth Edition: 2010 Reprinted: 2011, 2012, 2013, 2014 Fourteenth Revised Edition (New Syllabus): 2015 Repri11ted: 2016 PUBLISHER

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CONTENTS

Preface to the Thirteenth Edition

vii

Preface to the Eleventh Edition

vii viii "ix-x

Preface to the First Edition Revised Syllabus

xi-xxi

Important Formulae and Results

BUSINESS MATHEMATICS AND STATISTICS MODULE I (50 Marks) ALGEBRA Unit 1.

Permutations

2.

Combinations

5-25 26-49

3.

Logarithms

5()..69

4.

Elements of Set Theory

70-106 ... 107-126

5.

Binomial Theorem

6.

Compound Interest and Compounding

7.

Annuities:. Sinking Fund

127-142 ... 143-172

STATISTICS

8.

Introduction: Definition of Statistics: Importance and Scope, Types, Sources and 175-190

Collection of Data

9.

Summarization of Data: Classification and Tabulation; Frequency Distribution

191-219

10.

Diagrammatic Representation of Statistical Data

. .. 220-261

11.

Measures of Central Tendency: Mean, Median and Mode

.. . 262-317

12.

Measures of Dispersion

... 318-350

MODULE II [50 Marks} 13.

Moments, Measures of Skewness and Kurtosis

... 353-372

14.

Correlation and Regression

... 373-412

15.

Probability Theory

... 413-458

16.

Interpolation

... 459-481

17.

Index Numbers

... 482-543

18.

Time Series Analysis

... 544-590

19.

Measures of Association of Attributes

... 591-601 i-viii

Appendix: Logarithms and Antilogarithms Calcutta University B Com Examination Questions

lvJ

1-

"This page is Intentionally Left Blank"

Preface to the Thirteenth Edition The present volume is a thoroughly revised thirteenth edition of our textbook Business Mathematics and Statistics strictly conforming to the new syllabus prescribed by the University of Calcutta effective from the Academic Session 2010-11. In this edition, a large number of problems set in B.Com. General and Honours Examinations of Calcutta, Burdwan and North Bengal Universities held during 2006-2009 have been solved in each chapter of the book. This book is meant for all B.Com. first year Pass and Honours students. Any suggestion/criticism from the readers will be gratefully received. R. K. Ghosh

Kolkata

S. Saha

15 August 2010

Preface to the Eleventh Edition 'The present volume is a thoroughly revised edition of our textbook Business Mathematics and Statistics strictly according to the new (revised) syllabus prescribed by the University of Calcutta to be effective from the academic session 2006-07. Dne to the changes in the new syllabus some of the topics of the earlier edition (e.g., Partial Differentiation, Differentials: Total Differentials and their Applications, Theory of Probability and Concept of Random Variable and Probability Distribution) had to be dropped. According to the new syllabus, one new topic (e.g., Discounting) had to be inserted to Part II: Mathe:µiatics of Finance. Also several typical problems from the current examination papers have been added in each chapter of the book. Any suggestion or criticism from our beloved students and their esteemed teachers will be gratefully received. R. K. Ghosh

Kolkata 9 May 2006

S. Saha

[vii J

Preface to the First Edition Nowadays the knowledge of MATHEMATICS AND STATISTICS is considered essential for a scientific approach to business problems and also for profitable business decisions. The importance of Mathematics is found to be reflected in the revised curriculum of the commerce course in different universities and professional institutes. The present treatise is intended as a textbook of Business Mathematics and Statistics for all the aforesaid courses. In particular, the syllabi prescribed by the University of Calcutta for the two-year B.Com. Course under the 10 + 2 + 2 system have been followed very closely. In preparing the text we have taken into consideration the fact that many students who take up a business career have had a far less thorough grounding in mathematics than those entering into pure science courses. The book, therefore, has been written in a language that is easy to follow; theory and applications 'run side by side; illustrated diagrams are plenty. An adequate number of carefully chosen examples are worked out and hints for solving harder problems are given at the end of each exercise. In every exercise, gradation of problems has been made with utmost care. In order to give our readers an idea of the standard expected by different examining bodies we have incorporated a number of questions set previously for the B.Com., Cost and Chartered Accountancy Examinations. The book is divided into three parts-Algebra, Elements of Co-ordinate Geometry and Business Statistics, covering the entire syllabus of the B.Com. Pass Courses of different universities of West Bengal and other States of the country. In each part, new and difficult concepts have been dealt with the greatest possible clarity, taking illustrations from real-life situations. In -short, every attempt has been made to create interest in the subject among the young learners. How far the authors have been successful in their endeavour shall be judged by the readers themselves. The authors deeply regret the inconvenience caused to the students for the unavoidable delay in the publication of the book. The major constraint was the uncertain working hours of the press due to load-shedding. The authors do not believe that the book is free from printing mistakes though all care has been taken to avoid them. Suggestions for further improvement of the book and correction of printing errors, if any, will be most gratefully received. The authors wish to record their deep obligation and gratitude to the late Professor Ajit Kumar Basu with whom the first author jointly wrote the book Commercial Mathematics (Ghosh and Basu). It will not be surprising to find some influence of that boo1 n; (b) mep + m 2_ 10

> 0.3.

or, 10 log 10 2

>

3, i.e.

[ICWAI Dec. 1976]

> 3 log 10 10 r;r,

Iog 10 (2 10 )

> log 10 (103 ),

21. If a,b,c are in G.P. and x,y,z are in A.P., show that (y - z)loga + (z - x)logb + (x - y) log c = 0, all the logarithms being. with same base. [C. U. B.Com. 1980, '96] [Hints. ac = b2 and y - x = z - y or, x - y = y - z. L.H.S. = (x-y)(loga+logc) +(2y-x-x) Iogb = (x-y) Iogb 2 -2(x-y) logb = O = R.H.S. (Proved).]

B 1. Find the value of each of the following:

(a) log 7 343; (b) log 5 v'5125; (c) log 3 tog 4 log 3 81. 2. Find the value of each of the following:

(a) log 25 125; (b) log 10 3000, when log 10 3 = 0.4771. 1 1 1 3. Prove that · + · + = 1. loga (abc) logb (abc) loge (abc) 4. Prove that logy(x 2 )

x logz(y 3 )

x logx(z 4 ) = 24.

5. If a= b2 = c3 = d4 , prove that loga(abcd) =

25 . 12

[Hints. b=a 1 12 ,c=a 1 13 ,d=a 1 14 and abcd=a.1+!+!+! =a~.]

60

BUSINESS MATHEMATICS AND STATISTICS

6. If a2 + b2 = 14ab, prove that log (a; b)

=~(log a+ logb).

7. If a 2 + b2 = 27ab, prove that log (a-b) - = 1 (loga + logb). 2 5 a+ b) 1 a b = "2(1oga + logb), show that b +-;;: = 7. 8. If log ( - 3 9. If log ( -a - 4

b) = 1 (loga + logb), prove that a

2

2

+ b2

= 18ab.

. . log Y27 + log v'8 - log vTI5 3 = 2· 10. Without usmg tables, show that log - log 5 6 ~x

~Y

~z

.

11. I f -- = - - = - -, prove that x,y,z are m G.P. 2 1 3 [C.U. B.Com. 1995; N.B.U. B.Com.(H) 2007) logx logy Jogz [Hints. Let = = = k. Then Jogx 1

2

3

= k,logy = 2k,logz = 3k.

:. Jogx + logz = k + 3k = 4k = 2 · 2k = 2logy, or, Jog(xz) = Jogy 2 ;

:. xz = y 2 . This shows that x, y, z are in G.P.]

12. If log(x 2 y3 ) = a and log (

~)

= b, find log x and logy in terms of a and b. [N.B.U. B.Com. 1996; V.U. B.Com. 1998)

[Hints.

2 Jog x + 3 logy == a and Jogx-logy = b

(1) (2)

1 By (1) - (2) x 2, we get 5logy =a- 2b, or, logy= -(a- 2b). 5

By (1) + (2) x 3, we get 5Jogx =a +.3b, or, Jogx =

13. Prove that log 3

(

}a../ava. ··oo) ~

1.

14. Show that the value of log 10 2 lies between [Hints. 2

3

. Aga.m, 24

1

5(a + 3b).]

~

and

~-

[V.U. B.Com. 1996)

< 10, or, log 10 23 < log 10 10, or, 3 Jog 1o 2 < 1, or, log 1 o 2

> 10, or,

log 10 24

1 < 3·

1 l > log 10 10, or, 4 log 10 2 > 1, or log 10 2 > 4·

15. If x = 1 + loga be, y = 1 + logb ca, z = 1 +loge ab, then prove that xy + yz + zx = xyz. 16. If x = loga be, y = logb ca and z =loge ab, then show that x + y + z + 2 = xyz. 1 1+ x

1 l+y

1 l+z

1

1 l+y

[Hints. See worked-out Ex. 20. - - + - - + - - = 1, or, - - + - -

or, (x + y + 2)(z + 1) = z(l + x)(l + y), etc.]

I+x

=1-

1 -l+z

61

CHAP. 3: LOGARITHMS

ANSWERS A 1.

2.

(a) (b)

(b) l. 5'

4·

..'

~

3'

3.

(c) -5; 3 (d) 2· (a) 81;

(c) 8· ' (a) 3;

(b)

l

4.

5.

.

112'

9.

(c) 3v"2.

(a) (b) (a) (b) (b)

4· ' !!.

12. 24.

3'

17. 8.

No. 3.30103.

4, 8.

B 1.

3.4

(b) 3.4771.

(c) 0.

(a) 3;

(b) 2;

2.

(a) ~;

12.

t (a+ 3b); t (a -

2b).

Common Logarithms

For numerical calculations, logarithms are calculated with respect to the base 10. Such logarithms are called Common Logarithms. The base 10 is usually ommitted. For example, we write log 3 to mean log 10 3. We stress upon the fact that we can find logarithms of positive numbers only; thus expressions like log(-5) will not come under our discussions.

Natural Logarithms If the base is e, where e = 2.71828 (approx.) is a constant, then logarithm is called natural logarithm denoted by ln. Thus

ln x

= loge x, ln 1 =loge 1 = 0

lne = logee = 1, loge(en) = nloge e = n x 1 = n. Now we shall discuss common logarithm in details.

Characteristic and Mantissa It can be proved that the common logarithms like log 34.49 = 1 + .5377; log .002 = -3 + .3010. In fact, the common logarithms of a number consists of two parts, one integml and the other decimal, of which the integral part may be zero, or any integer, (positive or negative) and the decimal part is less than one and is always positive.

Definition. The integral part of a common logarithm is called the characteristic and the decimal part, which is always positive, the mantissa of the logarithm. Suppose log 33.9 = 1.5289, then 1 is the chamcteristic and .5289 is the mantissa of the logarithm. If log 0.338 = -1 logarithm.

+ .5289,

then -1 is the chamcteristic and .5289 is the mantissa of the

62

BUSINESS MATHEMATICS AND STATISTICS

The negative sign of the characteristic is usually written over the number. Thus, we write log 0.338 as i.5289. This implies that the characteristic is -1, but that the mantissa is +.5289. Note: I.5289 and -1.5289 are not the same. I.5289 etc. are read as bar 1, bar 2, bar 3, etc. respectively.

3.5

= -1 + .5289 and

-1.5289

= -1- .5289.

I,2,3,

Rules for Determining Characteristics

·We give below two rules for finding the characteristics of common logarithms:

A. Logarithms of Positive Numbers greater than or equal to 1 Rule 1. For a number N > 1, the characteristic of log N is positive and is one less than the number of digits in the integral part of N. Example. Number N

7283 3

Characteristic of log N

347.5 2

48 1

3 7.893 0 0

23.85 1

B. Logarithms of Positive Numbers lying between 0 and 1 The numbers of this class begin with a decimal point, so that they have 0 in the integral part, and in each number the decimal point is followed by either no zero or a number of zeros; e.g., 0.3784, 0.025, 0.0007896, etc. Rule 2. For a positive number N < 1 (i.e., for the positive number that begins with a decimal point), the characteristic of log N is negative and is numerically one greater than the number of zeros immediately after the decimal point of N. Example. Number N

0.3485

Characteristic of log N

3.6

0.0038

0.030708

I

Table of Logarithms : Determination of Mantissa

The mantissa of the common logarithm of a number cannot be determined, like its characteristic, by inspecting the number. It is to be found from a table called Table of Logarithms or Log-Table.

A. Arrangement of the Table In the extreme left-hand column of the table we find bold-type figures ranging from 10 to 99. The top-most row contains the figures 0, 1, 2, 3, · · · , 8, 9. Below each of these figures we observe numbers containing four digits (or in some tables, more than four digits). These are the mantissa with decimal points at the beginning being dropped. Each mantissa is understood to have a decimal point preceding it. This forms the main body of the table. There is a side-table on the right called Mean-Difference Table whose significance we shall explain later. A table which gives mantissa with four figures is called a Four-Figure Log-Table. Five-Figure Tables or Seven-Figure Tables, ~tc. are also in use for more precise calculations.

63

CHAP. 3: LOGARITHMS

B. The following Rule will be very often Helpful Rule 3. Logarithms of numbers having the same significant figures in the same order have the same mantissa, the mantissa being the same as that of the log of the whole number obtained on ignoring the decimal point or the non-significant zeros of the number.

Thus to find log 7.835 or log 783.5 or log 7835000 or log .0007835, we first find the characteristic in each case by inspection. The characteristics will be different in each of these cases. But by Rule 3, all of them have the same mantissa as that for log 7835. So if, by using a log-table, we find that the mantissa for log7835 is .8941, then log log

7.835 = 0.8941, 783.5 = 2.8941,

log 7835000 = 6.8941, log .0007835 = 4.8941.

C. How to use a Log-table to find the Mantissa To find the mantissa for 7835, we first search for the figure 78 in the extreme left-hand column of a log-table; now glance horizontally to the right to the column headed by 3 of the top-most row. Note the figures 8938 (we refer to a four-figure table). For the fourth figure 5 look to the mean-difference table and glance down the column headed by 5 and on the same horizontal line as 78; let the figure be 3. Then log 7835 has the mantissa .8938 +3 .8941

Thus log 7835 = 3.8941.

At the end of the book both four-figure and five-figure log-tables are given. The students may use any one for their calculations.

D. When the Number Contains five Digits or more When the number contains more than four significant figures, we can approximate it up to 4 significant figures, for rough calculations. For more precise results, the methods illustrated in the following example and in the section (E) are followed. Example. Find log 4.6889 and log 4688.9 using the four-figure log-table.

log 4.68 Add Mean Difference for Add Mean Difference for

= 0.6702 8 =

7

9=

8

. . log 4.6889 = 0.6710 log 4688.9 = 3.6710.

E. Method of Interpolation The method is based on an important principle of Mathematics,· called the Principle of Proportional Parts which we state here without proof:

64

BUSINESS MATHEMATICS AND STATISTICS

If the value of a quantity depending on a variable quantity x be tabulated for different values of x at a regular small intervals, then for a very small change in x, the corresponding small change in the tabulated quantity is proportional to the change in x. Example. Given log 34938 = 4.5432980 and log 34939 = 4.5433105, find by interpolation log 349387, correct to 7 decimal places.

log 34939 = 4.5433105 log 34938

= 4.5432980

Difference for 1 = 0.0000125 : . Difference for .7

= 0.00000875 Usually written as :

Now log 34938 = 4.5432980 Add Difference for .7 = .0000087 .. log34938.7 = 4.5433068 .·. log349387 = 5.5433068

1 = 125

Difference for

5

: . Difference for .7 = 87 .5 log 34938 = 4.5432980 87 Add Difference for .7 =

= 4.5433068

: . log 34938. 7

3. 7

5

Antilogarithms

In addition to finding the logarithm of a given number it is also necessary to reverse the process, that is, to find the number called antilogarithm, when the logarithm is given. For this purpose we use the table of antilogarithms, whose arrangement is similar to that of the table of logarithms. For example, suppose we require to find antilog 1.9058, i.e., to find the number whose logarithm is 1.9058. Initially ignore the characteristic and start with the mantissa part .9058. We look up the antilog table and search for .90 in the extreme left-hand column and then glance horizontally to the column headed by 5 and note the figure 8035. Now from the meandifference table we obtain the figure 15 which lies below 8 along the row of 90. Adding the two figures we obtain the number 8050. Ignoring the characteristic, Antilog Add Mean Difference for Antilog

905

= 8035

8 =

15

9058 = 8050

: . antilogL9058 = 80.50. Now the characteristic 1 indicates that there should be 2 integral places; hence, the required number should be 80.50; thus antilog 1.9058 = 80.50. Obtain the following:

antilog 0.2403, antilog i.6334, antilog 12.8934.

65

CHAP. 3: LOGARITHMS

If the given mantissa contains more than four figures we may ignore them, or approximate the mantissa up to 4 decimal places, for rough calculations, but for more precise results, the method illustrated in the following example is followed: Example. Find N, where log N = 2.92479. N = antilog 2.92470

- Ignoring characteristic, antilog 924

= 8395

Add Mean Difference for

7 =

14

Add Mean Difference for

9=

1

7

antilog 92479 = 8411 Now, the characteristic 2 indicates that the number must have 3 integral places; hence N = antilog 2.92479 = 841.1. Note: The mantissa of a logarithm is to be always positive, while its characteristic may be positive, negative or zero. Finding the mantissa from a table of logarithms and the characteristic by inspection, the value of log0.07213 is found to be 2.8581, which means that the characteristic is negative being = -2, but the mantissa is positive being = +.8581, so that log0.07213 = -2 + .8581 = -1.1419, which is the actual value of the logarithm. But this actual value, in which the decimal part is negative cannot be used in reference to Log or Antilo,g Tables, and the decimal part should be made positive before using the Tables. Thus, if log N is found to be equal to -2.5673, it is to be transformed as below, before comiulting a Table of Antilogarithms, for finding N: log N = -2.5673 = -3

3.8

+3 -

2.5673

= -3 + .4327 = 3.4327.

Illu,strative Examples

Example 1. Find the value of (using logarithm) 11 7 2 9 36 22 17 x 9 x 21 x 29 x 61 x 71. Solution. Let x = the required value. Then logx = (logll-log17)+(log7-log9)+(log2-log21) + (log 9 - log 29) + (log 36 - log 61) + (log 22 - log 71) log 11 log 7 log 2 log 9 log 36 log 22

= 1.0414 = 0.8451 = 0.3010

= 0.9542 = 1.5563 = 1.3424

6.0404

log 17 log 9 log 21 log 29 log61 log 71

= 1.2304 = 0.9542 = 1.3222 = 1.4624 = 1.7853 = 1.8513

8.6058

: ; logx = 6.0404 - 8.6058 = -3 + 9.0404 - 8.6058 = 3.4346.

:. x = antilog 3.4346 = .002720. .Bus. Math. & Stat. [C.U.}-5 .

66

BUSINESS MATHEMATICS AND STATISTICS

~~~~~~~~~~~~--~~~~~~~~~~~~--'-~~~~~

. d h l 1 0.4932 xx653. 7 . l . Example 2. Fm t e va ue o 0. 456, using ogarithm. 07213 8 . Solution. Let x

0.4932 x 653.7

= 0. 07213 x 8456 =

N ( ) D say .

. . log x = (log 0.4932 +log 653. 7) - (log 0.07213 +log 8456) logN

logD

logN

logD

log 0.4932 = I.6930

log 0.07213 ( +) log 8456

( +) log 653. 7 = 2.8154

= 2.5084

= 2.8581 = 3.9272 = 2.7853

:. logx = 2.5084 - 2.7853 = I.7231.

:. x

= antilog

I.7231 = 0.5285.

Note: Before taking antilog care should be taken to make the decimal part positive (if it is not so) by properly adjusting the integral part. 78 41 3 · ) Vi"42.3 4 (0.1562) .

Example 3. Evaluate with the help of logarithm

. (78.41) 3 v'142.3 . N ( ) Solut1on. Let x = ~ D say . {/(0.1562)

.

logD

logN 1

3 log 78.41 = 3 x 1.8944 = 5.6832 ( +)

! log 142.3 = 4x 2.1532 = log D

1

-

4log0.1562=4x1.1937

i{-4 + 3.19~7)

1.0766

=

6.7598

= I.7984

= i. 7984

logx = logN - logD = 6.9614

:. x = antilog 6.9614 = 91, 49, 000 (approx.).

Involution and Evolution Example 4. Compute the (i) cube, (ii) cube root of 0.01865.

Solution.

(i) Let x = (0.01865) 3 ; :. logx

= 3log0.01865 = 3 x 2.2706 =

6.81.L8.

:. x = antilog 6.8118 = 0.000006483. (ii) Let x = ~0.01865; : . log x =

1· 1 3 1og 0.01865 = '3

:. x= antilog

i.4235

-

x 2.2706 =

= 0.2652.

'13 (-3 +

1.2706)

-

= 1.4235.

67

CHAP. 3: LOGARITHMS

Example 5. Using log table, find the value of (i) 1{/1129, (ii)

if;J;,.

Solution. (i)

= (1129)1/18. Then logx = i18 log1129 = 118 x 3.0527 = 0.1696. 1

{/IT29 = x (say), i.e., x

:. x = antilog 0.1696 = 1.478. (ii) Let x

= {j (i.i3s) = ( i.i3s) 1/7 .

. . logx = log =

c.~35 y/7 =~log c.~35 ) =~(log 1 -

71(0-0.0917 ) =

log 1.235)

0.0917 - - 7 - = -0.0131=-1+1- 0.0131=1.9869.

:. x = antilog i.9869 = 0.9703. Example 6. Find the number of digits in 720 . Solution. Let x = 720 ; :. log x =log 720 = 20 log 7 = 20 x 0.8451 = 16.9020. Since the characteristic of log 720 is 16, the number of digits in 720 is 17. Example 7. Solve: (1

+ i) 15 =

2 65

· 2~~8\838.

Solution. 15 log(l + i) = log 2.654 + log 1.838 - log 2.083 = 0.3696. log 2.654 = 0.4239 (+) log 1.838 = 0.2644 0.6883 (-) log 2~083 = 0.3187 0.3696 Example 8. Solve : 3.:r- 1 = 4.5 1 -

or, log(l + i) = 115 x 0.3696 = 0.0246 or, 1 + i = antilog 0.0246 = 1.058 or, i = 0.058.

3 x.

Solution. (x - 1} log 3 = log4 + (1 - 3x} log 5 or, x(log 3 + 3 log 5) = log 4 + log 5 + log 3 = log 60, or, x(log 375} = log 60 log 60 1. 7782 or, x = log 375 = 2.5740 = 0 ·6910· 7782 . use 1ogant . hm.' Thus I norder to eval uate 1.. we may agam 2 5740 log 77 = 0.2480 Add Mean Difference for 8 = 20 log 2.57 = 0.4099 Add Mean Difference for 2= 5 Add Mean Difference for 4 = 7 log 1.7782 = 0.2501 log 2.574 = 0.4106 :. log 1.7782 = 0.2501 :. log 2.5740 = 0.4106 (-) log 2.5740 = 0.4106 = i.8395

:i.

antilog i.8395

= 0.6910 = required value.

68

BUSINESS MATHEMATICS AND STATISTICS

_ _ _ _ ____.,EXERCISES ON CHAPTER 3(11)

1 - I- - - - - -

1. Given log 2 = 0.3010300, log 3 = 0.4771213, log 7 = 0.8450980, find the logarithms of: (a) (0.0125) 115;

(b) {0.405) 116:

(c) {0.0025) 11 19 .

2. Given log 6 = 0. 7781513 and log 8 = 0.9030900, find log 36.

3.

(a) Find the value of

V::

(up to 3 decimal places).

[C.U. B.Com. 1976)

(b) Find, with the help of logarithmic table, the value of log{(2. 7) 3 x (0.81) 415+(90) 514}. 4. Given log 2 = 0.3010300.

{a) Find the number of digits in (i) 264 ; {ii) 525 . (b) Find the number of zeros that follow the decimal point and precede the first significant figure in the reciprocal of 240 . 5. (a) If log 3 = 0.4771, find the number of digits in 343 . [CA Nov. 1976) (b) How many zeros are there between the decimal point and the first significant figure in (0.5) 100 ? [ICWAI June 1976] (c) Find the square root of 3. Use log table. [N.B.U. B.Com. 1975) 6. Find the values of the following with the help of log-table: (a) {b) (c) (d) (e) (f)

42.97 x 0.00258; 0.8176 X 13.64; [C.U. B.Com. 1972) 0.2543 + 0.09027; 0.001834 + 0.04316; 0.678 x 9.01+0.0234; 9.753 x 10.34 x 0.9252. 1.453 x 3.143 '

(g) ~;

1

{k) {1+0.12) 20 ' {l) {m)

[C.U. B.Com. 1977)

{h) {789.45) 118 ; (i) ?'45.37 x (0.7692) 4;

U)

[C.U. B.Com. 1m)

v(36~21); 5

[B.U. 1974; '76)

y'2.4i5.

(0.824) 4 ' {n) (68.28) 3 x {0.00843) 2 x {0.4623). (412.3) x {2.184)5 '

{o)

{1.0~5)20;

?'QAI65 x ~.

?'0.6438 x ~0.0186' [C.U. B.Com. 1974]

1. Find the approximate values of the following:

{a) 36.085 x (1.005) 4; [C.U. B.Com. 1974)

(

b) ?"19.41x4.62 x {1.783)- 213 . V'l.436 '

400 x {1.04) 30 . (1.04)30 - 1 ' (d) 30 [{1 +0.035) 15 -1]; {e) 1275 {1 - {1.0575)- 15 }. ( ) c

8. Find the 5th root of 349388 correct to 3 decimal places. (Given log 34938 = 4.5432980, log J4939 = 4.5433105, log 12842 = 4.1086327, log'l2843 = 4.1086665).

69

CHAP. 3: LOGARITHMS

9. Solve:

(f) (1 + x) 3 = 1.586(1 + x)- 10 ; (g) (1 - x) 12 = 0.5187;

(a) (o\)x =6.25; (b) (c) {d) (e)

2x = 100; 3x=2; 5x-l = 400; 42x-l = 5x+2;

(h)

(i)

20 = 2; (21)n 1 9 )n ( 10 2· =

(Use log3 = 0.4771213,log7 = 0.8450980 and log2 = 0.3010300, whenever necessary.) 10.

(a) Find by the help of logarithmic tables the values of x and y, correct to two places of decimals, if 2x = 3Y and 2Y+l = 3x- l. (b) Solve the equation 2x · 7Y = 80000, and 3Y = 500, having given log 2 = 0.30103, log 3 = 0.47712 and log 7 = 0.84510; the values of x and y are to be found correct to 4 decimal places.

11. Given log2592 = 3.413634 and log432 = 2.635483, find log2,log3. 12. Find with the help of logarithmic tables, the values of x, correct to 2 places of decimals, from the following equations:

(a) (i) 63-4x . 4x+5 = 8; (1.1') 5s-3x __ 2x+2.1 (b) 73x+2 + 4x+2 = 73x+l + 22x+6; (c) 55-3x + 4f+3 = 57-3x _ 2x+s.

(N.B.U. B.Com. 1975J

ANSWERS (a.) I.6193820 = -0.3806180; (b) I.9345759 = -0.0654241; (c) 4.8197044 == -3.1802956. 2. 1.5563026.

1.

3.

(a.) 0.109;

(b) -1.2218.

8. 12.843.

4.

(a)

(b) 12.

9.

1.

5. 6.

(i) 20; (ii) 18;

(a) 21; (a) (b) {c) (d) (e) (t) (g) {h)

0.1109; 11.15; 2.817; 0.04249; 261; 20.43; 1.2915; 2.302;

(b) 30;

(c) 1.732.

(i) (j) (k) (I) (m) (n) (o)

(a) 36.81 or 37; (b) 8.018; (c) 578.9;

(a.) (b) (c) (d) (e)

2; 6.643; 0.631; 4.723; 3.958;

(d) 20.19; (e) 722.3.

(f) 0.036;

(g) 0.875; (h) 14.2; (i) 6.9.

1.248; 0.415;

10.

0.5988;

(a.) x (b) x

0.1038;

= 2.71, y = 1.71; = 0.4070, y = 5.6568.

11. 0.30103, 0.47712.

2.588; 0.0004059; 1.37.

12.

(a)

(i) 1. 77; (ii) 1.206.

(b) 0.03; (c) 1.21.

Chapter ,4

Elements of Set Theory

4.1

Introduction

Set theory is the most basic concept in any discipline of studies-be it Science or Humanities or Commerce. It is, therefore, quite natural that we begin with the discussions on elements of sets. What is a Set? Not easy to give a precise definition of a set. Our purpose will be se.rved if we state: A set is a well-defined collection of distinct objects. Objects may be of any kind, e.g., Numbers, Letters of the English/Bengaii alphabets, Rivers, People, etc. We shall be more interested with Sets of Numbers. The objects which form a set are called elements or members of the set. The term well~defined collection in the definition of a set means a collection made according to some characteristic property: Given any object whatsoever, using that characteristic property, one can decide whether the given object should belong to that collection or it should not belong to the collection. The objects of the collection should be written within the braces { e.g.,

}.

(i) {l, 2, 3, 4, 5} is a set of first five natural numbers.

(ii) {a, e, i, o, u} is a set of five vowels of the English alphabet. It is usual to denote a set by capital letters like A, B, C, · · ·, S, T, X, ···,and the elements of/a set by small letters a, b, c, · · ·, x, y, z, '.··,or, by Greek letters a, /3, /, · · ·. The following symbols are used in Mathematics: Symbols E

Their meanings

belongs to or an element of e.g., a E A means the element a belongs to the set A. does not belong to or not an element of e.g., a¢ A means the element a does not belong to the set A.

s.t.

such that 70

71

CHAP. 4: ELEMENTS OF SET THEORY

w.r.t. iff

with respect to

:3

if and only if there exists

V

for all (e.g., Vx E S, for every element x belonging to the set S)

I\

and or; e.g., a I\ b means a and b but a Vb means a or b. implies that

V

:::} or ___,

[Let e.g., P and Qare two statements. When we write P:::} Q, it will mean "If P, then Q." If x

4.2

+2 =

4, then x = 4 - 2 = 2].

How to Write a Set

There are usually two methods of writing a set• Tabular or Roster Method. members,

In this method we define a particular set by listing its

e.g., let A consist of the numbers 1, 3, 5 and 10, then we write

A= {l, 3, 5, 10}, i.e., the elements are separated by commas and enclosed in brackets { }. We call this method Tabular method or Tabular form of a set or Roster method of writing a set. • Property-stated Method. We may define a particular set B by stating properties which its elements must satisfy: e.g., let B be the set of all even positive integers, then we use a letter x, to represent an arbitrary element and we write

B = {x

Ix

is an even positive integer}.

[rea and A as subsets of A. Procee~ing

in this way, we see that if A contains n elements, then P( A) contains 2n elements.

Thus, if a finite set A has n elements, its power set contains 2n elements. Illustration 18. If A= {2, 3}, then P(A) = 2A = [¢, {2}, {3}, {2, 3}], where

is the null set.

Illustration 19. If A= {1,2,3}, then P(A) = 2A = [, {I}, {2}, {3}, {1,2}, {1,3}, {2,3}, is the null set. [C.U. B.Com.(H) 2001)

{I, 2, 3}], where

Note: A power set is a set of sets.

4.4

Operations on Sets

By operations on sets, we mean the process of constructing new sets from given sets by combining the latter in some suitably well-defined fashion as explained below:

74

BUSINESS MATHEMATICS AND STATISTICS

Venn Diagram: Pictorial Way of Explaining Operations on Sets Operations on sets or any property or theorem relating to sets can be well understood with the help of a diagram known as Venn-Euler diagram or simply Venn diagram. Venn diagram was first introduced by Euler and subsequently this was developed by John Venn, a British Mathematician. In this diagram the universal set U is denoted by a rectangular region and any subset of U by a region enclosed by a closed curve (or a circle) lying within the rectangular region. These closed curves (or circles) representing the subsets of U will intersect each other if they have some common elements among them. We give below five operations on sets:

Operation I: Union or Join of Sets The union or join of two sets A and B, written as AU B, is the set of all elements which belong either to A or to B or to both A and B. Symbolically, AU B = {x : x E AV x E B). Here, 'V' means 'or'. AU Bis read 'A union B' or 'A cup B'. Illustration 1. If A= {1, 3, 4, 5} and B = {2, 3, 4, 6, 8}, then AU B = {1, 2, 3, 4, 5, 6, 8}. Venn Diagram for A U B

In Venn diagram of Fig 4.1, we have shaded AU B, i.e., the area of A and the area of B. Flg4.1

It follows from definition that A U B = B U A and both A and B are always subsets of AU B, i.e., AC AU Band BC AU B.

Operation II: Intersection or Meet of Sets The intersection or meet of two sets A and B, written as An B, is the set of all elements which are common to both A and B. Symbolically, An B = {x: x EA/\ x EB}. Here,'/\' means 'and'. An B is read 'A intersection B' or 'A cap B'. Illustration 2. If A= {2, 3, 5, 7} and B = {1, 3, 4, 6, 7, 9}, then An B = {3, 7}. Venn Diagram for A n B

In Venn diagram, we have shaded An B, i.e., the area common to both A and B of Fig 4.2. Fig 4.2

It follows from definition that A n B = B n A and each of A and B contains An B as a subset, i.e., A:) (An B) and B j (An B).

Disjoint Sets. If two sets A and B have no elements in common, i.e., if no element of A is in B and no element of B is in A, then A and B are said to be disjoint or mutually exclusive sets. Clearly, An B =,when A and Bare disjoint. [C.U. B.Com.(H) 1991]

CHAP.4:ELEMENTSOF~S_E_T_T_H_E_O_R_Y~~~~~~~~~~~~~~~~~~~~-7~5

Illustration 3. Two sets A= {2, 5, 7} and B = {1, 3, 6, 8} are disjoint, since they have no common elements. Here, An B = 200 = 100 (1+~0~)n=100(1.024)n·]

15. A machine is depreciated at the rate of 103 on reducing balance. The original cost was Rs.10,000 and the ultimate scrap value was Rs. 3,750. Find the effective life of the machine. 16. In how many years will the population of a village change from 15,625 to 17,576 if the rate of increase is 43 per year? (Given 17,576 = 263 and 15,625 = 25 3 ). 17. A savings scheme makes the investment double in 7 years. Find the rate of interest accrued if compounded annually. [Given log 2 = 0.30103, log 1.10409 = 0.043] (C.U.B.Com. 2005)

[Hints: A= P(l +i)n. Here, A= 2P,n = 7,i

=?

139

CHAP. 6: COMPOUND INTEREST AND COMPOUNDING

2P

= P(I + i) 7 => (1 + i) 7 = 2 => 7 log(l + i) =

log 2 = 0.30103 => log( I + i) = 0.043 => log(l + i) = log{l.10409) => 1+i=1.10409 => i = 0.10409.

Hence, the rate of interest= r = lOOi = 100 x 0.10409 = 10.409, or, 10.413.]

B

1. {a) Find the amount that Rs.100 will become after 20 years at compound interest at 5 per cent, calculated annually. {b) Find the present value of Rs.10,000 due in 12 years at 6% p.a. compound interest. (Given {1.06) 12 = 2.012] [C.U. B.Com. 2006] 2.

(a) The sum of Rs. 956 amounts to Rs.1,099.40 in 3 years, at simple interest; what sum of money invested at compound interest, payable yearly, will amount to Rs. 1,481.97 in 3 years, the rate of interest p.a. in both the cases being the same? (b) A man can· buy a flat for Rs.1,00,000 cash, or for Rs. 50,000 down and Rs. 60,000 at the end of one year. If money is worth 10% per year, compounded half-yearly, which plan should he choose? [C.U. B.Com. 1995] [Hints: 1st Plan: Cash-down price= Rs.1,00,000. 2nd Plan: Cash-down amount = Rs. 50,000. Amount paid at the end of one year = Rs. 60,000. If Rs. P be the present value of Rs. 60,000 to be paid at the end of 1 year, then 2

60000 = P (1 +

~) n=P 2

(1 +

~) 200

2

= P{l.05) 2 , or, P =

~0000) (1.05 2

= 54422

Cash-down price will be Rs. 50,000 + Rs. 54,422 = Rs. 1,04,422. Hence, the first plan should be chosen (·:it is cheaper by Rs~ 4,422).J

3. (a) A sum of money invested at compound interest, payable yearly, amounts to Rs. 2,704 at the end of the ·second year and to Rs. 2,812.16 at the end of the third year. Find the rate of interest and the sum. (b) A sum of money invested at compound interest amounts to Rs. 21,632 at the end of the second year and to Rs. 22,497.28 at the end of the third year; find the rate of interest and the sum invested. 4. A man left for his three sons aged 10, 12 and 14 years Rs. 10,000, Rs. 8,000 and Rs. 6,000 respectively. Tne money is invested in 3%, 6% and 10% compound interest respectively. They will receive the alllount when each of them attains the age of 21 years. Find, using a five-figure log-table, how much each would receive. 5 .. What is the rate per cent p.a., if a sum doubles itself in 17 years at compound interest? 6. A sum of money is lent at 83 p.a. compound interest. If the interest for the second year exceeds that for the first year by Rs. 32, find the original principal. 7. (a) A man wants to invest Rs. 5,000 for four years. He may invest the amount at 103 p.a. compound interest accruing at the end of each quarter of the year or he may invest it at 10~% p.a. compound interest, interest accruing at the end of each year. Which investment will give him slightly better return?

140

BUSINESS MATHEMATICS AND STATISTICS

(b) Mr Brown was given the choice of two payment plans on a piece of property. He may pay Rs. 10,000 at the end of 4 years, or Rs. 12,000 at the end of 9 years. Assuming money can be invested annually ,at 4% per year, compounded annually, what plan should Mr Brown choose? 8. (a) The population of developing countries increases every year by 2.3% of the population at the beginning of that year. In what time the population doubles itself? Answer to the nearest year.

(b) In a certain population the annual birth and death rates per 1,000 are 39.4 and 19.4 respectively. Find the number of years in which the population will be doubled assuming that there is no immigration or emigration. 9. (a) A machine is depreciated at the rate of 20% on the reducing balance. The original cost was Rs. 1,00,000 and the ultimate scrap value was Rs. 30,000. Find the effective life of the machine. (b) A machine depreciates at the rate of 10% of its value at the beginning of a year. The machine was purchased for Rs. 44,000 and the scrap value realized when sold was Rs. 25,981.56. Find the number of years the machine was used. 10. An air pump used to extract air from a vessel removes one-tenth of the air at each stroke. Find what fraction of the original volume of air is left after the 12th stroke. 11. A machine depreciates @ 10% p.a. for the first two years and then 7% p.a. for the next three years, depreciation being calculated on the diminishing value. If the value of the machine be Rs. 10,000 initially, find the average rate of depreciation and the depreciated value of the machine at the end of the fifth year. (C.U. B.Com. 2007) 12. A machine depreciates@ 8% of its value at the beginning of a year. If the machine-wa8 purchased for Rs. 15,000, what is the minimum number of complete years at the efid of . which the worth of the machine will not exceed ~th of its original value?

c 1. A sum of Rs.10,000 is investediri a firm at 5% compound interest. Find the interest at the end of two years.

2. A sum of Rs.1,000 is invested for 5 years at 12% interest per year. What is the simple interest? If the same amount had· been invested for the same period at 10% compound interest per year, how much more interest would he get? (ICWAI Pre. June 1987) 3. The difference in simple and compound interests on a certain sum of money in two years at 15% p.a. is Rs. 144. Find the sum. (ICWAI Pre. June 1987) 4. The difference in the simple interest ~nd the eompound.1nterest on a certain sum of money at the end of two years at a rate of ~'%.is·Rs; 25. Determine the sum. 5. A certain sum is invested in a firm at 4% ·cqmpound interest. The interest for the 2nd year is Rs. 25. Find the interest for the 3rd year, (Hints: If the sum invested be Rs. 100, then interest for the 1st year

= Rs. 4.

141

CHAP. 6: COMPOUND INTEREST AND COMPOUNDING

Principal for the 2nd year = Rs. 104 and interest for the 2nd year = Rs. 4.16. Principal for the 3rd year= Rs.108.16 and interest for.the 3rd year= Rs.4.3264 If interest for 2nd year = Rs. 4.16, then interest for 3rd year = Rs. 4.3264 If interest for 2nd year = Rs. 25, then interest for 3rd year = Rs. \~~~4 x 25 = Rs. 26.)

6. The difference between the simple and compound interests at the same rate for 2 years on a certain amount is ( 4 ~0 ) of the amount. Find the rate of interest. [ICWAI Pre. June 1986J

7. Determine the rate of interest for a sum of money to become 2~ times of itself in 2 years, if the rate of interest is to be compounded annually. (ICWAI Pre. Dec. 1986] 8. Mr Rama invested equal amounts, one at 6% simple interest and the other at 5% compound interest. If the former earns Rs. 486.50 more as interest at the end of two years, find the total amount invested. 9. The compound interest on a cert~.in sum of money invested for 2 years at 5% is Rs. 238. What will be the simple interest on it at the same rate and for the same period? (ICWAI Pre. Dec. 1986]

10. A borrower pays interest on his loan at the rate of 4% in quarterly instalments. He wishes to pay monthly in the future. What should be the new nominal rate so that the lender will receive an equivalent amount? (ICWAI Pre. Dec. 1989] 11. The population of a town is 1,25,000. If ti:ic annual birth rate is 3.3% and the annual death rate is 1.3%, calculate the population of the town after 3 years. (ICWAI June 1992] (Hints: Since annual birth and death rates are 3.3% and 1.33 respectively; :. population increases every 3 year by (3.3-1.3), i.e., 2%. Using A= P(l +i)n, we get A= 1,25,000 (1+ 1 ~) ; etc.)

12. Two partners A and B together lend Rs.12,615 at 5%, compounded annually. The amount A gets in 2 years is the same as B gets at the end of 4 years. Determine the share of each in the principal. (ICWAI Pre. June 1991] (Hints: If A lends Rs. x, then B lends Rs. (12,615 - x). . . x(l + 0.05) 2 = (12,615 - x)(l + 0.05) 4 , or, x(l.05) 2 = (12,615 - x)(l.05) 4 , or, x = (12;615 - x)(l.05) 2 , Of, x { 1 + (1.05) 2 } = 12,615(1.05) 2 .]

13. Find the rate per cent compound interest at which Rs. 576 will amount to Rs. 625 in two (ICWAI Found. Dec. 1995] years. 14. What principal will amount to Rs. 551.25 in 2 years at 5% compound interest? (ICWAI Found. June 1996]

15. The population of a town is Rs. 30,000. If it increases annually at the rate of 10%, what will be the population of the town at the end of 3 years? (ICWAI Found. June 1995)

142

BUSINESS MATHEMATICS AND STATISTICS-

ANSWERS

A 1.

(a) Rs. 1,576.25;

(b) 2.

Rs. 709.90.

(a) Rs.906; (b) Rs. 6,319.90; (c) Rs.192.

3. Rs.15.25. 4.

(a) 15.8 years;

(b) 5.

14.2 years.

(a) 14.2 years; (b) 7.1 years.

(b)

6. 47.17 years. 7. 28.06 years. 8. (a) 4.53; (b) 12.83; (c) 23. 9. Rs. 4,936.31. 10. (a) 5.063; (b) 11.83. 11. Rs.10,000. ua. (a) Rs. 1,785. 71;

13.

Rs.1,778.

(a) 5.73;

(b)

14.43;

(c) 8. 14.

(a) 15 years (approx.);

(b)

29.2 years (approx.).

15. 9.3 years. 16. 3 years. 17. 10.413.

B (a) Rs.265.50; (b) Rs.4,970. 2. (a) Rs.1,280; (b) First Plan. 3. (a) 43 p.a.; Rs. 2,500; (b) 43 p.a.; Rs. 20,000. 4. Rs. 13,813, Rs. 13,517; Rs. 11,691. 1.

s.

(b) 35 yeen. ~ --

4.2%.

9.

6. Rs.5,000. 7.

8.

(a) 2nd investment will give him slightly better return;

(a) 5.4 years; (b) 5

yeal'§~;;,ly).

10. 0.28213 of the original volume.

(b) 2nd plan.

11. 8.23, Rs. 6,515.

(a) 31 years;

12. 11 years.

c 1. Rs.1,025.

6. 53.

11. l,32;651.

12. Rs. 6,615 and Rs. 6,000.

2. Rs.10.51.

7. 503.

3. Rs.6,400.

8. Rs.55,600.

13. 4.23.

4. Rs.10,000.

9. Rs. 232.20.

14. Rs.500.

5. Rs.26.

10. 3.63.

15. 39,930.

Chapter 7

Annuities: Sinking Fund

7.1

Annuity

An annuity is a fixed sum paid at regular intervals under certain conditions. The interval may be either a year or a half-year or a quarter year or a month, etc. If nothing is mentioned about the periodr we shall take it to be a year. An annuity payable for a fixed number of years is called Annuity Certain. But, if the annuity is to continue for ever, it is called a Perpetuity. When the payments are made at the end of each period, an annuity is called Immediate or Ordinary Annuity or simply Annuity; but, if the payments are made at the beginning of each period, the annuity is called an Annuity Due. An annuity is called a Deferred Annuity if the payments are deferred, (i.e., delayed) for a certain period or number of years. When the annuity is deferred for n years, it is said to commence after n years, and the first instalment is paid at the end of (n + 1) years. In case of Deferred Perpetuity, the payments will commence after the lapse of the deferred period and will then continue for ever. Present Value of an annuity is the sum of the present values of all the payments. A Free-hold Estate is an estate which yields a perpetual annuity called rent, and a Lease- ...·· hold Estate is that which yields rent for a fixed period for which the estate is held. The value of a free-hold estate is equal to the present value of the perpetuity (or rent). Unpaid Annuity is an annuity left unpaid for some given period. In this case, the amount . of such annuity is the sum of all the instalments left unpaid together with compound interest on each instalment for the corresponding unpaid period.

Amount of an Annuity Definition 1. Amount of an annuity is the total of all the instalments left unpaid together with the compound interest of each payment for the period it remains unpaid. Formula I. The amount A of an annuity (or immediate annuity) P left unpaid for n years is 143

144

BUSINESS MATHEMATICS AND STATISTICS

given by

A=~ {(l+i)n-1}, i

where i is the interest on unit sum for 1 year.

Proof Since the annuity is left unpaid for n years, the first payment P which is due at the end of the first year will earn compound interest for (n - 1) years and, therefore, it will amount to P(l +it-I. Similarly, the second payment P will earn compound interest for (n - 2) years and it will amount to P(l + i)n- 2 , and s~n. Finally, the last payment P will earn no interest and it will amount to P only. Hence, A

-

sum of the amounts of all the payments

P(l +it-I+ P(l + =

it- 2 + · · · + P

[which is a G.P. with C.R. (I~i))

(~f}

P(l +it-I {l -

1 - l+i

P(l

+

P (l

or, A

=

i + p -;- { ( 1 i

.)n-I { (1 +it - 1} x 1 + i (1 + i)n 1 +i - 1

i

+it +it

i)n { (1

1}

(1

+ i) n

1}.

-

Corollary 1. If the payments are made p times a year, the annual rent is Rs. P (i.e., the periodic payment is Rs.~) and i is the interest per unit sum p.a., then the amount A of the annuity P for n years is given by A=

if:

pj

{ (

1+

~· ) np -

~{

}

1 =

(

1+

~· )

np

}

- 1 , i.e., A=

ip

{ (

1

i ) np +P -

1

}

.

If an annuity is payable half-yearly and interest is also compounded half-yearly, then the amount A is given by A=

i/p2 {

(

1+

~· )

2n

}

- 1 =

i2P { (. 1 + ~· ) 2n -

}

1 .

If an annuity is payable quarterly and interest is also compounded quarterly, then the amount A is given by

4pi { ( 1+~·) A=

4 n

-1 } .

Similarly, for monthly annuity, l2P i ) A= - { ( 1+i

12

2 I n

-1 } .

145

CHAP. 7: ANNUITIES: SINKING FUND

Example 1. A man decides to deposit Rs.10,000 at the end of each year in a bank which pays 10% p.a. compound interest. If the instalments are allowed to accumulate, what will be the total accumulation at the end of 9 years? [C. u. B.Com. 2001] Solution. Let Rs. A be the total accumulation at the end of 9 years. Then we have

A=

~i {(1 +it -

1}.

Here, P = Rs.10,000, i = 11~0 = 0.1, n = 9 years, A= ?

A=

10

~.~oo {(1+0.1) 9 -

1} = 100000 { (1.1)

9

-

1}.

(1)

Let x = (1.1) 9 ; :. logx = 9logl.1=9 x 0.0414 = 0.3726. :. x = antilog (0.3726) = 2.2583. From equation {1),

A= 100000{2.2583- l} = 100000x1.2583=125830. Hence, the requited total accumulation= Rs.1,25,830. Example 2. A person invests Rs. 1,000 every year with a company which pays interest at 10% p.a. He allows his deposits to accumulate with the company at compound rate. Find the amount standing to his credit one year after he has made his yearly investment for the tenth time. Solution. For amount of an annuity P, we have

Here, A = Rs. 1,000, i = 11~0 = 0.1, n : 1

= 10, A=

?

1000 { (1 +0.1) 10 -1 } = 10000 { (1.1) 10 -1 } . A= Q.1

10

Let x = (1.1) , then logx = lOlogl.1=10 x 0.04139 = 0.4139. :. x = antilog (0.4139) = 2.5936. :. From equation (1), A = 10, 000 x (2.5936 - 1) = 1.5936 x 10, 000 = Rs.15,936.

Interest for one year more after the tenth yearly investment = =

10% of Rs.15,936 Rs.1,593.60.

Hence, the required amount = Rs. 15,936 + Rs. 1,593.60 ·= Rs. 17,529.60. Note: Students may use calculator to find the value of (1.1) 10 instead of log-tables. R11ia. lAath ·,:R. Ct!!tt

rr

11 1

1n

(1)

146

BUSINESS MATHEMATICS AND STATISTICS

7.2

Present Value of an Annuity

Present value of an annuity is the sum of the present values of all payments (or instalments) made at successive annuity periods.

Formula II. The present value of V of an annuity P to continue for n years is given by

= ~ { 1-,- (1 + i)-n},

V

where i = interest on unit sum for 1 year. Proof.

Present value of the 1st payment P due 1 year

"

"

" " 2nd

"

"

"

p

"

,,

" 3rd

"

p hence= - - . 1 +i

2 years

"

P " 3 years

"

p

= (1 + i) 2 p

- (1 + i) 3

Present value of the nth payment P due n years hence = (l-.::)n. : . V = sum of the present values of all the payments

p

p

--+ 1+ i

+ i)

(1

2

+

p (1

p

+ i)

3

+ .. ·+--~ (1 +it

~ {1- (I-hr} l~i

1-

=

~ { 1-

(1

+ i)-n}.

Thus

~ { 1-

V =

(1

+ i)-n}.

Corollary 1. If the payments are made p times a year and the annual rent is Rs. P, then the present value V of the annuity P to continue for n years is given by p {

V=i

1-

(

.)-np} ,

1+~

where i is the interest on unit sum for 1 year. The present value V of an annuity P, payable half-yearly, quarterly and monthly, are respectively given by (i)

n} , n} . 1+-

2P { 1- ( 1+~·)V=i

12

(iii)

12P { 1- ( V= .i

i )-

1? .....

i)- n} , 4

2

(ii)

4p { 1- ( 1+4 V=i

147

CHAP. 7: ANNUITIES: SINKING FUND

Problems on Loans When a person borrows some money and agrees to pay in equal instalments, then the money borrowed is equal to the P.V. of the proposed annuity.

Example 3. Mr S. Roy borrows Rs. 20,000 at 4% compound interest and agrees to pay both the principal and the interest in 10 equal annual instalments at the end of each year. Find the [C.U. B.Com. 2006] amount of these instalments. Solution. Let Rs. P =amount of each instalment; then the sum of Rs. 20,000 is the P.V. of an annuity of Rs. P to continue for 10 years at 4% p.a. Now, V =

![ { 1 -

(1

+ i)-n}

:. 20,000 =

and here, V =Rs. 20,000, n = 10 and i = 0.04.

~ { 1- (1 + i)-n} = 0 ~4 { 1- (1.04)- 10 } p

p

0.04 (1 - 0.6757) = 0.04 x 0.3243. p = 20000 x 0.04 = 8000000 = 2466.851.

..

0.3243 3243 :. Required amount of each instalment= Rs. 2,466.50. 8000000 [P = = 2467, by four-figure log-tables as shown below. 3243 log P = log 8000000 - log 3243 = 6.9031 - 3.5109 = 3.3922;:. P = antilog (3.3922) = 2467.]

Example 4. A man borrows Rs. 1,000 on the understanding that it is to be paid back in 4 equal instalments at intervals of six months, the first payment to be made six months after the money was borrowed. Calculate the value of each instalment, if the money is worth 53 p.a. Solution. Here, Rs.1,000 =present value of an immediate annuity of Rs. P (say), payable twice a year, for 4 periods and rate of interest i = 1g0 = 0.05. 4 p { 1 - ( 1 + -20.05)- } = 0.05 p { 1 - ( 1.025)-4} . :. 1000 = 0.05

Let x = (1.025)- 4 • : . log x = -4 log 1.025 = -4 x 0.0107 = -0.0428 = I.9572. :. x = antilog I.9572 = 0.9061. From equation (1), 1000 = ~(1-0.9061) = p x 0.0939 = p x 939 0.05 0.05 500 500 x 1000 :. P = =Rs. 532 (approx.). 939 :. Required value of each half-yearly instalment p

= "2 =

Rs. 532 =Rs. 266 (approx.). 2

(1)

148

BUSINESS MATHEMATICS AND STATISTICS

Example 5. A man borrows Rs. 2,500 at 2~% p.a. compound interest and agrees to pay both principal and interest in 18 equal annual instalments. Find the amount of each of these payments, the first being paid at the end of the first year. Given (1.025) 18 = 1.55966.

Solution. Let Rs. P =amount of each payment; then the sum of Rs. 2,500 =the P.V. of an immediate annuity of Rs. P to continue for 18 years at 2~% p.a.; hence, here, V = Rs. 2,500, n = 18 and i = :. From Formula II (Art. 7.2), V = P { (1 i

fat = ~oi = 0.025. + i)n _.:. 1 } (1 +it

'

we obtain, 2500

p

= 0.025

{ (1.025)

18

-

(1.025) 18

1}

p 1.55966 - 1 p 0.55966 = 0.025 . 1.55966 = 0.025 . 1.55966;

2500 x 0.025 x 1.55966 0.55966

.. p

=

25 x 25 x 155966 559660

100 x 100 x 155966 - 17417 ( ) 8954560 · approx. · .·. Required amount of each payment = Rs. 174.17 (approx.).

Cash-down Price Involving Present Value Example 6. A wagon is purchased on instalment basis, such that Rs. 5,000 is to be paid on

the signing of the contract and four-yearly instalments of Rs. 3,000 each, payable at the end of the first, second, third and fourth years. If interest is charged at 5% p.a., what would be the cash-down price'?

Solution. If V be the present value of the annuity of Rs. 3,000 for 4 years at 53 p.a. compound interest, then the cash-down price of the wagon= Rs. (V + 5,000). / We have V = ~ { 1 - (1 + i)-n}. Here, P = 3,000, n = 4, i = 1 ~0 = 0.05, V = ?

:. v = ~~g~

{

1 - (1+0.05)-

4

}

= 60000 { 1 - (1.05)-

4

Let x = (1.05)- 4 ; . . log x

= =

-4log1.05 = ~4 x 0.0212 = -0.0848 -1 + 1 - 0.0848 = i.9152.

:. x = antilog (I.9152) = 0.8226.

}.

(1)

149

CHAP. 7: ANNUITIES: SINKING FUND

From equation (1),

v = 60000(1 -

0.8226) = 60000 x 0.177 4 = 10644.

Hence, the required cash-down price of the wagon = Rs. 10,644 + Rs. 5,000 = Rs. 15,644. Example 1. A dealer advertises that a tape-recorder is sold at Rs. 450 cash down followed by two yearly instalments of Rs. 680 and Rs. 590 at the end of first year and second year respectively. 11 the interest char:ged is 18% p.a. compounded annually, find the cash price of the tape-recorder. (ISC 1997]

Solution. If present value of the first yearly instalment be Rs. 100, then interest for the first year= 18% of Rs.100 = Rs.18. :. First instalment= Rs.100 + Rs.18 = Rs.118.

First Instalment

Present Value

Rs.118

Rs.100

Rs.680

100 Rs. 118 x 680 =Rs. 576.27

Let the present value of the second instalment Interest for the first year Principal for 2nd year Interest for the second year (= 18% of Rs. 118) Second instalment Second Instalment Rs. 139.24 Rs. 590.00

= Rs.100.00 =Rs. 18.00 = Rs.118.00 =Rs. 21.24 "= Rs. 139.24

Present Value Rs. 100 Rs. 1ig.~4 x 590 = Rs. 423. 73

Hence-, the required cash price of the tape-recorder =Rs. 450 +Rs. 576.27 +Rs. 423.73 = Rs.1,450. Example 8. A person takes a loan on compound interest and returns it in 2 equal annual instalments. If the rate of interest is 16% p.a. and the yearly instalment is Rs.1,682, find the principal and the interest charged-W'lth each instalment.

150

BUSINESS MATHEMATICS AND STATISTICS

Solution. Let P.V. of 1st Instalment Then interest for 1 year at 16% First Instalment

=;;='Rs. 100.00 = Rs. 16.00 =Rs. p6.00

Let P.V. of 2nd Instalment Then interest for the 1st year Principal for the 2nd year Interest for the 2nd year Second Instalment

= Rs.100.00 =Rs. 16.00 = Rs.116.00 =Rs. 18.56 = Rs.134.56

First Investment Rs.116 Rs.1,682

Principal (P.V.) Rs.100 Rs. ~~~ x 1,682 = Rs.1,450 Principal (P.V.) Rs.100 1,682 _ Rs 1 250 Rs • lOOx 134.56 • '

Second Instalment Rs.134.56 Rs.1,682

Principal of the First Instalment= Rs.1,450 and interest charged= 1,682 - 1,450 =Rs. 232. Principal of the Second Instalment = Rs. 1,250 and interest charged= 1,682 - 1,250 =Rs. 432. Example 9. A man purchased a house valued at Rs. 3,00,000. He paid Rs. 2,00,000 at the time of purchase and agreed to pay the balance with interest at 12% p.a. compounded half-yearly in 20 equal half-yearly instalments. If the first instalment is paid after the end of six months from the date of purchase, find the amount of each instalment. Given, log 10.6 = 1.0253 and log 31.19 = 1.494. (C.U. B.Com. 1994, '97]

' Solution. By the question, present value of the amount paid in 20 equal half-yearly instalmen~ =Rs. (3,00,000 - 2,00,000) = Rs.1,00,000. Let Rs. P be the half-yearly inst.1·,· ~rit. Since inte,., ~s compounded half-yearly, we have

i)- n} . 2

2P { 1- ( l-2 V=T Here, V = 100000, i = :. From equation (1),

1 1; 0

= 0.12 and n = 10 [·: 20 half-years= 10 years]

:.~2 { 1- (1 + 0 ·~ 2 )- x o}, 2

·100000 =

(1)

1

or, 12000 = 2P { 1- (1.06)-

Let x = (1.06)- 20 • Then logx = -20log1.06 = -20 x 0.0253 = -0.506 = l.494. :. x = antilog(I.494) = 0,3119.

20

}

(2)

151

CHAP. 7: ANNUITIES: SINKING FUND

From equation {2), 6000 = P{l - 0.3119}

= P x 0.6881

6000 = 8719.66 (approx.) 6881

:. P = 0.

Hence, the required amount of each instalment= Rs. 8,719.66.

Single Equivalent Sum of Money Involving Present Value Example 10. A professor retires at the age of 60 years. He will get the pension of Rs. 42,000 a year to be paid in half-yearly instalment for the rest of his life. Reckoning his expectation of life to be 15 years and that interest is at 10% p.a., payable half-yearly, what single sum .is equivalent to his pension? [C.U. B.Com. 1996) Solution. Let Rs. V be the single sum equivalent to his pension. We have

i)- n} . 2

2P { 1- ( 1+ V=T 2 Here, P = 42g00 = 21, 000, i = :. From equation (1),

v=

2 x ;,~ooo

1 1 0~ =

{l)

0.1 and n = 15, V = ?

-2xl5} = 420000 { 1 - (1.05)-30}. {1 - (1 + O~l )

(2)

Let x = {1.05)- 30 . Then log x = -30 log 1.05 = -30 x 0.02119 = -0.6357 = I .3643. : . x = antilog (I.3643) = 0.23137. :. From (2), V = 420000(1 - 0.23137) = 420000 x 0.76863 = 3,22,824.60. Hence, the required single sum equivalent to his pen~ion = Rs. 3,22,824.60 or, 3,22,825. Formula III. Present Value of a Perpetuity: V =

t·

In case of a perpetuity, annuity is to continue for ever and n is indefinitely large and, therefore, (1 +it can be made as large as we please and hence, (1 + i)-n can be made as small as we like, i.e., (1 + i)-n may be ultimately taken to be zero. Hence, from the Formula II, we have V =

t· ,

Endowment Fund and Free-hold Estate Example 11. A person desires to create an endowment fund to provide for a prize of Rs. 300 every year. If the fund can be invested at 10% p.a. compound interest, find the amount of the endowment. Solution. The required amount of the endowment is the present value V of the perpetuity of Rs. 300.

152

. BUSINESS MATHEMATICS AND STATISTICS

We have

v= Here, P = Rs. 300, i = Hence,

i

1 1 0

=

p --;-. i

1 10 •

300 V = l/lO =Rs. 3,000.

Example 12. The annual rent of a free-hold estate is Rs. 1,000. What is its current value, if the compound interest rate is 4% p.a. ? Solution. A free-hold estate enjoys a rent in perpetuity. If V be the current value of the free-hold estate, then V = here, P = Rs.1,000, i = 1 ~0 , V = ?

f;

1000

. . v = 4/100 = Rs. 25,000. Hence, the required current value of the free-hold estate is Rs. 25,000.

7.3

Deferred Annuity

We state below two formulae, one for the present value of a deferred annuity and the other for the present value of deferred perpetuity. The proof of each of these two formulae is exactly similar to that of formula II for the present value of an annuity.

Formula I. (The Present Value of a Deferred Annuity) The Present Value V of a deferred annuity P to begin at the end of m years and to continue for n years is given by V _ P { (1 +it - 1} or P {l _ - i (1 + ir+n ' ' i (1

1

+ i)m+n

}- P i

{i _(1 +

1

. } ir ·

Formula II. (The Present Value of a Deferred Perpetuity) The Present Value V of a deferred perpetuity P to begin after m years is given by

Example 13. Mr X borrows a certain sum of money at 8% p.a. compound interest and agrees to pay both the principal and interest in 10 equal yearly instalments of Rs. 1,200 each. If the first instalment is to be paid at the end of 5 years from the date of borrowing and the other yearly instalments are paid regularly at the end of the subsequent years, find the sum borrowed by him. Solution. Since the first instalment is to be paid at the end of 5 years, the annuity in this case is deferred by 4 years. The present value (i.e., the amount of loan) V of a deferred annuity is given by P { {1 + i)n - 1 } V-- i f1 + i) m+n .

153

CHAP. 7: ANNUITIES: SINKING FUND

. Here, P = 1,200, n = 10, m = 4, i =

. u= •

Let x

•

>

1 ~0

= 0.08, V

= ? 10

10

1200 { (1 + 0.08) - 1} = 15000 { (1.08) - 1} . 0.08 (1+0.08) 10 +4 (1.08) 14

(1)

= (1.08) 10 and y = (1.08) 14 . log x = 10log1.08 = 10 x 0.0334 = 0.3340.

x

= antilog (0.3340) = 2.158.

logy = 14log1.08 = 14 x 0.0334 = 0.4676.

y

=

antilog (0.4676)

= 2.935.

: . From equation ( 1) , 2 158 1 58 15000 { · - } = 15000 x 1.1 2.935 2.935

v =

15000

1,158

x 2935

= 3474000 = R 587

5 918 23

s. '

.

.

Sinking Fund It is a fund created by investing annually a fixed sum of money at compound interest to pay off a loan or a debenture stock on some future specified date or to accumulate to a desired amount for replacing an asset, like plant and machinery, etc. If P be the sum of money set aside annually and A, the liability to be redeemed after n years, then A is the amount of the annuity P payable for n years and hence the formula I for the amount A can be used in this case. Example 14. A machine costs a company Rs. 80, 000 and its effective life is estimated to be 20 years. A sinking fund is created for replacing the machine at the end of its effective lifetime when its scrap realizes a sum of rupees five thousand only. Calculate to the nearest hundred of rupees, the amount which should be provided every year for the sinking fund, if it accumulates . at 9% p.a. compounded annually. Solution. For accumulation in a sinking fund at compound rate, we have

A=~i {(1 +it-I}. Here, A = 80,000 - 5,000 = 75,000, i = 1 ~0 = 0.09, n = 20, P = ? :. 75000 = Let x

0 ~ 9 { (1+0.09)

20

-

1}, or, 6750 = P · { (1.09)

= (1.09) 20 • . . logx = 20logl.09 = 20 x 0.0374 = 0.7480 x = antilog (0.7480) = 5.598.

20

-

1}.

(1)

154

BUSINESS MATHEMATICS AND STATISTICS

:. From equation (1), 6750 = p. {5.598 -1} = p x 4.598 or,

p = 6,750 = 67,50,000 = 1468 03:::::: Rs.1 500

4.598

4598

.

'

,

(correct to the nearest hundred). :. Required amount to be provided every year= Rs. 1,500. Example 15. A machine costs a company Rs. 65,000 and its effective life is estimated to be 25 years. A sinking fund is created for replacing the machine at the end of its lifetime, when its scrap realizes a sum of Rs. 2,500 only. Calculate what amount should be provided, every year, out of profits, for the sinking fund, if it accumulates at 3~% p.a. compound interest. Solution. Cost of the Machine =Rs. 65,000 Scrap Value =Rs. 2,500 Balance =Rs. 62,500 Clearly, this sum will be required after 25 years, to purchase the machine; hence, if Rs. P be every year's accumulation, the annuity P should amount to Rs. 62,500 in 25 years at 3~% p.a.

Now, A=

If {(1 +it -1}, and here, A= Rs. 62,500, n =

25, i = ~ = ~o't, = 0:035.

p { (1.035) 25 - 1 } :. 62500 = 0.035

Let x = (1.035)

25

;

(1)

then, log x = 25 log 1.035 = 25 x 0.0149 = 0.3725.

:. x = antilog0.3725 = 2.358. Now, from equation (1), p

p

0.035 (2.358 - 1) = 0.035 x 1.358. . p - 62500 x 0.035 - 62500 x 35 .. 1.358 1358 . log P = log 62500 + log 35 - log 1358 = 4. 7959 + 1.5441 - 3.1328 = 3.2072.

:. P = antilog 3.2072 = 1612. :. Sum to be provided for sinking fund in every year= Rs.1,612 (approx.).

1.4

Amortization

In annuities, repayment of loan with compound interest is usua.Uy made in equal periodical instalments which include repayment of principal as well as interest. But to facilitate account-' ing, it is sometimes necessary to split up each instalment paid into repayment of principal and payment of interest on the outstanding balances. The principal is thus steadily amortized (see Example 16).

155

CHAP. 7: ANNUITIES: SINKING FUND

Example 16. A loan of Rs. 1,000 is to be paid in 5 equal annual payments, interest being at 6% p.a. compound interest and first payment being made after a year. Analyze the payments into those on account of interest and on account of amortization of the principal. Solution. For present value V of an annuity,

where P is the annual payment in rupees. Here, V = 1000, n = 5, i = 1 ~0 = 0.06, P = ? :. 1000 =

0

~6 { 1 -

(1+0.06)- 5 }, or, 60 = P · { 1 - (1.06)-

5

}.

(1)

Let x = (1.06)- 5 ; .. Iogx = -5logl.06 = -5 x 0.0253 = -0.1265

= I.8735.

:. x = antilog (i.8735) = 0.7473. :. From equation (1), 60 = p. {1 - 0.7473} = p x 0.2527 .or,

60

p = 0.2527 =

600000 2527 =Rs. 237.40 (approx.).

Amortization Table End of year 1st 2nd 3rd 4th 5th "Total

Annual payment p (Rs.)

Interest due (Rs.)

Amortization (Rs.)

237.40 237.40 237.40 237.40 237.40

60.00 49.36 38.08 26.12 13.44

177.40 188.04 199.32 211.28 223.96

1,187.00

187.00

1,000.00

Explanation: Amortization at the end of 1st year

237.40 =

Amortization at the end of 2nd year

Principal earning interest (Rs.) 1,000.00 822.60 634.56 435.24 223.96

g x 1000

1 0

Rs. 177.40.

237.40 - (60 -

lgO X

177.40)

237.40 - 49.36

Rs. 188.04 and so on.

7.5

Miscellaneous Problems

Example 17. A person invests Rs.1,000 every year with a company which pays interest at 10% p.a. He allows his deposit to accumulate with the company at compound rate. Find the amount standing to his credit one year after he has made his yearly investment for the tenth time.

'

156

. ·." BUSINESS MATHEMATICS AND STATISTICS

If {(1 +it - 1}.

Solution. If A be the amount of the annuities for 10 years, then A = Here, P = 1,000, i = l~o = 0.1, = 10, A= ?

n

:. A=

1

g.~o

{

(1+0.1)

10

-

1} = 10000 { (1.1)

10

(1)

1}.

-

Let x = (1.1) 10 ; then, logx = lOlogl.1=10 x 0.0414 = 0.4140. :. x = antilog (0.4140) = 2.594.

:. From equation (1), A= 10000(2.594 - 1) = 10000 x 1.594 = Rs.15,940. Hence, the required amount standing to his credit = Rs. 15,940 + Int. on Rs. 15,940 @ 10% p.a.

=Rs. 15,940 +Rs. 1,594 = Rs.17,534. Example 18. Compute the amount and present value of an annuity of Rs. 480, payable at the end of each 3 months for 12 years at 6%, compounded quarterly.

Solution. Here, i = 0.06, n = 12 and annuity P = Rs. 480, payable 4 times a year.

p {(1

i

:. A =

i)

+4

8000 { (1.015)

4

4

n 48

- l -

480 {(1 0.06) + -4- 0.06

}-

xl2

- l

}

1} = 8000 x 1.029 = 8232. [(1.015) 48 = 2.1029, calculated below]

:. Required Amount= Rs. 8,232 (approx.). Present value V

=

~ { 1-

~ )- n} = ~.~~ { (1 4

(1+

1.015)-

48

}

8000(1 - 0.4929) = 8000 x 0.5071 = 4056.80 (approx.). [(1.015)- 48 = 0.4929, calculated below] :. Required present value= Rs. 4,056.80 (approx.). [Let x = (1.015) 48 and y = (1.015)- 48 .. .. .. ..

log x = 48log1.015 = 48 x 0.0064 = 0.3072. x = antilog 0.3072 = 2.029. logy = -48 log 1.015 = -48 x 0.0064 = -0.3072 = I.6928. y = antilog I.6928 = 0.4929.]

Example 19. Fin~ the present value of an annuity of Rs. 300 p:a. for 5 years at 4%. Given log 104 = 2.0170333, log0.0821923 = 2.9148335.

157

CHAP. 7: ANNUITIES: SINKING FUND

Solution. Here, P =Rs. 300, n = 5, i = 1 ~0 = 0.04. ,', V

=

~ {1 -

i.~~

(1 + i)-n} =

{

1 - (1.04)-

5 }

=

i.~~ (1 -

0.821923)

[(1.04)- 5 = 0.821923, calculated below] 300 0 80 7 - 3 x 1780. 77 - 5342.31 4 4 0.04 x ·17 7 Rs.1,335.58 (approx.).

[Let x = (1.04)- 5 .. logx = -5logl.04 = -5 x 0.0170333 = -0.0851665 = I.9148335 = log0.821923 . .. x = 0.821923.) Example 20. What is the amount and present value of an annuity certain of Rs. 500 for

15 years, reckoning compound interest at 4!% p.a.? Given, (1.045) 15 = 1.93528.

Solution. Here;P = Rs. 500, n = 15, i = ,', A

=

~i {(1 +it-1} =

TJo

= t0t = 0.045.

500 500 15 {Cl.045) -1} = (1.93528-1) 0.045 0.045

500 x 0.93528 = 5 x 93528 = Rs. l0, 392 .00 0.045 45 and

v

=

~{ i

15

(1 +it - 1} = 500 { (1.045) - 1} = 500 x 1.93528 - 1 c1 + i r 0.045 ci.045) 15 0.045 1.93528

500 x 0.93528 = 500000 x 93528 = 100000 x 10392 = R . . 5 369 77 0.045 1.93528 45 193528 193528 s ' . Example 21. The accumulation in a Provident Fund is invested at the end of every year to earn 10% p.a. A person contributes 12~% of his salary to which his employer adds 10% every

month. Find how much the accumulations will amount to at the end of 30 years of his service, for every 100 rupees of his monthly salary. Give the answer to the nearest rupee.

Solution. If the monthly salary of the person be Rs. 100, then the total monthly contributions to Provident Fund = 12 ! + 10 = Rs. 22 ! . Total yearly contributions to Provident Fund= Rs. 22! x 12 =Rs. 270. If A be the total accumulation at the end of 30 years, then

Here, P = 270, i = 1~~ = 0.1, n = 30, A= ? 270 { (1+0.1) 30 - 1} = 2,700 { (1.1) 30 - 1 } ' . :. A= DI

(1)

158 Let x

BUSINESS MATIIEMATICS AND STATISTICS

= (1.1) 30 ; then log x = 30log1.1 = 30 x 0.0414 :. x

= 1.2420.

= antilog(l.2420) = 17.46 .

.·. From equation ( 1),

A= 2700{17.46-1} = 2,700 x 16.46 = Rs.44,442. Example 22. A person retires at the age of 58 and earns a pension of Rs. 6,000 a year. He

wants to commute one-fourth of his pension to ready money. If the expectation of life at this age be 12 years, find the amount he will receive when money is worth 4% p.a. compound.'(It is ass1tmed that pension for a year is due (J;t the end of the year.)

Solution. tth of the pension = ~ >< Rs. 6,000 = Rs.1,500. For the present value of annuities

V Here, P

=

~ { 1-

(1

= Rs.1,500, n = 12, i = 1 ~ 0 = 0.04, V =? ,', V = 1500 0.04 { 1 - (1+0.04 )-12}

Let x

+ i)-n} .

= 37,500 { 1 -

(1.04 )-,12} .

(1)

= (1.04)- 12 ; then logx

= -12logl.04 = -12 x 0.0170 = -0.2040 = f.7960. :. x = antilog(f.7960) = 0.6252.

:. From equation (1), V = 37,500(1 - 0.6252)

= 37,500 x 0.3748 =Rs. 14,055.

Example 23. A person purchases a house worth Rs. 70,000 on a hire purchase scheme. At the

time of gaining possession he has to pay 40% of the cost of the house and the rest amount is to be paid in 20 equal annual instalments. If compound interest is reckoned at 7 % p.a., what should be the value of each ins,talment'?

!

Solution. 40% of Rs. 70,000 = Rs. 28,000. Balance = Rs. 70,000 - Rs. 28,000 = Rs. 42,000. We have V = ~ { 1 - (1 + i)-n}, Here, V

71

= Rs. 42,000, n = 20, i = jfo 42000 = or,

3150

=

= 0.075, P

=

o.~75 { 1 -

(1+0.075)-

?

P · { 1- (1.075)-

20

}.

20

}

(1)

159

CHAP. 7: ANNUITIES: SINKING FUND

Let x = (1.075)- 20 ; then log x = -20log1.075 = -20 x 0.0314 = -0.6280 = I.3720.

:. x = antilog (i.3720) = 0.2355. :. From equation (1), 3,150 = P{l - 0.2355} = P x 0.7645 or,

p = 3150 = 31500000 = Rs. 4 120 ( l ) 0.7645 7,645 ' near Y ·

Hence, the required value of each instalment= Rs. 4,120. Example 24. The price of a tape-recorder is Rs.1,561. A person purchased it by paying a cash of Rs. 300 and the balance, with due interest, in 3 half-yearly equal instalments. If the dealer charges interest at the rate of 103 p.a. compounded half-yearly, find the value of each instalment. Solution.

Price of the tape-recorder Cash amount Balance to be paid by instalment We have

v= Here, ~ =

f

1 2 x 00

p {

i/2

1-

(

= Rs. 1,561.00 =Rs. 300.00 = Rs. 1,261.00

i). -n} .

1+ 2

= 0.05, V = 1,261, n = 3. (Here, n = no. of half-years.]

1261 = or, or, or,

0~5 { 1 -

(1+0.05)-

1,261 x 0.05 = P { 1 -

3 }

=

0~5 { 1 - (1.~5)3}

1.l 5~ 625 }

0.157625 1,261 63.05 = P x 1. _ = P x , (see Example 4] 157625 9 261 p = 63.05 x 9,261 =Rs 463 05 1,261 . . .

:. Required value of each instalment= Rs. 463.05. Example 25. A person borrowed some money and returned it in 3 equal quarterly instalments of Rs. 4,630.50 each. What sum did he borrow if the rate of interest was 20% p.a. compounded quarterly? Find also the total interest charged.

160

BUSINESS MATHEMATICS AND STATISTICS

Solution. We have

v

= -

p {

i/4

Here, P = Rs. 4,630.50, i/4=5/100=0.05, n = 3 (n =no. of quarters).

. )-n}

(

1- 1+.: 4

460~~~50 {1- (1+0.05)-3} 92610 { 1 -

= 4635050 { 1- (1.05)-3}

(1.~5)3} = 92610 { 1 - 1.15~625}

92610 x 0.157625 = 92610 x 157625 = 92610 x 1261 9261 1.157625 1157625 Rs.12,610. The total interest charged = =

Rs. 4,630.50 x 3 - Rs. 12,610 Rs. (13,891.50 - 12,610) Rs. 1,281.50.

Example 26. A man buys an old piano for Rs. 500, agreeing to pay Rs. 100 down and the balance in equal monthly instalment of Rs. 20 with interest at 6%. How long will it take him to complete payment? ' Solution. Amount paid in cash = Rs. 100 . .·. The balance amount = Rs. 500 - Rs. 100 = Rs. 400. Let the required time be n years. We have v = ip { 1- ( 1+12i )-12n} .

Here, V = 400, P = 20 x 12 = 240, i = 1g0 = 0.06, n = ? 400 =

~.~~ { 1 -

( 1+

2

0~~ )-l n}, 6

12 or, 24 = 240 { 1 - (1.005)- n}

or,

224~

or,

1 9 12n 10 12n = 10' or, (1.005) = -9 . (1.005)

= 1 - (1.005)-12n, or, (1.005)-12n = 1 - 110

12n log 1.005 = log 10 - log 9,

or,

12n x (0.0021) = 1 - 0.9542 = 0.0458

or,

12 = 0.0458 = 458 = 21. 81. n 0.0021 21

Hence, the required time_= n years= 12n months= 21.81months=22 months (approx.). '

Example 27. A machine costs a company Rs. 1,00,000 and its effective life is estimated to be 20 years. If the scrap is expected to realize Rs. 5,000 only, find, correct to the nearest rupee, the sum to be invested every year at 5% p.a. C.I. for 20 years, to replace the machine which is expected to cost then 25% more over its present cost. Assume that the proceeds from the sale of the scrap would be utilized for meeting the cost of the machine.

161

CHAP. 7: ANNUITIES: SINKING FUND

Solution. We have

A= Here, A P=?

= 1,00,000 + :. 120000 =

~i {(1 +it .

1}.

253 of 1,00,000 - 5,000 = Rs. 1,20,000; n

0~5 { (1+0.05)

20

-

= 20, i = 5/100 = 0.05,

1}, or, 6000 = P { (1.05)

20

-

1}.

(1)

' 20 Let x = (1.05) ; then

log x

= 20 x log 1.05 = 20 x 0.0212 =

0.4240;

: . x = antilog (0.4240) = 2.655. :. From equation (1),

p. (2.655 - 1) = p x 1.655

6000 or, P

=

6000 = 60,00,000 =Rs 3 625 38 L655 1, 655 . ' . Rs. 3,625, correct to the nearest rupee.

Hence, the required sum to be invested = Rs. 3,625.

Deferred Perpetuity Example 28. For endowing an annual scholarship of Rs. 12,000, a man wishes to make three equal annual contributions. The first award of the scholarship is to be made three years after the last of his three contributions. What would be the value of each contribution, assuming interest at 2.53 p.a. compounded annually? (Assume that the first contribution is to be made now and the of/ter two at an interval of one year thereafter.) Solution. The first scholarship is payable at the end of the 5th year and then it is contributed for ever. Thus, we have a perpetuity of Rs.12,000 deferred by 4 years. If.the annual contribution be Rs. x, then the present value of endowments is given by

v = x + T{1 Here, n

c1

+ i)-n}.

= 2, since the first payment is made now and i = f0~ = 0.025.

:. v =

x+

.; { 1 - (1+0.025)0 25

2

}

= x + .; { 1 - (I.025)- 2 }. 0 25

Let z = (1.025)- 2 ; then log z

= -2log1.025 = -2 x 0.0107 = -0.0214 = I.9786. :. z = antilog (I.9786) = 0.9519.

'. S!Js. Math. & Stal (C.U.}-11

(1)

162

BUSINESS MATHEMATICS AND STATISTICS

:. From equation (1)

v

=

x x x + 0.025 {l - 0.9519} = x + 0.025 x 0.0481

=

x ( 1+

:~~) = x . ( ~~~) .

For the present value of deferred perpetuity,

Here, P = 12,000, m = 4, i = 0.025, V = ?

v=

12000 x 1 = 12000000 = 480000 0.025 (1.025) 4 25. (1.025) 4 (1.025) 4 .

. . log V = log 480000 - 4log1.025 = 5.6812 - 0.0107 = 5.6705.

(2)

:. V = antilog (5.6705) =Rs. 4,68,200.

Hence, from equations (1) and (2), we get 731) - 4 68 200 - 4,68,200 x 250 x ( 250 - ' ' ' or, x 731 · . . log x = log 468200 +log 250 - log 731 = 5.6704 + 2.3979 - 2.8639 = 5.2044. :. x = antilog (5.2044) = Rs.1,60,100.

Example 29. What sum should be invested every year at 8% p.a. C.I. for 10 years, to repfoce plant and machinery which is expected to cost then 20% more than its present cost of Rs. 50,000? [C.U. B.Com• 1986]

Solution. 20% of Rs. 50,000 = 10,000. : . The cost of the machinery after 10 years = Rs. 60,000. This is the amount of an annuity of Rs. P for 10 years @ 8 p.a. C.I. Using the formula,

we get 60000 =

0 ~8 { (1.08)

10

-

1}, or,

or,

60000 x 0.08 = P, 2.158 -1 . [(1.08) 10 = 2.1&8, logarithmic calcuiations given below) 4800 P = 1. =Rs. 4,145 (approx.). 158

163

CHAP. 7: ANNUITIES: SINKING FUND

Logarithmic Calculations Let x = (1.08)

10

and P = 4800/1.158. log x = 10log1.08 = 10 x 0.0334 = 0.3340. x = antilog0.3340 = 2.158. log P = log4800 - log 1.158 = 3.6812 -- 0.0640 P == antilog3.6172 = 4145 (approx.).

= 3.6172

Example 30. A man wishes to buy a house valued at Rs. 50,000. He is prepared to payRs. 20,000 now and the balance in 10 equal instalments. If the interest is calculated at 83 p.a., what should [C.U. B.Com. 1987] he pay annually? [Given, (1.0~) 10 = 0.4634.] Solution. Here, the balance Rs. 30,000 is paid in the form of an annuity of Rs. P (say)-for 10 years @ 83 p.a. Using the formula

v=

~ { 1 - (1 ~it}'

we get 30000 =

0~8 { 1 -

(1.

1 08

) 10 } , or, or,

30000 =

P =

0 ~8 x {l - 0.4634}

30000 x 0.08 0. =Rs. 4,473 (approx.). 5366

Try a Similar Problem Example 31. A man wishes to buy a Government House valued at rupees one lac. He will have to pay 20% initially and the balance with compound interests in 20 equal annual instalments. If the rate of interest be 123 p.a., then what will be the amount of his ann'Ual instalments? 20 [Given: (1.12)- = 0.1037.J [Ans. Rs.10,710.70.] [C.U.B.Com. 1991]

7.6

Amount of an Annuity Due

In annuity due, payments are made at the beginning of each period. In this case, the first payment P will earn compound interest for n years, if the yearly payments continue for n years and so P will amount to P(l +it. Similarly, the second amount P will earn compound interest 1 for (n -1) years and it will amount to P(l + it- and so on. Finally, the last payment P will earn interest for 1 year and it will amount to P(I + i). Hence,

A

1

2

=

Amount= P(l +if+ P(I + it- + · · · + P(i + 1) + P(l + i)

=

P·(I+i) [1+(1+i)+(l+i) + .. ·+(l+if-

=

P-(l+i)·

2

1 ]

l·~~l:i;):~l} = ~(l+i)[(l+if-1].

164

BUSINESS MATHEMATICS AND STATISTICS

Thus,

A=

~i (1 + i) [(1 +it -

1].

Note: Notice that this formula can be easily obtained from A = ~ ((1 + i)n - 1] multiplying by ( 1 + i), since all the instalments (i.e., payments) are shifted by 1 year from the end to the beginning of the year.

Corollary 1. If the payments are made p times a year, each Rs. P and i is the interest per unit sum p.a. compounded p times a year, then the amount A is given by

The amount A of annuity due (P) payable half-yearly, quarterly and monthly are respectively 2Pi ( 1+~") • A=

4Pi ( 1+~') • A=

Example 32. A bank pays interest at the rote of 10% p.a. compounded yearly. If Rs. 2,000 is

deposited to the bank at the beginning of each year, end of 5 years?

wha~.

will be the total accumulation at the

Solution. For amount A of annuity due, we have A= f(l+ i) [(1 +it -1]. Here, P = Rs. 2,000, i = 1~~ = 0.1, n = 5 years, A= ? 2 0.1

:. A= 000 .(1+0.1)[(1+0.1) 5

-

1] = 20000 x f.1[(1.1) 5

-

1] .

Let x = (1.1) 5 ; :. logx = 5logl.1=5 x 0.0414 = 0.207; . . x = antilo~ (0.207) = 1.611. :. A= 20000 x 1.1[1.611 - 1] = 22000 x 0.611 = Rs.13,442. Example 33. A bank pays interest at the rote of 8% p.a. compounded half-yearly. Find how much should be deposited in the bank at the beginning of each half-year in order to accumulate Rs. 8,000 in 3 years. (CBSE 2002)

Solution. For amount A of annuity due payable at the beginning of each half-year,

165

CHAP. 7: ANNUITIES: SINKING FUND

Here, A

= Rs. 8,000, i = 8000 = or,

640

g = 0.08, n = 3, P =

1 0

J!!_

0.08

(1

+

?

0.08) [(1 0.08)3x2 - 1] 2 + 2

= 2P(l + 0.04) [(1.04) 6

-

1J = 2P x 1.04 [(1.04)

6

-

1J .

(1)

Now, let x = (1.04)6. . . log x = 6 log 1.04 = 6 x 0.0170 = 0.1020. :. x = antilog(0.1020) = 1.265.

:. From equation (1), 640

P

= 2P x 1.04[1.265 - 1] = 2P x 1.04 x 0.265 = 0.5512P. 640

= 0. 5512 = Rs.1,161.10

or,

Hence, the half-yearly payment to be made to the bank

7. 7

Rs.1,161.

= Rs.1,161.

Present Value (V) of an Annuity Due (P)

In annuity due, payments are made at the beginning of each period. In this case,

v

The present value (V) of annuity due (P) payable half-yearly, quarterly and monthly are respectively 2

p ( l+~·) •V=i 4

[ 1- (

1+~·)-

p ( 1+±· ) [ 1- ( 1+±· )•V=-i-.

2

nl ;

4

nl ; 12

• v

12P ( 1 + li2· ) [ 1 - ( 1 + li2· )= -i-.

nl .

Example 34. Find the present value of an annuity due to Rs. 700 p.a., payable at the beginning

of each year for 2 years, allowing interest at 6% p.a., compounded annually. [Take (1.06- 1 0.943.]

=

166

BUSINESS MATHEMATICS AND STATISTICS

Solution. The Present Value (V) of an annuity due (P) is given by V =

I[ (l +i)

(1 - (1 + i}-n] ~

Here, P = Rs. 700, i = 1 ~ 0 = 0.06, n = 2, V = ?

.'. v

=

~.~~(1+0.06) [1- (1 +0.06)- 2] 700

~

100

[1.06 - 0.943] =

70 00

~

=

~.~~

[(1.06}- (1.06)-

1 ]

x 0.117 = 1365.

Hence, the required present value= Rs.1,365. Example 35. A loan is to be paid in 6 equal annual payments, each payment of Rs. 2,500 being made at the beginning of the year. If interest is charyed at 8% p.a. compounded yearly, find the amount paid for the loan. Solution. For the amount of an annuity due payable yearly, A=

f (1 + i) [1 -

(1 + i)-n] .

Here, P = 2500,i = 1 ~0 = 0.08,n = 6,A = ?

:. A = =

~~i~ (1+0.08) [1 -

(1+0.08)-

31250 x 1.08 [1 - (1.08)- 6 ]

6

)

•

(1)

Let x = (1.08)- 6 . . . logx = -6logl.08 = -6 x 0.0334 = -0.2004 = I.7886. :. x

= antilog (I.7886) = 0.6304 .

. . A= 31250 x 1.08[1 - 0.6304] = 31250 x 1.08 x 0.3696 = 12474. Hence, the required amount paid for the loan= Rs.12,474.

---------11 EXERCISE ON CHAPTER 1 !1-------(Annuities)

A 1. Find the amount of an immediate annuity of Rs. 100 p.a. left unpaid for 10 years, allowing 5% p.a. compound interest. [(1.05) 10 = 1.629] [C.U. B.Com. 2007]

2. Find the present value of an annuity of Rs. 500, payable at the end of every year for 10 years at 53 p.a. compound interest. [C.U. B.Com. 2007] 3. Find the amount and the present value of an annuity of Rs. 150 for 12 years, reckoning interest at 3!% p.a. [Given, (1.035) 12 = 1.511066] 4. Find the amount of an annuity of Rs. 100 in 20 years, allowing compound interest at 4!%. (Given, log 1.045 = 0.191163 and log 24.117 = 1.3823260]

CHAP. 7: ANNUITIES: SINKING FUND

5.

167

(a) A man decides to deposit Rs. 300 at the end of each year in a bank which pays 33 p.a. compound interest. If the instalments are allowea to.accumulate, what will be the total accumulation at the end of 15 years? (b) A man decides to deposit Rs. 10,000 at the end of each year in a bank which pays 103 p.a. compound interest. If the instalments are allowed to accumulate, what will be the total accumulation at the end of 9 years? [C.U.B.Co~.(H) 2001) (c) A man invests Rs. 2,000 every year with a company which pays interest at 10% p.a. He allows his deposits to accumulate with the company at compound rate. Find the amount standing to his credit one year after he has made his yearly investment for the 8th time. [Hints: See worked-out Example 2.)

6.

(a) A TV is purchased on instalment basis, such that Rs.1,000 is to be paid on signing of the contract and the balance in 6 yearly instalments of Rs. 600 each, payable at the end of each year. If interest is charged at 8% p.a., what would be the cash-down price? (b) A motor cycle is purchased on instalment basis such that Rs. 3,400 is to be paid on the signing of the contract and four-yearly instalments of Rs. 2,400 each payable at the end of the first, second, third and fourth years. If interest is charged at 8% p.a., what would be the cash-down price? [Given that (1.08) 4 = 1.36] (c) A TV set is purchased on instalment basis such that 60% of its cash-down price is to be paid on signing of the contract and the balance in six equal instalments ~yable half-yearly at 10% p.a. compound interest (compounded half-yearly). If the cash-down price of the set qe Rs. 4,200, find the value of each instalment. [C.U. B.Com.(Hl).1986)

7.

(a) What is the value of perpetuity of Rs. 250 at 8~% p.a.? (b) Mr S. Dey desires to create an endowment fund to provide for a prize of Rs. 500 every year. If the fund c:an be invested at 8% p.a. compound interest, find the amount of ·the endowment.

8.

(a) Shri A. B. Roy borrows Rs. 20,000 at 4% p.a. compound interest and agrees to pay both the principal and the interest in 10 equal annual ins.talments at the end of each year. Find the amount of each instalment. (b) A person borrowed a sum of money at compound interest and returns it in 2 equal annual instalments. If the rate of interest is 203 p.a. and the yearly instalment is Rs. 1,980, find the principal and the interest charged with each instalment. [Hints: See worked-out Example 3.)

9. A company borrows Rs.10,000 on condition to repay it with compound interest at 5% p.a. by annual instalments of Rs.1,000 each. In how many years will the debt be paid off? 10. What sum should be invested every year at 53 p.a. compound interest for 20 years, to replace plant and machinery, which is expected to cost then 25% more over its present cost of Rs. 60,000?

168

BUSINESS MATHEMATICS AND STATISTICS

11. A person desires to endow a bed in a hospital the cost for which is Rs. 6,000 p.a. If the money is worth 10% p.a., how much should he provide in perpetuity? He further desires to provide for the building of a ward at a cost of Rs. 1,00,000 and also to provide for periodical repairs thereto entailing 5% of the cost of the ward per year on the average. What should be the total value of the endowment? 12. A limited company intends to create a depreciation fund to replace, at the end of the 25th year, assets costing Rs. 1,00,000. Calculate the amount to be retained out of profits every year, if interest rate is 3% p.a. 13.

(a) A sinking fund is created for the redemption of debentures of Rs. 1,00,000 at the end of 25 years. How much money should be provided out of profits each year, for the sinking fund, if the investment can earn interest@ 4% p.a.? (b) A sinking fund is created by a company for replacing some machinery worth Rs. 1,00,000. after 25 years. How much should be set aside from profit each year, if the rate of compound interest be 9% per annum? [Given, log 1.09 = 0.037 4 and log 8.590 = 0.935] [C.U. B.Com. 2008]

(Hints: In Formula, A= :.100000 =

f {(1 + i)n -

o.~9 {(1+0.09) 25

-

1}, A= 1,00,000; i =

1}, or, 9000 = P{(l.09) 25

-

1 ~0

= 0.09, n = 25, P =?

1}.

'

Let x = (1.09) 25 ; :. logx = 25log1.09 = 25 x 0.0374 = 0.935 . . ', x = antilog(0.935) = 8.590. So, 9000 = P(8.590 - 1) = (7.59)P. :. P = ~~g~ = 1185.77, or, Rs. 1,186 (approx.).]

14. A man retires at the age of 60 years and his employer gives him a pension of Rs. 1,200 a year in half-yearly instalments for the rest of his life. Reckoning his expectation of life to be 13 years and that interest is 4% p.a., payable half-yearly, what single sum is equivalent to his pension?

15. A machine costs a company Rs. 97,000 and its effective life is estimated to be 12 years. If the scrap realizes Rs. 2,000 only, what amount should be retained out of profits at the end of each year to accumulate by compound interest at 5% p.a.? 16. The accumulations in a Provident Fund are invested at the end of every year to earn 10% p.a. A person contributes 12~% of his salary to which the employer adds 10% every month. Find how much the accumulations· will amount to at the end of 30 years of his service, for every 100 rupees of his monthly salary. (Give the answer to the nearest rupee.)

17. A person borrowed some money and paid back in 3 equal annual investments of Rs. 2,160 each. What sum did he borrow, if the rate of interest charged was 20% p.a. compounded annually? Find also the total interest charged. 18. A man buys a house for Rs. 60,000 on condition that he will pay Rs. 30,000 cash-down and the balance in 10 equal annual instalments, the first to be paid one year after the date of purchase. Calculate the amount of each instalment, compound interest being computed at the rate of 5% p.a. Given, {1.05)- 10 = 0.6139. [C.U. B.Com.(H) 1988] 19. A professor retires at the age of 60 years. He will get his pension of Rs. 42,00Q a year to be paid in half-yearly instalments for the rest of his life. Reckoning his expectation of life to be 15 years and that interest is at 10% p.a., payable half-yearly, what single sum is equivalent to his pension? [C.U. B.Com. 1996, 2006]

169

CHAP. 7: ANNUITIES: SINKING FUND

20. A firm borrows Rs.1,000 on condition to repay it with C.I. at 43 p.a. by annual instalments of Rs. 100 each. In how many years will the debt be paid off? 21. A sinking fund is created for replacing some machinery worth Rs. 54,000 after 25 years.

The scrap values of the machine at the end of the period is Rs. 4,000. How much should be set aside from profit each year for the sinking fund, when the rate of the C.I. is 53? 22. A man decides to deposit Rs.1,000 at the beginning of each year in a bank which pays 103 p.a. compound interest. If the instalments are allowed to accumulate, what will be the total accumulation at the end of 5 years? 23. A bank pays interest at the rate of 83 p.a. compounded half-yearly. If Rs. 500 is deposited to the bank at the beginning of every 6 months, what will be the total accumulation at the end of 8 years? 24. Find the present value of annuity due of Rs. 700 p.a., payable at the beginning of each year for 2 years, allowing 63 p.a. compound interest. [Take (1.06)- 1 = 0.943.] 25. A loan is to be paid in 8 equal half-yearly payments, each of Rs. 4,000, to be made at the beginning of every 6 months. If interest is charged at 103 p.a., payable half-yearly, find the amount of the loan. 26. A man buys a house for Rs. 6,00,000 on condition that he willpay Rs. 3,00,000 cash-down and the balance in 10 equal annual instalments, the first to be paid one year after the date of purchase. Calculate the amount of each instalment, compound interest being [C.U. B.Com. 2005)· computed @ 53 p.a. [Given, (1.05)- 10 = 0.6139]. [Hints.: Balance amount= Rs. 6,00,000 - Rs. 3,00,000 =Rs. 3,00,000. We have, V =

t {1 -

Now, 300000 =

0 ~5

{

(1 + i)-n}. Here, V = 3,00,000, n = 10, i = 0.05, P = ? 1 - (1+0.05)- 10 }

or, 15000'= P { 1 - (1.05)- 10 } = P{l - 0.6139} = P(0.3861) or, P

= ~-~~~~ = Rs. 38,850.04 {approx.).] B

1. A person invests Rs. 500 at the end of each year with a bank which pays interest at 103 · p.a. compounded annually. Find the amount standing to his credit one year after he has made his yearly investment for the 12th time. 2. The accumulations in a Provident Fund are invested at the end of every year to earn 103 p.a., CJ. A man contributes 11~3 of his salary to which his employer adds 8~3 every month. Find how much the accumulations will amount to at the end of 20 years of his service for every Rs. 100 of his monthly salary. 3. Calculate the present value of an annuity of Rs. 5,000 p.a. for 12 years, the 43 p.a. compounded annually.

int_er~st

being

4. A person retires at the age of 60 years and earns a pension of Rs. 7,200 a year. He wants to commute one-third of his pension to ready money. If the expectation of life at this age be 10 years, find the amount he will receive when money is worth 83 p.a. compound interest.

170

BUSINESS MATHEMATICS AND STATISTICS

5. Sri T. Guba purchases a house worth Rs. 60,000 on a hire purchase scheme. At the time of taking possession he has to pay 403 of the cost of the house and the rest amount is to be paid in 18 equal annual instalments. If compound interest is reckoned at 8!3 p.a., what should be the value of each 1nstalment? 6. A machine costs a company Rs. 80,000 and its effective life is estimated to 15 years. If the scrap is expected to realize Rs. 8,000 only, find the sum to be invested every year at 73 p.a. C.I. for 15 years, to replace the machine which is expected to cost then 203 more over its present cost. (Scrap would be utilized for meeting the cost of the machine.) 7. Which is cheaper-an annuity of Rs.1,500 to last for 8 years or the deferred perpetuity of Rs. 800 to commence 5 years hence at 63 p.a. compound interest? 8. A person borrows a sum of Rs. 5,000 at 43 compound interest. If the principal and interest are to be repaid in 10 equal annual instalments, find the amount of each instalment, first payment being made after 1 year. 9. A loan of Rs. 8,000 is to be paid in 5 equal annual payments, interest being at 5 per cent p.a. compound interest and first payment being made after a year. Find the amount of each of the payments. 10. Shri Biswas bought a house, payibg Rs. 20,000 down and Rs. 4,000 at the end of each year for ~5 years. What ought he to have paid for the house if he had bought it cash-down? Reckon interest at 53 p.a. compound. 11. A sinking fund is created for the redemption of debentures of Rs. 1,00,000 at the end of 20 years. How much money should be provided out of profits each year for the sinking fund, if the investment can earn interest 83 p.a.? 12. A company sets aside for a reserve fund the sum of Rs. 20,000 annually to enable it to pay off a debenture issue of Rs. 2,39,000 at the end of 10 years. Assuming that the reserve accumulates at 4% p.a. compound, find the surplus after paying off the debenture stock. 13. A man purchases a house and takes a mortgage on it for Rs. 60,000 to be paid off in 12 years by equal annual payments. If the interest rate is 5 per cent compounded annually, what amount will be re9uired to pay each year? 14. Find the sum of money received by a pensioner at 58 if he wants to commut~ his annual pension of Rs. 1,200 for a present payment when compound interest is reckoned at 43 p.a. and the expectation of his life is assessed at 10 years only. 15. For endowing an annual scholarship of Rs. 10,000, Mr Reddy wish~&_to- make five equal annual contributions. The first award of the scholarship is to be made two years after the last of his five contributions. What would be the value of each contribution, assuming interest at 53 p.a. compounded annually? [Assume that the first contribution is to be made now and the other four at an interval of one year thereafter.] 16. On his 48th birthday, a man decides to make a gift of Rs. 5,000 to a hospital. He decides to save this amount by making equal annual payments up to and including his 60th birthday to a fund which gives 3!3 compound interest, the first payment being made at once. Calculate the amount of each annual payment.

CHAP. 7:

ANNUITIES~

171

SINKING FUND

17. An article is sold for Rs. 485 in cash or Rs. 105 as cash-down payment followed by 3 equal monthJy instalments. If the rate of interest charged under the instalment plan is 16% p.a., find the monthly instalment. [Hints: P.V. of the 3 months' instalments

380 or,

= ~/ {10.16 12

380 x -0.04 3

=P { 1-

(1

+

= 485 -

105

= Rs. 380

3 0 16 0 16 · )- } , or, 380 x · 12 12

(3.04)3

3

}

=> -76

15

=P {1-

= P {1-

(i + 0·304 )-

3 }

27 .} = p x 1.094464 .l 28.094464 28.094464

18. A man buys a car for Rs. 16,000. He estimates that its value will depreciate each year by 20 per cent of its value at the beginning of the year. Find the depreciated value (Rs. x, correct to the nearest rupee) of the car at the end of five years. If the man sets aside at the end of each of the five years a certain fixed sum (Rs. y) to accumulate at 4 per cent compound interest in order to be able to buy at the end of five years another car costing Rs. 22,000 (after allowing the above depreciated value Rs. x for the old car in part exchange), find the value Rs. y of each payment. 19. A man aged 40 years wishes his dependents to have Rs. 40,000 at his death. A banker agrees to pay this amount to his dependents on condition that the man makes equal annual payments of Rs. x to the bank commencing now and going on until his death. What should be the value of x, assuming that the bank pays interest at 3% p.a. compound? From the table on the expectation of life, it is found that the expectation of life of a man of 40 is 30 years. 20. A man purchased a house valued at Rs. 3,00,000. He paid Rs. 2,00,000 at the time of purchase and agreed to pay the balance with interest at 12% p.a. compounded qalfyearly in 20 equal half-yearly instalments. If the first instalment is paid after the end of six months from the date of purchase, find the amount of each instalment. Given, logl0.6 = 1.0253 and log31.19 = 1.494. [C.U.B.Com. 1994, '97] 21. What su~ should be invested every year 8% p.a. compound interest for 10 years, to

replace plant and machinery which is expected to cost then 20% more over its present cost of Rs. 50,000? [log 108 = 2.0334, log2158 = 3.334] [N.B.U. B.Com. 1995) 22. The life-time of a machine is 12 years. The cost price of the machine is Rs. 1,00,000. The estimated scrap value and the increase in the cost of machine after 12 years are Rs. 30,-000 and 20% respectively. Find the amount of each equal annual instalment to be deposited at 12% interest p.a. compounded annually. Given: log 1.12 = 0.0864 and log 10.88 = 1.0368. [C.U. B.Com. 1993]

23. The cost price of a machine is Rs. 80,000. The estimated scrap value of the machine at the end of its lifetime of 10 years is Rs. 12,000. Find the amount pf equal annual instalment to be deposited at 9% p.a. compound interest annually just sufficient to meet the cost of a new machine after 10 years, ·assuming an increase of 40% of the price of the machine then. The first instalment is to be paid at the end of the first year. Given, (1.09) 10 = 2.368.

172

BUSINESS MATHEMATICS AND STATISTICS

ANSWERS

A 1. Rs. 1,258.

8.

3. Rs. 2,190.29; Rs. 1,449.50. 4. Rs. 3,137.12. 5.

(a) Rs. 5,560; (b) Rs. 1,25,830; (c) 25,168.

6.

(a) Rs. 3, 772; (b) Rs.11,341.18;

7.

(a) Rs. 2,470;

18. Rs. 3,885.

(b) Rs. 1,650; Rs. 1,375; Rs. 330; Rs. 605.

2. Rs. 3,862.

9. 14.2 years.

21. Rs. 1,046.90 or Rs. 1,047.

10. Rs. 2,266.

22.

11. Rs.60,000; Rs. 1,10,000.

12. Rs. 2,755. 13.

19. Rs. 11,400. 20. 13.1 years.

a~. 6, 721 (by Jog-table) or Rs. 6, 716 (by calculator).

23. Rs. 5,661.50 or Rs. 5,674.

(a) Rs. 2,408;

24. Rs. 1,369.62.

(b) Rs. 1,185.77 or Rs. 1,186.

(c) Rs. 330.90 or Rs. 331.

14. Rs. 12,075.

(a) Rs. 3,000;

16. Rs. 44,442.

25. Rs. 27,157.20 (by log-table) or Rs. 27,145.50 (by calculator).

(b) Rs. 6,250.

17. Rs. 4,550; Rs. 1,930.

26. Rs. 38,850.04.

15. Rs. 5,960.

B 1. Rs. 11,764.50.

2. Rs. 13,752. 3. Rs. 46.850. 4. Rs. 16,098. 5. Rs. 3,976.60. 6. Rs. 3,498. 7. 2nd, i.e., perpetuity of Rs. 800 is cheaper [·: P.V. of annuity of Rs. 1,500 is Rs. 14,825 and P.V.

of deferred Rs. 9,963].

perpetuity

8. Rs. 617.50. 9. Rs. 1,847. 10. Rs. 76,392. 11. Rs. 2, 188.18.

12. Rs. 500 (":A= Rs. 2,39,500).

is

15. Rs. 34,440. 16. R.s. 343.14. 17. Rs.130.11. 18. Rs. 5,243; Rs. 3,103. 19. Rs. 844.48. 20. Rs. 8,719.66 or Rs. 8,720. 21. Rs. 3,800.

13. Rs. 6, 767.

22. Rs.1,093.12 or Rs.1,093.

14. Rs. 9,717.

23. Rs. 6,578.2,5 or Rs. 6,578.

STATISTICS

"This page is Intentionally Left Blank"

Chapter 8

lntrodu.ction: Definition of Statistics: Importance' and Scope, Types, Sources and Collection of Data

8.1

Introduction: Origin and Development of Statistics

The word Statistics seems to have been derived from the Latin word Status meaning Political State. The Italian word Statista or the German word Statistik also bears similar meaning. In ancient times, the governments used to collect information regarding the population and property or wealth of the State. These information were necessary to achieve political purposes of rulers-to assess the manpower and to int~oduce new taxes or levies. In course of time the character of quantitative information extended, an&} details regarding births, deaths, marriages, etc. were necessaty. In modern times collection of data and analyzing them are not limited to any particular sphere of human activity. Quantitative information are not only necessary to run the Government or semi-government organizations like Corporations, etc. but they are the backbone of the study of Agriculture, Biology, Electronics, Medicine, Political Science, Economics, Psychology, Sociology, Business and Commerce-in fact, any branch of study, where numerical figures (or data as we call them) are involved. With the introduction of Theory of Probability since mid-seventeenth century Statistics has been placed on a very sound mathematical footing. The present treatise, however, will not go deep into such theoretical development of Statistics. Our main purpose will be to acquaint our readers with the definition of Statistics and the systematic method of collection and summar:ization of data, analyzing them and interpreting them to help business decisions.

175

17'6

BUSINESS MATHEMATICS AND STATISTICS

8.1.1

Definition of Statistics: Statistics as Statistical Data and Statistics as Statistical Methods

By Statistical Data we mean numerical statement of facts while Statistical Methods will refer to complete information of the principles and techniques used in collecting and analyzing such data. The word Statistics has been defined by several Statisticians in two different but at the same time two allied senses: (a) Plural sense--Statistics as Statistical Data and (b) Singular sense-Statistics as Statistical Methods.

Statistics as Statistical Data In this sense Statistics connects numerical statement of facts. To be more precise, it means that Statistics implies the statement of those classified fact~ that represent the conditions of the people in a state~especially those facts which can be stated in numbers or in any other tabulated and classified arrangement. This being too concise the more exhaustive definition on Prof. Horace Secrist may be quoted here: "By Statistics is meant the aggregate of facts, affected to a marked extent by a multiplicity of causes, numerically expressed, enumerated or estimated, according to reasonable standards of accuracy, collected in a systematic manner for a pre-determined purpose and placed in relation to each other." When the definition of Horace Secrist is elucidated it comes to mean: Statistics as aggregate of facts always means that a single fact or isolated facts or unrelated figures do not constitute Statistics. Thus, the case of a single death, however illustrious the person may be, a single street accident or the value of imports of one kind of article, however valuable, do not come under the purview of Statistics. Then Horace Secrist refers to Statistics as affected to a marked extent by a multiplicity of causes. This is also very remarkable. For example, Statistical data in relation to number of deaths at different ages in a particular country are sure to be affected, among other factors, by the climatic condition of the country, by the habits, customs, nature of occupation, illiteracy and ignorance of hygenetic rules, etc. of the peotQle inhabiting that country. Then Horace Secrist lays down that Statistics must be numerically expressed. This condition is self-evident. For example, statistical data in respect of rice production in West Bengal have no concern with the quality of rice but it occupies itself with the quantity of rice production in the State which must be expressed in numbers of quintals, years, etc. and must be presented in tabular form. Two other characteristics of Statistics as emphasised by Horace Secrist are-reasonable standard of accuracy and systematic collection for pre-determined purpose. These two characteristics are present in the very nature of Statistics. Statistical data must be precise and accurate. But when such precise and accurate data are not forthcoming, one must remain content with their reasonably accurate values. For example, statistics of lengths of different highways in different states of Union of India when recorded may reasonably

CHAP. 8: INTRODUCTION: DEFINITION OF STATISTICS

177

ignore centimetres and metres and that will not detract from the statistical data any of their importance. Then, as regards-the coll~~t!_op of data for a pre-determined purpose, it can be said that a p~ior knowledge of the purpose for which an investigation is conducted certainly facilitates the collection of data. Without this pre-determined purpose, mere collection of data will serve no end and is likely to hinder ultimate analysis with the help of collected data.

Statistics as Statistical Methods Statistical methods literally mean methods that are applied for collection, analysis and interpretation of numerical observations. By the efficient application of such methods (or tools), useful deductions are made and statistical relationships governing the data are suitably formulated. , These methods also facilitate emphasizing the underlying pattern and trend of such data and also their proper interpretation. It is for this reason that Saligman has defined Statistics as "the science which deals with the methods of collecting, classifying, presenting, analyzing, comparing and interpreting numerical data collected to throw light on any sphere of enquiry." Croxton and Cowden defined Statistics more concisely as "the science which deals with the collection, analysis and interpretation of numerical data."

8.2

Importance and Scope of Statistics

In ancient days, Statistics was regarded as the Science of Statecraft and was used by Governments (or Rulers) to collect data relating to population, property or wealth, military strength, crimes, etc. Such collections, were aimed at assessing manpower to safeguard the country and for introducing new taxes and other duties. With the passage of time and concept of Welfare State, the scope of Statistics has been widened to .social and economic phenomena. Again, with developments in statistical techniques, statistics is now used not only for mere collecting data, but also for their handling, analysis and drawing valid inferences from them. It is now widely used in almost all spheres of life. Statistics is used in all sciences-Social, Physical an!f Natural, and also in various diversified fielqs like Agriculture, Industry, Sociology, Econom1cS,Education, Business Management, Psychology, Accountancy, Insurance Planning, etc. etc. Statistics has now assumed such a dimension that statistical techniques are becoming more and more indispensable in our everyday human activity. As a subject, Statistics has also acquired a tremendous progress and even an elementary knowledge of Statistical Methods has become a part of general education in the curricula of most of the countries. Importance of Statistics is clearly stated in the following words of Carrol D. Wright of USA. "To a very striking degree, our culture has become a Statistical Culture. Even a person who may never have heard of an index number is affected by those index numbers which describe the cost of living. It is impossible to understand Psychology, Sociology, Economics, Finance or Physical Science without some general idea of the meaning of an average, of variation of concomitance, of sampling of how to interpret charts and tables." According to H. G. Wells: "Statistical thinking will one day be as necessary for effective citizenship as the ability to read and write." Bus. Math. & Stat. [C.U.J-1,.,

178

BUSINESS MATHEMATICS AND STATISTICS

Importance and Scope of Mathematics in Business and Business Decisions Mathematics is an integral part of education of students in Business, Economics, Statistict and Social Sciences. Mathematical Principles are now used in all spheres of making business decisions. Formulae of Compouni interests may be used to determine growth in investment and de~ predation in machinery of a company. Annuities are used in creating Sinking Fund to replace machinery in future or to fix an EMI in repayment of loans. Linear (or curvilinear) trend in Business may be represented by a line (or a curve). This line (or curve) may be fitted to the past (given) few years business data by the principle of least squares involving Maxima and Minima of Mathematics. This fitted line (or curve) may be used for forecasting in making business decisions on production, knowing increase or decrease in sales. Graphs and charts are used to depict production, sales, profits and earning per share in the Annual Report of a company. These help shareholders to discuss the Profit and Loss Accounts in the Annual General Meeting about the growth of the company and for taking new decisions in producing various new products. Differential Calculus is used to find MC and MR (i.e., marginal cost and marginal reverme) when cost (C) and revenue (R) are known. Integral Calculus is used to find C ahd R .when MC and MR are known. Linear Programming is used to maximize profit and minimize cost of a company on producing and selling varioQs items, the decision variables being men, materials, machines and lands involved in the objective function of the Linear Programming Problem (LPP)./

8.3

Application of Statistics to Business and Industry

With the gradual industrialization and expansion of the business world, businessmen find Statistics an indispensable tool. Now-a-days the success of a particular business or industry very much depends on the accuracy and precision of statistical analysis. Wrong forecasting due to inaccurate statistical analysis may lead to disaster. Before taking a new venture or for the purpose of improvement of an existing one the Business Executives must have a large number of quantitative facts (e.g., price of raw materials, price and demand of similar products prevailing in the present market, various taxes to be paid, labour conditions, taste of the consumers, sales records, etc). All these facts are to be analyzed statistically before stepping in for a new enterprise or before fixing the price of a commodity. Statistical methods are now used for exploring possibilities to advertising campaigns, for adjustment of production methods and as an aid to establish standards. Business activity follows a definite trend-boom periods being followed by periods of depression. Statistical techniques determine such business cycle and help in forecasting future markets. Market research and market surveys by statistical sampling methods are now extremely useful for any businessman. In industry, Statistics is widely used in quality control. In production engineering, to find whether the product conforms to specifications, statistical tools like inspection plans, control charts, etc. are of great use. Wide_ applications of Statistics can be found in Insurance, where the premium rates are fixed on the basis of mortality, average length of life, possibilities of investments, etc.

CHAP. 8: INTRODUCTION: DEFINITION OF STATISTICS

179

Statistical Techniques Commonly Used in Business Activities (i) Various statistical measures like average, dispersion, skewness, correlation are necessary to bring out the characteristics of the available data. (ii) Time Series Analysis helps in isolating various components (trend, seasonal and cyclical). \ They are useful for forecasting and consequent planning.

(iii) Regression analysis establishes relationship between production and sales, demand and per capita income, input and output, etc. They are useful for prediction. (iv) Sampling methods (like random sampling or stratified sampling) are used for conducting business and market surveys or for checking the accuracy of records. (v) Statistical Quality Control is used to find whether the manufactured goods conform to required specifications. Sampling Inspection in this connection is indispensable in any manufacturing concern. (vi) For the calculation of mortality rates Vital Statistics and Demography are useful. These rates and calculation of probabilities of death are used in insurance for the determination of premium rates. (vii) Even for appointment of personnels the efficiency of candidates are statistically determined by using Test Scores. (viii) Index numbers (of price, of production, of cost of living, etc.) are constructed by the use of statistical formulae. They are extremely useful in Business and Commerce, and also helpful to the Government for making economic decisions.

8.4

Limitations of Statistics and also Its Characteristics

(i) Statistics is not suited to the study of qualitative pl,lenomenon. Statistics is applicable to the study of those objects of enquiry which are capable of quantitative measurement. As such qualitative objects like honesty, poverty, culture, etc. are not capable of statistical analysis unless one reduces qualitatiJe expressions to precise quantitative terms. Intelligence of a group of candidates can be studied on the basis of their Test Scores. (ii) Statistics does not study individuals. Statistics can be used only to analyze an aggregate of objects. No specific recognition is given to the individual items of-a series. We study group characteristics through statistical analysis.

1

(iii) Statistical decisions are true only on an average and also the average is to be taken for a large numb~l"_Qf observations. For a few cases in succession the decision may not be true. (iv) Statistical decisions are to be made carefully by the experts. The use of statistical tools by untrained persons may lead to false conclusions. Statistics are like clay of which one can make a god or a devil as one pleases. Misuse of Statistics has, in fact, created some distrust on the subject. That is why we often hear comments like 'Statistics can prove anything in this universe;' or, 'There are three types of lies-lies, damned lies and

180

BUSINESS MATHEMATICS AND STATISTICS Statistics.~

Our final observations are: The Science of Statistics is the most useful servant but only of great value to those who understand its proper use. From the discussions given above the reader can easily summarize what we call Characteristics of Statistical Analysis: (a) In Statistics all information are to be expressed in quantitative terms. Even in the study of a quality like intelligence of a group of students we require scores or marks secured in a test. (b) Statistics deals with a collection of facts, not an individual happening. (c) Statistical data are collected with a definite object in mind, i.e., there must be a definite field of enquiry. (d) In every field of enquiry there are large number of factors, each of which contributes to the final data collected. So Statistics may be affected by a multiplicity of causes. (e) Statistics is not an exact science. Conclusions are usually derived from samples and hence exactness cannot be guaranteed. (f) Statistics should be so related that cause-and-effect relationship can be established. (g) A statistical enquiry passes through four stages:

ICOLLECTION OF DATA I !

ICLASSIFICATION AND TABULATION OF THE DATA I !

IANALYSIS OF THE DATA I ! INTERPRETATION OF THE DATA

8.5

A Few Terms Commonly Used in Statistics

Data A collection of observations expressed in numerical figures obtained by measuring or counting.

Population To a statistician the word population is simply a useful means of denoting the totality of the set of objects under consideration. Population (or Universe) is the aggregate or totality of statistical data forming a subject of investigation, e.g., • the population of books in the National Library, • the population of the heights of Indians,

CHAP. 8: INTRODUCTION: DEFINITION OF STATISTICS

181

• the population of Nationalized Banks in India, etc. A Sample is a portion of the population which is examined with a view to estimating the characteristics of the population, i.e., • to assess the quality of a bag of rice, we examine only a portion of it. The portion selected from the bag is called a sample, while the whole quantity of rice in the bag is the population, • to estimate the proportion of defective articles in a large consignment, only a portion (i.e., a few of them) is selected and examined. The portion selected is a sample. When information is collected in respect of every individual item, the enquiry is said to be done by Complete Enumeration or Census; e.g., during the Census of Population (which is done every ten years in India), information in respect of each individual person residing in India is collected. This method gives information for each and every unit of the population with greater accuracy. But this method involves huge amount of money, long time and much effort. In most cases of statistical inquiry, because of limitations of time and cost, only a portion (i.e., a sample) of the available source of information is examined, and the data collected from them. This process of partial enumeration is known as Sample Survey. The results and findings are then generalized and made applicable to the whole field of inquiry. This is known as Sampling. The process of sampling is based on two important principles of Statistics: 1. Law of Statistical Regularity, which states that a moderately large number of items selected at random from a given population exhibits nearly the same composition and characteristics of the population. 2. Law of Inertia of Large Numbers, which states that, other things remaining the same, the larger the size of the sample, the more accurate is the result obtained. This is nothing but a corollary to the Law of Statistical Regularity.

Characteristic A quality possessed by an individual person, object or item of a population is called a charac-. teristic of the individual. Height of a person, Age, Nationality are examples of characteristics.

Variable and Attribute A characteristic may be measurable or non-measurable. Measurable characteristics are those which are expressible numerically in terms of some units. A measurable characteristic is also called a Variable or a Variate, e.g., Age, Height, Income are variables. A non-measurable characteristic is a qualitative object, e.g., Religion, Nationality, Occupation. We call such characteristics Attributes. A variable takes different values and these values can be measured numerically in suitable units. An attribute cannot be measured numerically, but can be classified under different heads or categories (e.g., persons classified as Single, Married, Widowed, Divorced).

182

BUSINESS MATHEMATICS AND STATISTICS

Continuous Variable and Discrete Variable A variable which can assume for its value any real quantity within a specified interval is a continuous variable. In measuring height of a group of people we may observe that the height of an individual may have any value between 147.3 cm and 182.8 cm. Clearly, the height can be measured to arbitrary degree of accuracy. We say that height is a continuous variable. Similarly, the weight of an object is also a continuous variable which can be recorded to any degree of approximation we desire. Other examples are temperature, time, size of agricultural holding of a family, etc. A di~~ete variable can, however, take only isolated values. In most cases a discrete variable takes only integral values, e.g., the number of workers in a factory, the number of defectives produced, the prices of articles, readings on a taxi-meter, bus fares in a certain route, the number of births in a certain number of years, the number of telephone calls on different days, etc.

EXERCISE ON CHAPTER 8(1) 1. What are the two senses in which the word Statistics is generally used? [C.U. B.Com. 1993]

2. What are the different statistical. techniques used in Commerce, Business and Industry? [C.U. B.Com. 1994, '95]

3. What are the characteristics of Statistics? What are its limitations? [C.U. B.Com. 1993]

4. Explain with examples the distinction between: (a) an attribute and a variable, (b) a continuous variable and a discrete variable. Classify the following characteristics as either an attribute or variable; if variable, mention whether continuous or discrete: Family size; family income (p.m.); mother tongue of a student; size of agricultural holding of a family; division obtained by a student at the Higher Secondary Examination.

8.6

Types and Sources of Data

The raw material (i.e., numerical data) of Statistics ordinates from the operation of counting (or enumeration) or measurement. The person who conducts statistical enquiry, i.e., counts or measures the characteristic under study is called investigator. The process of counting or measurement and systematic recording of results is called collection of data. The first task of a Statistician is to collect and assemble his data. He may prepare the data himself or borrow them from other sources (government, semi-government or non-official

CHAP. 8: INTRODUCTION: DEFINITION OF STATISTICS

183

records). We may Gall these, two types of data-one primary and the other sec-Olldary. For collection of primi:irry da:ta we/may use complete Enumeration or Sampling. lri the:present chapter we shall study the whole procedure of collection of such data. We may observe here that utmost care and importance should be placed in the collection of these data-in fact, they are the raw materials and all subsequent analysis will be based on these materials. If these 'raw materials are not reliable, then the conclusions based on .them will lead to useless and false conclusions. Sp reliability of raw data, as we ca~l them, snould be our first and foremost consideration. ••

8.6.1

.

i

,..-/

Primary and Secondary Data

Primary Data are collected for a specific purpose directly from the field of enquiry-in this sense these data are original in nature. The primary data are published by authorities who themselves are responsible for their collection. ,Usually trade associations collect data from their member concerns, Government organizatio.ns collect data from its subordinate offices. They are considered primary data. But individuals or any organization also can collect primary data from the actual field of enquiry by appointing trained investigators. Secondary Data are numerical information which have been previously collected as primary data by some agency for a specific purpose but are now compiled from that source for use in a different connection. In fact, data collected by some agency when used by another or collected for one purpose when used for another, may be termed secondary data. The same data is primary for its collecting authority but secondary to another agency who uses them. Examples: Sources of Primary and Secondary Data Primary Data 1. Reserve Bank of India Bulletin (Monthly) [issued by the Reserve Bank of India, Bornbay] 2. Indian Textile Bulletin (Monthly) [issued by the Textile Commission, Govt of India] 3. Annual Report of the Railway Board [Ministry of Railways, Govt of India] 4. Indian Coal Statistics (Annual) (Chief Inspector of Mines, Ministry of Labour, Govt of India] 5. Jute Bulletin (Monthly) [Indian Central Jute Committee]

Secondary Data 1. The Statistical Abstract of India (Annual) [Central Statistical Organization (CSO), New Delhi] 2. Annual Statement of the Foreign Trade of India [DGCIS Kolkata] 3. Monthly Abstract Statistics [CSO Delhi]

4. International Labour Review (Monthly) [ILO Geneva]

On Secondary Data The chief sources of secondary data are: (a) Publications of Central and State Governments, or Foreign Governments and international bodies like ILO, UNO, UNESCO, WHO, etc. (b) Publications of various Chambers of Commerce, Trade Associations, Cooperative Societies, etc.

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BUSINESS MATHEMATICS AND STATISTICS

{c) Reports of Committees and Commissions of Enquiry appointed time to time for specific purposes of Enquiry. {d) Reports and Research Papers published by Research Scholars, Labour and Trade Unions, etc. One must be very careful before using secondary data because of its many limitations. Secondary data may contain the following errors: (i) Transcribing Errors (i.e., errors occurring due to wrong transcription of the primary data). (ii) Estimating Errors (the data published in many secondary data are mere estimates and not_facts). ·

(iii) Error due to Bias; sometimes fictitious figures are put in secondary data. Before using secondary data the user should scrutinize the following points and then decide how far such data will be useful for his purpose: • the scope and object of enquiry for which the data were originally collected; • the methods of collection; • the time and area covered; • the reliability of the authority who collected the original data; • terms used and unit of measurements considered while the data were collected.

Relative Advantages of Primary Data (a) Primary data contain more detailed information and some information are suppressed or condensed in secondary data. (b) Secondary data may contain errors like transcribing errors, etc. while primary data cannot have such errors. (c) In the primary data precise definition of the terms used and the scope of the data are mentioned. (d) Primary data often include the method of procedure or any approximation used so that one can know its limitations but secondary data usually lack such information. Observations: In spite of these advantages of primary data, it is suggested that secondary data should be used particularly when a large number of related items are needed. One main reason for accepting secondary data is that it is much less expensive.

CHAP. 8: INTRODUCTION: DEFINITION OF STATISTICS

8. 7

185

Collection of Prirnary Data

An investigator may collect primary data: Either • by direct personal observation, or • by indirect oral investigation, or • by sending questionnaires by mail, or • by sending schedules through paid investigators.

Comments on These Methods (a) A research worker may collect data by direct experimentation. An investigator may himself meet persons who can supply his requisite information. The results obtained are quite reliable for further analysis but it involves time and money. (b) In this case indirect sources are tapped for collecting information, e.g., Enquiry Commissions obtain information through persons who are likely to have the requisite information. The reliability of these data very much depends on the integrity of the persons selected. (c) A set of questions is prepared keeping in mind the object of enquiry; these questions are sent by mail to selected persons with a request for their replies by return mail. Though this method can cover large areas and is also comparatively cheaper but the responses are, in most cases, not satisfactory and as such expected amount of data are seldom available. (d) 'Schedules sent through paid investigators' method is most widely used, particularly in market research. Schedules are prepared; paid investigators are trained; they carry with them the schedules and meet people concerned. The investigators fill up the schedules on the spot on the basis of the replies they receive from the informants. The data are thus collected. (e) This method is popular and yields better results. Much, however, depend on the tactfulness and personality of the investigators. This method is adopted during the decennial census of the population in our country.

Two Procedures for Collection of Data By Complete Enumeration (or Census) By Sample Survey. When information is collected in respect of every individual person or item of a given population, we say that the inquiry has been done by complete enumeration or census. This process is called Census Survey. But in most cases of statistical inquiry, because of limitations of time and money, a portion only of the population is examined and the data collected therefrom. This process of partial enumeration is known as sample survey. The results and findings are, however, made applicable to the whole field of inquiry. This is known as Sampling.

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BUSINESS MATHEMATICS AND STATISTICS

Illustration 1. During the census of population of a country (in India census is taken after every ten years) each and every individual person residing in that country is examined and the data are collected. This is an example of complete enumeration. A census of manufacturing industries in India must include each and every manufacturing industry and none is excluded. Usually, a complete enumeration of these industries are made with regard to the wages, attendance and absenteeism, hours of work, labour disputes and time lost, accidents and compensation, etc. in respect of the employees, etc. In verifying the accuracy of entries in Books of Accounts, only a small number of entries are . checked (Test Audit). This then is an example of Sample Survey. Other fields, where sample survey is used; estimation of the crop yield of a country, counting the number of trees in a forest, etc.

A Comparative Study of the Two Methods • In the census method the entire population is investigated and hence a very large number of investigators (or enumerators) are to be employed. Considerable time are necessary for collection of data and processing the large quantity of information collected. Thus, census method involves considerable time, money and labour. But on the positive side this method is expected to yield accurate information (provided, of course, enumerators do their work honestly). In fact, if intensive investigations are needed, complete enumeration method cannot be avoided (however expensive it may be). Again, if the field of survey is small, this method should be used. • Sample survey method, on the other hand, has the following advantages; that is why sampling (particularly, random sampling techniques) is getting more popular in recent times: (a) In sample survey only a part of the population is to be investigated and hence it takes less time, less money and less labour. (b) In census method a large group of investigators are employed-a thorough and intensive training of such a large group of people can hardly· be possible. But in sample survey method only a small group of estimators are employed-they are given specialized training. Hence, though little costlier the output in this method is much more. (c) Sample survey even with a l.imited budget can cover more geographical area. (d) In sample survey, sampling error cannot be avoided, but in census there is no question of sampling errors (since the whole population is investigated). However, large magnitudes of non-sampling errors are involved in the census methodtheir magnitudes remain undetermined and they greatly influence the accuracy of the results. (e) Sampling errors can be theoretically calculated and hence, their magnitudes are known but non-sampling errors do not follow any law of probability and hence their magnitudes cannot be precisely determined. (f) Sometimes complete enumeration if not feasible or practicable. For instance, to examine the quality of rice in a particular consignment, i~ is not possible to investigate

CHAP. 8: INTRODUCTION: DEFINITION OF STATISTICS

187

every grain of rice-one must take a small portion of the contents and thereby decide the quality of the whole population. It is true that suitable method of sampling should be used so that the conclusion about the sample can be extended to the whole population.

Collection of Statistical Data: Different Stages Statistical Enquiry: We use this term to mean some statistical investigation wherein relevant information is collected, analyzed and interpreted by the application of statistical techniques. Statistical enquiry consists of the following stages: -

Planning.

-

Framing questionnaire and devising other forms to be used for the enquiry.

-

Actual collection of data.

-

Editing the information collected through questionnaire.

-

Analysis of data and interpretation of results.

-

Preparation of report.

Short Explanations of Different Stages Planning: Before making a venture for the actual investigation, planning is the most essential step. While making the planning the following points are to be considered: • Object and scope of enquiry. • Source of data-primary or secondary [If time and money are available, efforts :should be made for collecting primary data]. • Type of enquiry should be decided upon-complete enumeration or sample survey. • Units in terms of which requisite information are to be collected should be clearly defined (Such units are called Statistical Units). • Degree of accuracy [Maximum accuracy subject to limitations and available resources should be aimed at].

Questionnaire: In a statistical enquiry the necessary information are generally collected in a printed sheet in the form of a questionnaire. Such a sheet contains a set of questions which the investigators are supposed to ask the informants and note down their answers against each question. The framing of the questions requires skill and foresight. The success of the ~nquiry depends very much on thoughtful drafting of the questionnaire. In setting up the questions the following points are to be considered:

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BUSINESS MATHEMATICS AND STATISTICS

• The schedule of questions must not be lengthy. Many questions may arise but not all of them can be included in the questionnaire; with a lengthy questionnaire the respondents become bored and may not disclose accurate information. Only essential points are to be included in the questionnaire. • Questions must be simple, brief and unambiguous--otherwise, accurate information may not be available. If necessary, several complicated questions are to be broken up into smaller parts so that they can be easily answered by the respondents. Questions should be so framed so as to elicit only one of the two possible answers-YES or N 0. • Personal questions (income, property, etc.) are to be avoided, if possible, or to be put in indirect form. People are reluctant to disclose such facts. No questions should be asked as to hurt the feelings or sentiments (religious or otherwise) of the informants. • Questions should be of objective nature-there should be less scope for matters of opinion. • The units in which the information are to be collected should be precisely mentioned; e.g., State your age: ...... years ...... months. Annual Output: Rs. . .... . • The arrangement of questions should be made in such a way that an easy and systematic flow of answers may come out. After the preparation of a questionnaire it is desirable to make a PILOT SURVEYto try on a few individuals. It will be helpful in finding out the shortcomings of the questionnaire and hence necessary modifications can be made before the actual survey. Collection of Data: • Questionnaires may be sent to the respondents by post and request may be made to send the completed form also by post (necessary postage being enclosed). This method has many advantages: It is least expensive and vast areas can be covered. Moreover time required for collection v1 the data will be comparatively short and the respondents feel more free to answer questions which they are ordinarily reluctant to answer before investigato~s. But this mailing system of the questionnaire has the greatest disadvantage-LOW PROPORTION OF RESPONSE and consequently the information may not represent the representative view. Another minor point is that due to misunderstanding of the meanings of some questions by the respondents, correct information may not always be available. • Another method is to employ suitable persons as investigators and to train them for collection of data. These investigators then interview the informants, ask questions as per the questionnaire and note down their answers. This method is commonly used. Though reliable information are expected through this method but it requires huge organizational efforts including necessary fund. Editing the Returns: Soon after the actual collection of data starts, arrangements should be made to receive the completed forms and scrutinize them. Some of the completed forms may contain inconsistencies or omissions. These forms are to be referred back for correction.

189

CHAP. 8: INTRODUCTION: DEFINITION OF STATISTICS

When all returns have been received and checked, the information are classified and tabulated. If necessary, the work of tabulation and sorting may be done through specially devised machines like electric tabulators. However, when the study does not involve a large number of returns, manual sorting and tabulations are used. Analysis of, Data and Their Interpretations: The main object of statistical analysis is to abstract significant facts from a large collection of data. With the ~lp of the data and using statistical theories various statistical measures are computed. They help to make statistical conclusions about the present position and also to forecast the future. Preparation of Report: It is usual to publish a report after the completion of the enquiry. The report should contain a complete description of all the stages of enquiry, definitions of terms used, coverage degree of reliability of the data. and final results of the enquiry. Such a report is helpful for planning future enquities on related problems.

--------i

EXERCISE ON CHAPTER 8(11)

1. Explain the terms: Primary data and Secondary data. Give some illustrations. [B.U.B.Com.(H) 1990, 2002)

2. Describe various methods of collecting primary data and comment on their relative advantages. [C.U.B.Com.(H) 1991; Guwahati U. 1996; Bangalore U. 1997) 3. Define 'secondary data'. State the chief sources and point out the precautions necessary before using them. [Madras U. B.Com. 1997; Kanpur. U. B.Com. 1997} 4. Statistical data are usually of two types-(i) Primary, (ii) Secondary. Explain in brief. State the various methods of collecting primary data and the chief sources of secondary data (three of each). [C.U. B.Com. 1987) 5.

(a) Distinguish between a census and a sample enquiry, and briefly discuss their com(C.A. 1992; D. U. B.Com. 1995, '96; C. U. B.Com. 1989) parative advantages. (b) Distinguish between Population and Sample, with the help of an example. [C.U. B.Com. 1995, '96)

6. Whati&meant by Statistical Enquiry? Describe different stages of a statistical enquiry. 7.

(a) What is a questionnaire as used as statistical enquiry? Describe the important considerations in the framing of a questionnaire. What special measures are necessary / in case of enquiries by mail? (b) What is the difference between census survey and sample survey?- Which one of them is more suitable when the available data is very large? Give reasons.

8.

(a) Describe the characteristics of a good questionnaire. (b) Write short notes on Techni'-i ue of Drafting Questionnaires.

le) Explain the Law of Statistical Regularity.

[D.U. B.Com. 1996]

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BUSINESS MATHEMATICS AND STATISTICS

9. Write short notes on the following: (a) Collection of data, (b) Various stages of sample survey. 10. Distinguish between Complete Enumeration and Sample Survey. How far is the latter more advantageous than the former and why? Name different types of sampling in Statistics. 11. (a) Enumerate the various steps in conducting a sample survey.

[K.U. B.Com. 1994]

(b) Explain the following statistical terms: (i) Primary data, (ii) Population,

(iii) Sample survey, (iv) Questionnaire.

12. What are the methods by which primary data can be collected? Write brief account of [M.U. B.Com. 1999) each of them pointing out their merits and demerits.

Chapter 9

Summarization of Data: Classification and Tabulation; Frequency Distribution

9.1

Introduction

In Statistics we often use a term called Characteristic. An attribute (or a quality) possessed by an individual person, object or item of a population is called a characteristic of the individual. Height, age, nationality are examples of characteristics. A characteristic may be measurable or non-measurable. Measurable characteristics are those which are expressible numerically in terms of some units. A characteristic of this type is also called a variable, or sometimes a variate. Age, height, income, etc. are examples of this type of characteristics. Non-measurable characteristics cannot be expressed numerically. Nationality, occupation, etc. are examples of non-measurable characteristics.

Classification All statistical investigations are carried on with a definite purpose in view and the collection of data, described in the previous chapter, is only the first step in this direction. To make the collected data really useful, they must be classified or grouped under appropriate heads and systematically arranged according to some common characteristic possessed by the individuals, about whom the data have been collected. The process of arranging or bringing together all the enumerated individuals/items under separate heads or classes according to some common characteristics possessed by them is known as Classification. It facilitates comparison and drawing of inferences by dropping unnecessary details and very clearly showing different points of agreement and disagreement. 191

192

BUSINESS MATHEMATICS AND STATISTICS

During the population census, apart from the number of members in each family, various other information, e.g., age, sex, occupation, nationality, etc. of all the people in the country are collected. The total population of the country are then classified: according to sex, into Males and Females. according to marital status, into Single (never married) and Married. -

according to the place of residence, into Rural and Urban.

-

according to livelihood categories, into Agricultural classes, Production other than cultivation, Commerce, Transport and Other services.

-

according to age,, into 0-5 years, 6-10 years, 11-15 years and 16-20 years, etc.

-

according to residence in States, into West Bengal, Assam and Bihar; etc.

In the classifitations illustrated above, the common characteristics are sex, marital status, place of residence, livelihood category, age and residence in States. When the items/individuals are classified according to some common non-measurable characteristics possessed by them, they are said to form a Statistical Class and when they are classified according to some common measurable characteristics possessed by them, they are said to form a Statistical Group.

9.2

Types of Classification

We may broadly consider four types of classification: (i) On a Qualitative Basis: Classification according to some non-measurable characteristics, such as sex, nationality, occupation, etc. are examples of this type. Such classifications may also be termed classification by attributes .

.(

Illustration 1. Total population of the country may be classified according to nationality-into Indian, Chinese, Ceylonese, English, Dutch, etc.; according to mothertongueinto Bengali, Hindi, Gujarati, Malayalam, Punjabi, etc.; similarly, insurance may be classified into life insurance, fire insurance, marine insurance, accident insurance, etc. all of which are examples of classification by attributes.

(ii) On a Quantitative Basis: The classification of population according to age, industries according to the number of persons employed, banks according to the amount of paid-up capital and reserves, etc. are included under this type. Here the basis of classification is some measurable quantity. Such classifications may also be termed classification by variables .

./

Illustration 2. The age of a person is measurable in years, say 31 years, and is thus a variable and the population may be classified by the variable age. Similarly, prices, wages, barometer readings, etc. are examples of variables.

CHAP. 9: SUMMARIZATION OF DATA

193

(iii) On a Time (Chronological or Temporal) Basis: In this type, the enumerated individuals or items are classified according to the time at whieh they were measured. Thus, the production of a factory may be shown by weeks or months or quarters. Similarly, ships first registered at Indian Ports may be classified according to their year of registration. Statistical data arranged chronologically constitute what is called a ·Time Series. (iv) On a Geographical Basis: The classification of the total population of a country according to states or districts falls under this type. The total value of imports of merchandize may, similarly, be classified according to the place or country from which imports have been made.

9.3

Presentation of Statistical Data

There are three different ways, in which statistical data may be presented: (i) Textual Presentation, (ii) Tabular Presentation, and (iii) Graphical Presentation. Textual Presentation: In this way of presentation numerical data are presented in a descriptive form. Two illustrations presenting information in a textual form are given below: ./

Illustration 1. In 1965 out of a total of 2000 workers in a factory 1550 were members of a trade union. The number of women workers employed was 250, out of which 200 did not belong to any trade union. In 1970 the number of union workers was 1725 of which 1600 were men. The number of non-union workers was 380 among which 155 were women .

./ . Illustration 2. Numerical data with regard to industrial diseases and deaths therefrom in Great Britain during the years 1935-39 and 1940-44 are given in a descriptive form: During the quinquennium, 1935-39, there were in Great Britain 1775 cases of industrial diseases made up of 677 cases of lead poisoning, 111 of other poisoning, 144 of anthrax and 843 of gassing. T__he number of deaths reported was 203. Of the cases /qr all the four diseases taken together, that for lead poisoning was 135, for other poisoning 25, and that for anthrax was 30. During the next quinquennium, 1940-44, the total number of cases reported was 2807. But lead poisoning cases reported fell by 351 and 'imthrax cases by 35. Other poisoning cases increased by 748 between the two periods. The number of deaths reported decreased by 45 for lead poisoning, but decreased only by 2 for anthrax from the pre-war to the post-war quinquennium. In the later period, 52 deaths were reported for poisoning other than lead poisoning. The total number of deaths reported in 1940-44 including those from gassing was 64 greater than in 1935-39. [C.A. Final May 1958)

The disadvantages to be faced with such a mode of presentation are: • that it is a lengthy text, Bus. Math. & Stat. [C.U.}-13

BUSINESS MATHEMATICS AND STATISTIC~,

194

• that there has been much of repetition in words, • that comparison between the corresponding figures in the two periods is, difficult, and • that it is difficult to grasp, from a lengthy text the important points i1 there be a number of them, making it all the more difficult, to arrive at any conclusion. Tabular Presentation: Tabulation may be definecj to be the orderly and systematic presentation of numerical data in rows and columns designed to clarify the problem under consideration and to facilitate comparison between the figures. Tabular presentation or Tabulation is also a form of presentation of qtJ,antitative data in· a condensed form so that the numerical figures are ca.pable of easy reception through the eyes. The numerical-descriptions of Examples 1 and 2 given above have been condensed in the form of Tables shown below: It is clear from the illustrations 1-4 that the lengthy text have been condensed into a small table, while facilitating better comparison between the corresponding ngures in the two time periods. Much of repetition in words has been avoided with the help of the table. The tabular form of presentation of data is thus much superior in use to a text statement. The tables provide examples of time series, where classification on both qualitative iil.d quantitative bases occurs. Textual presentation in Illustrations 1 and 2 is given in Tabular forms in Illustrations 3 and 4. ./

Illustri'tion 3. Number of workers in a factory classified according to sex and their. membership of the Union.

I

TABLE 9.1: NUMBER OF WORKERS IN A FACTORY 1965

SI. No.

(1) 1 2 ./

Characteristic

(2) Trade Union Members Non-Trade Union Members Total

No. of Male Workers (3) i 150Q 250 1750

No. of Female Workers (4) 50 200 250

-

1970

No. of Male Workers (5) 1600 225 1825

No.- of Female Workers (6) 125 155 280

Illustration 4. -

TABLE 9.2:

DEATH~OM

INDUSTRIAL DISEASES IN qREAT BRITAIN 1935-39 1940-44 SI. Types of Disease No. ol No. ol ~Qf No. of No. of % of No. Cases Deaths' Dea~hs Cases Deaths Deaths (4) (2) (1) (3) (5} (6) (7) (8) 135 7.6 1 Lead Poisoning 677 326 3.2 90 2 Other Poisoping 111 25 1.4 52 859 1.9 144 1.7 3 Anthrax 30 109 28 1.0 4 Gassing 843 165 9.3 1513 249 8.8 Total 1775 20.0 2807 355 419 14.9

195

CHAP. 9: SUMMARIZATION OF DATA

Workings: 20% of 1775 = 355 and 355 - (135 + 25 + 30) = 165. 135 1775 x 100 = 7.6,

25 1775 x 100 = 1.4,

30 1775 x 100 = 1.7,

165 x 100 = 9.3, etc. 1775

A table consists of four parts: (i) Title, (ii) Stub, (iii) Caption and (iv) Body. Title: A title of a table is self-explanatory and clearly conveys in as few words as possible, the contents of the table; it is usually put at the head of the table concerned. Stub: In a table, the left-hand column, which contains the headings of the rows, with its heading is called Stub. Captions: These are the headings of the columns other than the Stub. Body: The whole portion of a table excepting its Title, Stub and Captions and consisting usually of the figures of the table is called Body of the table. Note: It is usual to indicate at the end of the table the source from where the information have been collected. , There is no ideal method of tabulation. Skill in tabulation is generally attained by years of experience. Nevertheless, the following general rules may be borne in mind while tabulating:

• A table must contain a title. This title should be clear and convey, in as few words as possible, the contents of the table. When several tables are used, each should be numbered, e.g., Table 1, Table 2, etc. to facilitate future references. For our purpose we have also given references to chapters. Thus, Table9.l refers to Table 1 of Chapter 9. The columns should, similarly, have definite and at the same time, sufficiently comprehensive headings. The columns should also be numbered, if their number in a table is 4 or more. • Units of measurement must be clearly shown. The units are generally shown at the top of columns, if the different columns show figures in different units; or sometimes at the top right-hand corner of the table, if all the measurements are in the same unit. •'1'he table should be well-balanced in length and breadth. If the table is unduly long, try a separate coarse grouping. If, however, this causes any serious loss of information, some breaks or extra spaces should be left at equal distances apart (compare Table9.9). • Columns of figures which are directly comparable should be kept as close as possible. • Light and heavy rulings may be used to distinguish the sub-columns and main columns. Double rulings may also be used instead of heavy ones. • The totals of columns may be shown at the bottom of the table between heavy rulings. Sometimes, totals of rows are also useful and should accordingly be shown. • Tables should not usually be burdened with unnecessary details. If necessary, such details may be shown in separate smaller tables. • In case, the data are procured from some sources, references must be made of the sources, at the right-hand corner below the tables.

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BUSINESS MATHEMATICS AND STATISTICS

• Where doubts exist in any figure, this should be explained at the bottom of the table with a suitable explanatory note. Graphical Presentation: The third form of presentation of quantitative data is made by graphs and charts.

9.4

Types of Tabulation

Tabulation may either be-(i) Simple or (ii) Complex. Simple Tabulation: In this type the number or measurement of the items are placed below the headings showing the characteristics .

./

Illustration 1.

I TABLE 9.3: PROGRESS OF CO-OPERATIVE SOCIETIES IN INDIA I Year

No. of Societies

No. of Members

1950-51 1951-52 1952-53 1953-54 1954-55 1955-56

1,81,189 1,85,650 1,89,436 1,98,598 2,19,288 2,40,395

1,54,83,159 1,57,83,571 1,63,78,364 1,73,24,744 1,82,98,645 1,76,21,978

In this table the characteristics of the progress of co-operative societies as measured by the number of societies and the number of members, have been shown year by year.

This is an example of a time series in which quantitative basis of classificat!on is also present. Complex Tabulation: In this type, each numerical figure in the table is the value of the measurement having the characteristics shown both by the column and the row headings.

./

Illustration 2. TABLE 9.4: INDIA'S BALANCE OF PAYMENTS (CURRENT ACCOUNT) FOR THE YEAR 1955 (Core of Rupees) Area Starling Area Dollar Area Other Areas Total

Receipts

Payments

Balance

503.8 178.8 205.6

447.7 137.4 262.8

+ 56.1 + 41.4

888.2

847.9

+ 40.3

- 57.2

In this table, the figure 503.8 (for example) which represents 503.8 crore of rupees, i.e., Rs. 503,80,00,000 is the value of India's current account receipts (indicated by the column heading) during 1955 from the sterling area (indicated by the row heading).

CHAP. 9: SUMMARIZATION OF DATA

197

The complex tabulation shown in Table 9.4 may also be called a two-fold tabulation; because the table shows on one side classification according to Receipts, Payments and Balance, and on the other side classification according to areas, viz., Sterling area, Dollar area and Other areas. Classification here, has been made on geographical basis combined with quantitative basis. Similarly, complex tabulation may show three-fold, four-fold, etc. tabulations.

9.5

Frequency Distributions

Variables: Continuous and Discrete A Variable is a symbol, such as x, y, h, n, X, Y, etc. which can assume any of a prescribed set of values (called the domain of the variable). If a variable can assume any value (any real number for our purpose) within two given values, ~call it a continuous variable; otherwise the variable is a discrete variable. Discrete variables assume isolated values (say, whole numbers only). Continuous variables may be measured to an arbitrary degree of accuracy. Illustration 1. The number n of children in a family which can assume values like 2, 3, 5,_ etc. but not 2.5 or 3.42 which is an example of a discrete variable. Illustration 2. The height h of an individual, which can be 123 cm or 123.5 cm or 123.489 cm or 123.4885 cm depending on the accuracy of measurement desired, is a continuous variable. Another example is the weight of an object which may be recorded in grams, tenths of grams or thousandths of grams. We may come across discrete variables like: Marks obtained by a,sample of students in an examination; Reading of a taxi meter; Number of births/deaths; Number of accidents; Number of defects found in a sample of production, etc. Data which can be described by a discrete variable are called discrete data; those which can be described by a continuous variable are called continuous data. In general, measurements give rise to continuous data while enumerations or countings give rise to discrete data.

Raw Data Statistical data may originally appear in a form (See Tabte9.5), where the collected data are not organized numerically. We call them raw data.

198 ./

BUSINESS MATHEMATICS AND STATISTICS

Illustration 3. TABLE 9.5: MARKS IN MATHEMATICS OF 50 STUDENTS (SELECTED AT RANDOM FROM AMONG CANDIDATES APPEARING IN B.COM. EXAMINATION)

37 47 32 26 21 41

38 41 50 45 30 51

40 46 40 52 38 37

36 38 50 45 32 26

38 31 47 41 48 40

37 33 41 44 47 38

36 48 50 39 41 46

40 37 43 16 45 32

50 52

This representation of the data do_E)S not furnish useful information to a statistician. A better . way may be to express the figures in ascending or descending order of magnitude, commonly termed as ARRAY. ./

Illustration 4. TABLE 9.6: ARRAY OF MARKS GIVEN IN TABLE 9.5 (ARRANGED IN ORDER OF ASCENDING MAGNITUDES)

16 21 26 26 30 31

32 32 32 33 36 36

37 37 37 37 38 38

38 38 38 39. 40 40

40 40 41 41 41 41

41 43 44 45 45 45

46 46 47 47 47 48

48 50 50 50 50 51

52 52

Range From the array it is easy to find what we call Range. Range of a given data is the difference (Largest Measure - Smallest Measure). In Table9.6, the largest measure is 52 and the smallest one is 16; hence the range = 52 -16 = 36. Though Array has certain advantages over Raw Data still this does not reduce the bulk of the data and when large masses of data are given the array is also cumbersome.

Frequency: Tally Sheet Frequency of a value of variable is the number of times it occurs in a given series of observations. In Table9.6, 38 occurs 5 times-we say that frequency of the value 38 is 5. A Tally sheet may be used to calculate the frequencies from the raw data. A tally mark (/) is put against the value when it occurs in the raw data. Having occurred four times, the fifth occurrence is represented by putting a Cross tally mark (\) on the first four tally marks, see Table9.7. This technique facilitates the counting of tally marks at the end.

199

CHAP. 9: SUMMARIZATION OF DATA

Illustration 5.

TABLE 9.7 Marks (x)

Tally Marks

16 21 26 30 31 32

I I II I I Ill I II /Ill

33

36 37 38 39 40 41 43 44 45 46 47 48 50 51 52

/'HJ

I Ill/ /'HJ

I I Ill II Ill II /Ill I II

Total Frequency

Frequencies (/)

1 1 2 1 1 3 1 2 4 5 1 4 5 1 1 3 2 3 2 4 1

2 50

Such a representation of the data is known as the Frequency Distribution. Here marks of students may be taken as a discrete variable x and the number of students against each value of the variable is the frequency of the value of the variable. (The word frequency thus refers to the information how frequently a value of a variable occurs.) Such representation of data, though better than Array, does not condense the data very much.

Grouped Frequency Distribution When large masses of raw data are to be summarized and the identity of individual observation or the order in which observations arise is not relevant for the analysis, we distribute the data into Classes or Categories and determine the number of individuals belonging to each class, called the Class frequency. A tabular arrangement of raw data by classes, where the corresponding class frequencies are indicated (tally marks should, in general, be shown in such presentation), is known as a

200

BUSINESS MATHEMATICS AND STATISTICS

Grouped Frequency Distribution (or sometimes, simply, Frequency Distribution). The name Grouped Frequency Distribution is given because the whole range of observations has been divided into smaller groups and the frequency of each such group is taken into consideration. In the data of Table9.7 the Range is 36 (nearly, 40). We take 8 classes or goups-first group containing marks ranging from 16 to 20, the next group 21 to 25 and so on, the last group of 51-55 ..We have then a Grouped Frequency Distribution as shown in Table9.8. Data organized and summarized as in the frequency Table 9.8 are often called Grouped Data. Though it destroys much of the details of raw data, it helps us to give a clear overall picture in a condensed form.

Class Interval or Class A symbol defining a class (such as 16-20, 21-25, 26~30, etc.) is called a class interval (or we may refer them simply~ a class). There are 8 classes in Table9.8 . .(

Illustration 6. TABLE 9.8: FREQUENCY DISTRIBUTION OF MARKS OF 50 STUDENTS IN MATHEMATICS SI. No.

Marks

Number of Students

1 2 3 4 5 6

16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55

1 1 3 5 16

7 8

10 11

Total

3 50

A .Few Points to Remember before the Construction of Classes • The classes should be clearly defined and should not lead to any ambiguity. • The classes should be exhaustive. (Each value of the raw data should be included.) • The classes should be mutually exclusive and non-overlapping. • The classes should preferably be of equal size or width. In special cases classes of unequal size may have to be used. • The number of classes should not be too large or too small. Normally, it should not· exceed 20 but should not be less than 5, depending on the number of observations in the raw data.

CHAP."9: SUMMARIZATION OF DATA

201

Class Limits In the construction of a grouped frequency distribution the class intervals must be defined by a pair of numbers such that the upper end of one class does not coincide with the lower end of the next class (see 16-20, then 21-25, etc.). The two numbers used to specify the limits of a class interval are called class limits-the smaller number is called the lower class limit and the larger one is the upper class limit. We note that both the class limits may coincide with the observed data. Class limits are used solely for the construction of grouped frequency distribution· from raw data.

Class Mark The Class Mark is the middle point of the class interval (and hence also called Mid-value of the Class): , Class Mark= (Upper Class Limit +Lower Class Limit) [See Table9.8: The class interval 16-20 has lower limit 16 and the upper limit 20; mid-value is 18.J Remember that class mark is used as a representative value of the class interval (useful iii. discussions on Mean, Standard deviation, etc.).

!

Class Boundaries When'w' deal with a continuous variable all data are recorded nearest to a certain unit. Let us consider the distribution height of individuals in centimetres. If we decide to record height · to the nearest centimetre and if then we have a class interval like 125-130, then all heights between 124.5 cm and 125.5 cm will be recorded as 125 cm. Similarly, all heights between 129.5 cm and 130.5 cm will be recorded as 130 cm. The extreme values (indicated by the number 124.5 and 1130.5) such that any observed value (any real number) included between them may be taken within the class interval are called Class boundaries or True class limits-the smaller number is the lower class boundary and the larger number is the upper class boundary. In practice, remember: Lower Class Boundary = Lower Class Limit -0.5 (if observations are recorded to the nearest integral unit) but = Lower Class Limit -0.05 (if observations are recorded to the nearest tenth of a unit) Upper Class Boundary

= Upper Class Limit +0.5 or Upper Class Limit +0.05 in

the two respective cases mentioned in 1. For observations taken to the nearest whole number, suppose the classes are 126-130, 131.135, etc. Table9.9. Lower Class Boundary = 126 - 0.5 = 125.5 } for the class 126-130. Upper Class Boundary = 130 + 0.5 = 130.5 For observations taken t9 the nearest tenth (i.e., correct to one place of decimal) suppose the class intervals are: 126.5-130.5, 130.6-134.6, etc. _ Then the lower class boundary = 126.5 - 0.05 = 126.45 } fi th or e c1ass 126 ·5 130 , ·5· and the upper class boundary = 130.5 + 0.05 = 130.55

202

BUSINESS MATHEMATICS AND STATISTICS

We give below a table, where class limits, class boundaries and class marks are indicated along with the class intervals and their frequencies . ./

Illustration 7. TABLE 9.9: HEIGHTS OF 100 STUDENTS Class Intervals Height (in cm)

Frequency No. of Students

126-130 131-135 136-140 141-145 146-150

5 18 42 27 8 100

Total

Class Limits Lower Upper

126 131 136 141 146

130 135 140 145 150

Class Boundaries Upper Lower

125.5 130.5 135.5 140.5 145.5

130.5 135.5 140.5 145.5 150.5

Class Marks

128 133 138 143 148

Note: The upper boundary of any class coincides with the lower boundary of the next class; but the upper limit of any class may be different from the lower limit of the next class.

Width (or Size) of a Class . Width of a class= Upper class boundary - Lower class boundary, e.g., the width of the class 126 - 130 is 130.5 - 125.5 = 5. In the construction of a frequency distribution it is preferable to have classes of equal width. Classes of unequal width are also in use (when some of the observations are isolated and far apart from the rest) specially to avoid empty classes (classes with frequency zero). Remember that if all class intervals of a frequency distribution have equal widths, this common width difference between two successive lower class limits (or two successive upper class limits) or difference between two successive lower class boundlU'ies (or upper class boundaries) or difference between two successive class marks. [See Table 9.9]

A Few More Terms Frequency Density: We have observed that in a frequency distribution the class intervals may or may not be of equal width. If the classes be a varying width, the different class frequencies will not be comparable. Comparable figures can be obtained by dividing the frequency of the class by its width. We call this ratio Frequency Density. Thus, Frequency Density of a class = Frequency of the class

+ Width of the class.

Frequency density is used in the construction of histogram, when the classes arE! of unequal width. Percentage Frequency of a class =

Frequency of the class 'I1 fr x 100%. ota1 equency

CHAP. 9: SUMMARIZATION OF DATA

203

.

Relative Frequency of a class is the frequency of the class divided by the total frequency of all classes and is generally expressed as a percentage. For example, the relative frequency of the class 126-130 in Table9.9 is 5/100 = 5%. Note that the sum of the relative frequencies of all classes is clearly 100% or 1. If the frequencies in Table9.9 are replaced 'by corresponding relative frequencies, the resulting table is called a relative frequency distribution or a percentage distribution.

We may have occasion to use the expressions like open class interval to mean that the class has either no upper class limit or no lower class limit. For example, we may refer to age groups of individuals, the class '60 years and over' or '25 years and below'-these are open class intervals.

Systematic Procedure for Construction of a Frequency Distribution from Raw Data Obtain the largest and smallest numbers in the raw data and thus find the Range. Divide the Range into a convenient number of class intervals (of equal size preferably). The number of class intervals should not exceed 20 but not less than 5 either. Class intervals are so chosen that the class marks (or mid-values) may, if possible, coincide with observed data or most of the observations may lie near class marks. But class boundaries should not coincide with actually observed data. Determine the number of observations falling into each class interval, i.e., find the class frequencies by means of a Tally Sheet. The frequency table is now constructed showing the class intervals and their frequencies.

9.6

Different Ways of Presenting Frequency Distributions

We give below a few specimens of frequency distributions for illustrating the different ways in which they are presented.

Illustration 1. TABLE 9.10: FREQUENCY DISTRIBUTION OF MONTHLY WAGES EARNED BY THE WORKERS OF A FACTORY Monthly Wages in Rupees 110-120 120-130 130-140 140-150 150-160 160-170 170-180 180-190

No. of Workers 25 40 65 85 80 70 35 20

204

BUSINESS MATHEMATICS AND STATISTICS

Here, the upper limit of one class is the lower limit of the next· class and an upper limit or a lower limit occurs in two classes; but the explicit understanding is that only the items measuring less than the upper limit of a class are included in that class while the upper limit of a class is to be considered an item of the next class. Thus, 120, 130, 140, etc. are respectively the items of the 2nd, 3rd, 4th, etc. classes. The limits look like overlapping ones, but actually they are non-overlapping.

Illustration 2. TABLE 9.11: FREQUENCY DISTRIBUTION OF HEAD CIRCUMFERENCE OF 1071 INDIAN BOYS 16 TO 19 YEARS OF AGE Head Circumference (cm) 50 but less than 51 51 but less than 52 52 but less than 53 53 but less than 54 54 but less than 55 55 but less than 56 56 but less than 57 57 but less than 58 58 but less than 59 59 but less than 60 60 but less than 61 Total

Number of Boys 4 23 59 108 224 257 230 110 38 16 2 1071

Here, the upper limit of a class does not belong to the class, but is an item of the next class. Clearly, 51 belongs to the second class, 52 to the third and so on. The limits are nonoverlapping. Sometimes, in place of the words, 'but less than' or 'and less than' the words 'and under' are used; t 1 ·1s we get '50 and under 51', '51 and under 52', etc.

Illustration 3. TABLE 9.12: FREQUENCY DISTRIBUTION OF MONTHLY INCOMES (IN Rs.) OF MIDDLE-CLASS FAMILIES IN A CERTAIN AREA OF KOLKATA

I

Monthly Income in (Rs.) 0- 50- 100- 150- 200- 300- 500- 750- 1000- 1500-2000 Total 18 62 185 261 364 299 170 65 48 Number of Families 38 11510

Here, the first class is the same as 'O but less than 50', the second class the same as '50 but less than 100', and so on, so that the upper limit of a class does not belong to the class, but is, as is clearly shown, an item of the next class. The limits, here, are non-overlapping. Note that the class intervals are unequal. In the cases when there are great fluctuations in data, when there is a sharp rise or fall in the frequency over a small interval, unequal class intervals will be more useful in the construction of frequency distributions than class intervals of equal widths.

205

CHAP. 9: SUMMARIZATION OF DATA_

Illustration 4.

TA:aLE 9.13: FREQUENCY DISTRIBUTION OF ANNUAL INCOME OF A GROUP OF MIDDLE-CLASS FAMILIES IN A CERTAIN AREA OF CALCUTTA Annual Above 1.2 Above 1.8 Above 2.4 Above 3.0 Above 3.6 Above 4.2 Above 4.8 Above 5.4 Above 6.0 Above 6.6

Income (Rs. '000) and not exceeding 1.8 and not exceeding 2.4 and not exceeding 3.0 and not exceeding 3.6 and not exceedingA.2 and not exceeding 4.8 and not exceeding 5.4 and not exceeding 6.0 and not exceeding 6.6 and not exceeding 1.2 Total

__/'

No. of Families 206 325 475 556 762 603 490 350 220 108 4095 ~

Here, the first class is 'Above Rs.1200 and not exceeding Rs. 1800'. Clearly, the upper limit of a class belongs to that class, but the lower limit does not belong to the class. The limits are thus non-overlapping. Sometimes, in place of the words Above ... and not exceeding ... ' the words 'Exceeding ... and not exceeding ... ' or 'More than ... and not more than' or clauses bearing like meanings are used.

Illustration 5.

.

TABLE 9.14: FREQUENCY DISTRIBUTION OF OUTPUT OF 180 WORKERS Output (units per worker) 500-09 510-19 520-29 - -530-39 54D-49 550-59 56D-69 - 570-79 Total

No. of workers 8 18 23 37 47 26 16 5 180

The frequency distribution is of a discrete variable. The classes are 500-509, 510-519, etc. The limits are non-overlapping.

206 /

BUSINESS MATHEMATICS AND STATISTICS

Illustration 6. Grouped Frequency Distribution with Open Ends. TABLE 9.15: FREQUENCY DISTRIBUTION OF MARKS OBTAINED IN COMMERCIAL MATHEMATICS BY 350 ST!JDENTS IN AN ANNUAL EXAMINATION Marks Obtained

No. of Students

Below 13 14-23 24-33 34-43 44-53 54-63 64-73 74 and above

6 21 54 71 89 62 38 9 350

Total

In the case of grouped distributions with open-end class at one extremity or at both extremities, it is difficult to choose the values of the variable which are representative of the open . classes, i.e., the mid-values of those classes. If the closed classes in the frequency distribution are of equal widths, it is most usual in such cases to take the widths of the open classes to be the same as the common width and determine the mid-value. The reason why grouped distributions are sometimes left with open end or ends is that there are only a small number of items which are scattered over a long interval, and the frequencies falling within class intervals at these ends with common width are few and far between. f

Illustration 7. Percentage Frequency Distribution: TABLE 9.16: PERCENTAGE FREQUENCY DISTRIBUTION FROM THE DISTRIBUTION OF HEIGHTS OF 200 STUDENTS Class int_ervals (Heights in cm}

Frequency

Percentage Frequency

126-130 131-135 136-140 141-145 146-150 151-155 156-160 161-165 166-170 171-175 176-180

2 9 16 26 33 41 36 21

1.0 4.5 8.0 13.0 16.5 20.5 18.0 10.5 5.5 1.5 1.0 100.0

11

Total

3 2 200

/

207

CHAP. 9: SUMMARIZATION OF DATA

9. 7

Cumulative Frequency Distribution

There is another type of frequency distribution, viz., Cumulative Frequency Distribution, in which as the name suggests, the frequencies are cumulated. This is prepared from a grouped frequency distribution, showing the end values, i.e., the class boundaries by adding each frequency to the total of the previous ones, or those following it .. The. former is termed cumulative frequencies from below, or less than cumulative frequencies and the latter cumulative frequencies from above, or more than cumulative frequencies. We now refer to Table9.17, where a cumulative frequency distribution of the data of Table9.16 is given.

Method of Compilation of the Cumulative Frequency Table After writing down the class boundaries, from Tabte9.16, in the left-hand column, it is argued, for less than column as: number of men of heights less than 125.5 cm is nil; hence 0 against 125.5, number of men of heights less than 130.5 cm is 2: hence 2 against 130.5; number of men of heights less than 135.5 cm is 2 + 9 = 11; hence 11 against 135.5; .. ." Again, for more than column, the frequencies are to be cumulated from bottom . .It is argued as: number of men of heights 180.5 cm or more is nil; hence 0 against 180.5 ; number of men of heights 175.5 cm or more is 2; hence 2 against 175.5; number of men of heights 170.5 cm or , more is 3 + 2 = 5; hence 5 against 170.5; ... The cumulative frequency distribution from below is useful for answering questions like: How many students have heights less than 150.5 cm. The answer is the frequency under column (2) of Table9.17 against the class boundary 150.5, viz., 86. Similarly, the number of students who have heights 145.5 cm or more is the corresponding frequency under column (3) of Table9.17, viz., 147, and so on.

Illustration 1. TABLE 9.17: CUMULATIVE FREQUENCY DISTRIBUTION OF HEIGHTS OF 200 STUDENTS (Data of Table9.16} Class Boundaries (Heights in cm) (1) 125.5 130.5 135.5 140.5 145.5 150.5 155.5 160.5 165.5 170.5 175.5 180.5

Cumulative Frequency Less than (from below) More than (from above) (2) (3) '

0 2 11 27 53 86 127 163 184 195 198 200

200 198 189 173 147 114 73 37 16 5 2 0

208

BUSINESS MATHEMATICS AND STATISTICS /

/

9.8

Some Illustrative Examples

Example 1. Draw up a blank table to show the number of students sex-wise, attending for the Higher Secondary (HS) as well as First Year, Second Year and Third Year classes of Bachelor Degree courses in the faculties of Arts, Science and Commerce of a college in a certain session.

Solution. TABLE 9.18: NUMBER OF STUDENTS APPEARING IN DIFFERENT CLASSES OF A COLLEGE ACCORDING TO SEX AND FACULTY IN A CERTAIN SESSION

I~

~

Class

~

Bachelor Degree Courses Second Year Third Year H.S. First Year Male Female Male Female Male Female Male Female Total

Faculty 1. Arts 2. Science 3. Commerce

Total

/

Example 2. Present the following information in a suitable tabular form, supplying the figures not directly given: In 1980, out of a total of 5000 workers in a factory 3900 were members of a trade union. The number of women workers employed was 650, out of which 500 did not belong to any trade union. In 1981, the number of Union workers was 4200 of which 3400 were men. The number of non-Union workers was 1400, among which 450 were women. [V.U.B.Com. 1994)

Solution.

TABLE 9.19: NUMBER OF WORKERS IN THE FACTORY ACCORDING TO SEX AND MEMBERSHIP OF TRADE UNION FOR THE YEARS 1980 AND 1981 Year

Tua~~ Memb Trade Union Members Non-Trade Union Workers

Total

1981 1980 No. of Total No. of No. of Total No. of Male Cols. Male Female Cols. Female Workers Workers (1) & (2) Workers Workers (3) & (4) (3) (4) (1) (2)

3750 600 4350

150 500 650

3900 1100 5000

3400 950 4350

800 450 1250

4200 1400 5600

209.

CHAP. 9: SUMMARIZATION OF DATA

Example 3. In a sample study about coffee-habit in two towns A and B the following informa-

tion were received: ' Town A: Females were 40%, total coffee drinkers were 45% and male coffee drinkers were 20%. Town B: Males were 55%, male non-coffee drinkers were 30% and female coffee drinkers

cc

.5

.e

~

Q) Q)

J ............

800 600

I

\

200 0

\

.5

lZ -200

11/

3

2

3

'"

J J{

\

1V

\ \

J

\

Q.

=> cc

I I

............... \

400

I"'

\ \

'/

.J ' 4

5

6

7

8

9

10

I

I

..)•

.. 11

12

13

14

15

Fig 10.2 Daily Profit and Loss for the first fortnight of a month.

We represent 'Days' along X-axis, 'Profits' along the positive Y-a.xlBalld.'Losses' along the negative Y-axis according to the following scales: 1 side of a small square on the graph-paper along the X-axis 1 day, 1 side of a small square on the graph-paper along the positive Y-axis= Rs. 200 profit, 1 side of the small square on the graph-paper along the negative Y-axis~ Rs. 200 loss.

=

224

BUSINESS MATHEMATICS AND STATISTICS

Note that the zero line is the horizontal X-axis. Profits are shown above the zero line and Losses are shown below the zero line. Fig 10.2 represents the required graph.

Inclusion of Zero in the Vertical Scale Example 3. Fig 10.3 represents the graph of the data in the following table. Scales are clearly shown in the figure:

,

TABLE 10.1: PRODUCTION OF PIG IRON AND FERRO-ALLOYS IN WEST BENGAL DURING 2004-05 Months

Production (in Tons) 2004 2005 '

January February March April May June July August September October November December

64,980

73,728

71,005 64,138

8;3,715 95;262

73,175

86,826 94,105

73,815 67,479 74,606

78,795 97,082

79,740

94,830

77,627

79,615

79,428

79,539

90,205

77,380

99,541

82,295

/

Thousand _ Tons 100 .--~.....-~~~-r-~~~..--~~--...-~~~~~~-r-~~~...,.-~~---. ~ 90

~ 80 r--~--r~~~i-~::.::t11::::::::,;t:::~~--t-\:-7~-r~~~"'i/"~~--'--::w:::::;;;;7T

.E

70

_g

50

~ 60._~.......~~--1-~~~~~~-1-~~--4~~~-+-~~-1-~~~

~40 1--~--t-~~~+-~~~+--~~--1~~~-+-~~~-+-~~~+-~~----l '15 30

~ ~

20

1--~--1-~~~-t--~~~+--~~--1~~~-t-~~~-1-~~~+-~~----l

10

0

'---'--'---'-~L-_.___.__._~..___.__.___..~..__.....__.___.~...__,__.___,.__-'-'-,...i_-'---'

JV

F MR A MY JN JL

2004

A

S

0

N

D

JV

F MR A MY JN JL

Years and Months

A

S

0

2005

Fig 10.3 Production of Pig Iron and Ferro-alloys in West Bengal during 2004-05.

N

D

..

/

CHAP. 10: DIAGRAMMATIC REPRESENTATION OF STATISTICAL DATA

225

Thousand Tons 100~~~-~-+-~-~-~~

90

l--+---+--+--l--t\--lt""rl-++f--\4---1

80

r-t--t~.,.t;.1-TT;f-t-_,__.,...o"1

70

~+t--,_-+--+---t--t---+---1

60

...................

.._._.._,__._..__.__.~~..._..__,_,_~..._._~

JFMAMJJASONDJFMAMJJASOND 2004

2005

Years and Months (Incorrect Chart) Fig 10.4 Production of Pig Iron and Ferro-alloys in West Bengal during 2004-05.

Observations

The inclusion of zero in the vertical scale, as has been done in Figl0.3 is very important. Most chart-makers neglect to include the zero point, as has been in Fig 10.4. The verti@l scale in Fig 10.4 starts from 60 thousand tons. Consequently, this chart gives a misleading impression of the production of Pig Iron and Ferro-alloys. For instance, production in December, 2004 appears to be 8 times that in January 2004, which is not at all so. In fact, it will be seen from the Table 10.1 that the figures were actually 64,980 tons in January and 99,541 tons in December, the latter being only times the former approximately.

q

Fig 10.4 should not, therefore, be used to indicate the actual magnitude of Pig Iron and Ferro-alloys production.

Sometimes, one wishes to emphasis the movements of the curve and is less interested in the magnitude of the data. Thus where it is desired not to include the complete scale, the fact may be indicated by breaks Figs 10.5 &. 10.6. In any case, the zero of the vertical scale must be shown.

Thousand Tons

100 ..--....---.--.--...-.-.,.----.--..,.----.

o-~~~~~~~~~~~~

JFMAMJJASONDJFMAMJJASOND 2004

2005

Years and Months Fig 10.5 Production of Pig Iron and Ferro-alloys in West Bengal during 2004-05.

226

BUSINESS MATHEMATICS AND STATISTICS Thousand Tons 100 ~~~~~~~~-,...-~~~~~~~~~

JFMAMJJASONDJFMAMJJASOND 2004

2005 Years and Months

Fig 10.6 Production of Pig Iron and Ferro-alloys in West Bengal during 2004-05.

Two or More Charts in the Same Graph It is not necessary that one diagram will contain only one curve. Two or more curves may also be drawn on the same sheet, representing two or more related series of values. Such curves are helpful for comparing the related series. Too many curves should not, however, be drawn on.the same chart, because in that case the plotting may overlap and the simplicity in presenting the data is lost. The different lines in this case should be drawn so as to be easily distinguishable. They should be differentiated from each other by solid, dotted, dashed and the like lines and properly designated. The placing of designation along the curves is desirable Fig 10. 7.

Example 4. Fig 10.7 shows the plotted data contained in the following table: TABLE 10.2: COMPENSATION PAID FOR ACCIDENTS TO WORKMEN IN INDIA UNDER THE WORKMEN'S COMPENSATION ACT, 1923 (Thousand Rupees) Year

1996 1997 1998 1999 2000 2001 2002 2003 2004 2005

Death 1179 1581 1871 2003 2208 2008 2341 2474 2159 2041

Amount paid for Permanent Disability Temporary Disability 1210 937 1615 1024 2026 1320 1439 2265 2293 1397 1342 2288 1482 2315 2443 1508 2190 1618 1760 1382

Total

3326 4220 5217 5707 5908 5638 6138 6425 5967 5183

Note: See that the points have been plotted in the chart at the middle of each interval representing the years. This is an alternative way of plotting of points on the vertical lines at equal distances apart, which may indicate the periods, as shown in the previous figures.

227

CHAP. IO: DIAGRAMMATIC REPRESENTATION OF STATISTICAL DATA Thousand Rupees

1996

'97

'98

'99

2000 '01 Years

'02

'03

'04

'05

Fig 10.7 Compensation paid for accidents to workmen in India under the Workmen's Compensation Act, 1923.

Graphical Interpolation When the graphs of continuously varying variables are drawn we may find the values of the variables at any intermediate stage from the graph itself. The process is known as Interpolation by Graphical Methods. We illustrate the process by the following example: Example 5. The table below gives the height and weight of a man at different ages:

TABLE 10.3 Age (years) Height (in cm) Weight (in kg)

4 75 20

6 90 25

8 105 35

10 125 40

12 140 45

14 150 55

16 175 60

18 185 65

20 195 70

Drow two line charts on the same graph to represent the above data. Assuming that the height and weight of the man to vary continuously, find by graphical interpolation the height and weight of the man when he was 9 years old.

228

BUSINESS MATHEMATICS AND STATISTICS 200 195 185 175

80

150 140

§ .5 :;::-

:§, Cl>

::r:

70 65 60 55

f-

125 115 105 100 90

p

~

,_ -

,Jr

45 :§_ 40 £, 37.5 ~ 35 >

a

75 25 50

20

f-

4

6

8

9

10

12

14

16

18

20

Years

Fig 10.8 Height and Weight of a man at different ages.

Fig 10.B gives the two line charts-one for the height and the other for the weight. We have represented the Ages along X-axis. \ Scales: 5 sides of the squares on the graph-paper\ along X-axis = 2 years 10 sides of the square on the graph-paper along Y-axis= 50 cm 10 sides of the squares on the graph-paper along Y-axis = 20 kg The height and the weight of the man when he was 9 years old are given by the points P and Q respectively. Now read off from the graph-paper the values corresponding to these points on the Y-axis. We see that the required height = 115 cm and the required weight= 37.5 kg.

Bar Charts or Bar Graphs Bar Charts (or Bar Graphs) are another most frequently employed diagrammatic representation of statistical data. They are particularly useful for comparing the values of a variable classified qualitatively (i.e., classified according to some non-measurable characteristic possessed by them). They are also used in presenting statistical facts involving time-factor (i.e., in timeseries). Bar Charts consist of a number of rectangular areas usually made up with deep black ink, having the appearance of solid bars. The bars originate from a horizontal base line and their lengths are proportional to the values they represent. The bars should be of uniform width and equally spaced. The Bar Charts are called Vertical Bar Charts (or Column Charts), if the bars are placed vertically. On the other hand, we have the Horizontal Bar Charts, where the bars are plac--ed horizontally. Column Charts (and NOT Horizontal Bar Charts) are used in time-series. Fig 10.9 illustrates a column chart and Fig 10.14 illustrates a horizontal bar chart.

Simple Column Charts In a column chart the space between consecutive columns should generally be one-half of the width of each column. These charts are generally used for depicting the values of a variable

229

CHAP. 10: DIAGRAMMATIC REPRESENTATION OF STATISTICAL DATA

over a period of time. Such charts are also used for data classified quantitatively, e.g., the population figures classified by age-groups, the number of coal-mines classified according to the number of persons employed, etc. The values of the variable over a period of time or the frequencies, in the case of data classified according to groups, are represented by the heights of the columns. Example 6. (i) The monthly production of Hind Motor Cars for the first six months of thf year 2005 is given below: '

,January-180 April-250

February- 240 May-200

March- 210 June- 160

Represent the production figures by Bar Charts. We represent 'Months' along the X-axis OX and the 'production of Cars' along the Y-axis OY. The above data have been represented by Bar Chart. [See Figl0.9(a)]. Scale: 1 side ofa small square along the Y-axis OY represents 5 cars. (Students are advised to use graph-pape'r.) y

250

t------------,,

f 200 ~ . (.) 15,0

0

.8 1100 E · ~ 5.o

0

'----,..--1'""'"""....,_

Jan

Feb

Mar Apr Months-

May

Jun

x

Fig 10.9(a) Bar Chart showing the monthly production of Hind Motor Cars. ·

lii) Below is a table giving the daily average number of workers employed in the factories in India. TABLE 10.4: DAILY AVERAGE NUMBER OF WORKERS EMPLOYED IN THE FACTORIES IN INDIA Year 1996 1997 1998 1999 2000

Number of Workers 22,74,689 23,60,201 24,33,966 25,04,399 25,36,544

-

Year 2001 2002 2003 2004 2005

Number of Workers 25,67,458 25,28,026 25,89,757 26,90,403 28,82,309

gives a simple column chart showing the number of factory workers in different years as given by the above table. The heights of the bars are proportional to the number of 'workers. Taking, say, 0.3" to represent 5 lac of workers we see that in 1996 the number of workers is 22.746 lac+; so we take a column of height o.3 2 · 746 = 1.36" (approx.) or x 22.746 = 4.55), i.e., 4.5+ divisions. Similarly, other columns are to be drawn. Fig 10.9(b)

(t

x;

230

BUSINESS MATHEMATICS AND STATISTICS Lacs 30 ~

25

~ 0

20

iiE

10

z

5

15

::>

0

'97

'98

'99 2000 '01

'02

'03

'04

'05

Fig 10.9(b) Daily average number of workers employed in the factories in India.

Example 7. (i) The profits and losses of a business concern for the years 2001-2005 are given below: TABLt 10.5 . Year Profit (in Rs.) 2001 3,000 4,000 2002 2003 2,500 2004 . 2005 6,000

Loss (in RS.)

2,000

Represent the above data by a Bar Graph. We have represented the above data by a column chart in Fig 10.10.(a). Taking 0.2" of the I vertical line to represent a profit or loss of Rs. 500 we see that in 2Q01 there is a profit of Rs. 3,000. Hence, the height of the column corresponding to the year 2001 = (0.2 x. 6) = 1.2" or 6 divisions, and so on for other cases. The bar representing the loss has been shown below the zero line or base line. Such bars drawn both above and below the base line, to represent profits and losses, excesses and deficits, etc. are called deviation bars. 6000 5000 4000

.:;::

3000

~ en ..9

2000

e

1000

0 - 1000 - 2000 Fig 10.10(a) Bar Chart showing the profits and losses of a business concern.

· The above data can also be represented by a line chart. This is left as an exercise to the students. (ii) Another example of Deviation Bars (deviated from the zero line) is given by FlglO.lO(b). The corresponding table is given below: ·

231

CHAP. 10: DIAGRAMMATIC REPRESENTAT ION OF STATISTICAL DATA

TABLE 10.6: INDIA'S TRADE WITH A FOREIGN COUNTRY (Value in Crore of Rupees) Year

Exports

Imports

Balance

+ 2001 2002 2003 2004 2005

149 141 146 160 208

126 148 176 168 187

-

23 7 30 8

-

21

Crore of Rupees

30 ,---------,..,,,,.,...,,,,..-------., 25 t------~ .~~~11-------1 20 1---- - - - 1; 15 f - - - - - - - f 10 t - - - - - - - 1 5 f - - - -·h

0 i--,;:;,:;.;=r'-'-"'=L.L -5 -10 -15

-20 -25

~-----------~

Fig 10.1 O(b) Deviation Bars showing India's Trade with a foreign country.

Multiple Column Charts (Compound

Coh~mn

These charts depict more thari one type of data at a time. In these diagrams we may have 2, 3 or 4 or even more bars constructed at the same time side by side to represent 2, 3 or 4 series of Yalues for comparison. For convenience the columns are differently shaded.

Example 8. The following figures relate the value of sugar

Charts) Cror~s

of Rup;~s 60 "r - - - - - - , ·!.-

U.P.

Bihar Other States

U.P.

50 1-'- '-'----

@Bihar

40

I

manufactured in the states of the Union of India in a certain year: States

I

t----

Other States

Value in Rupees

57,53,87,000 32,48,32,000 15,12,65,000

10

0

Represent the above data by a multiple column chart. We have represented the above data by a triple column chart in Flgl0.11.

Fig 10.11 Triple bar chart showing the manufacture of sugar in different states.

232 Example 9.

BUSINESS MATHEMATICS AND STATISTICS

Fig 10.12 is also another compound column chart depicting the data of the following

table:

TABLE 10.7: AVERAGE MONTHLY CASH EARNINGS OF TEA-PLANTATION WORKERS (SETTLED LABOURERS) IN ASSAM Monthly Cash Earning (Rs. P. hundred) Men Women

Year 1996-97 1997-98 1998-99 1999- 2000 200~01

2001--02 2002--03 2003- 04 2004-05

14.31 17.54 18.56 21.10 19.80 21.64 20.52 34.66 39.45

11 .02 14.60 14.22 15.33 16.90 19.35 18.13 29.64 34.41

-

Hundred of Rupees

40

en =Women

1996-97 -'98

- '99

-2000 -'01 -'02 , -Years-

-'03

-'04

-'05

Fig 10.12 Average monthly cash earnings of tea-plantation workers (settled labourers) in Assam .

Component Bar Charts In these, each bar is subdivided into certain parts. Fig 10.13(b) represents such a component bar chart. There are two bars-one of them represents the total cost and its component parts (viz., Cost of Direct Materials, Direct Labour, Direct Expenses, Overhead Charges) for the year 1999 and the other depicting the same items for the year 2000. The component parts are indicated by different hatchings and the total cost by the complete bars.

Example 10. (i) Following are the heads of in'(:omes of a Railway Company during 1999 and 2000: 1999 (in crore of Rs.) Coaching Goods Others

2000 (in crore of Rs.)

26

31

40

39 3.50

4.50

Represent the above data by a Bar Chart.

233

CHAP. 10: DIAGRAMMATIC REPRESENTATION OF STATISTICAL DATA

We see that the given data consis~s of three components for two consecutive years. In order that one may make a comparison of t he incomes under different heads it will be appropriate in this case to draw a component bar chart. This has been done in t he adjoining diagram. so .----------------------~

60 50

40

20

' Co~~hiri~ . ,.;

10

'·'

'-.¢~17,,,'

,___ _ _ __,

2000

1999

Fig 10.13(a) Component bar chart.

(ii) The following table gives the total cost in Rupees and its component parts in two consecutive years: TABLE 10.8 Direct Materials Direct Labour Direct Expenses Overhead Charges Total Cost

./ Unit

1999 50,000 55,000 15,000 25,000 1,45,000

2000 60,000 70,000 18,000 32,000 1,80,000

= Rs. 5,000 4 0 . - - - - - - - - - - - - - -- - - - --

25 20

15

5

1999

2000

Fig 10.13(b) Component .bar chart .

--,

234

BUSINESS MATHEMATICS AND STATISTICS

Fig 10.lJ(b) gives the component bar charts of the above data. Scales are clear from the 1 . figure. See that how effectively the chart visualizes a heap of figures.

Horizontal Bar Charts Like the column charts they ar:e also very simple to draw. The bars originate from a vertical base line (Fig 10.14) to the left and extend to the right. The horizontal lengths of the b~rs representing the values of the variable are read from a few vertical scale lines. Example 11. The horizontal bar chart corresponding to the following table is represented by Fig 10.14.

TABLE 10.9: NET AREA UNDER IRRIGATION IN INDIA DURING 2004-05 Net Area Irrigated (thousand acres)

Modes of Irrigation By Canals (i) Government (ii) Private By Tanks By Wells By Other Sources

19,356 2,863 9,889 16,562 5,587 54,257

Total

Area Irrigated (thousand acres)

0

5

10

15

20

By Govt. Canals By Pvt. Canals By Tanks By Wells By Other Sources Fig 10.14 Net Area under Irrigation in India during 2004-05 (Data of Table 10.9).

Sometimes, numerical values (Fig 10.15) are inserted at the extreme right ends of the bars. The attributes are listed to the left of the bars. Area Irrigated (thousand acres)

0

5

10

15

20

By Govt. Canals By Pvt. Canals By Tanks By Wells • • • • •• • • • • • • 16,562 By Other Sources • • • • 5 .587 Fig 10.15 Net Area under Irrigation in India during 2004-05 (Data of Table 10.9).

235

CHAP. 10: DIAGRAMMATIC REPRESENTATION OF STATISTICAL DATA

lf a bar is unusually long as compared to the others, it may be broken at the end and a numerical figure indic~ting its value is inserted in the broken portion (Fig 10.16). 47

Murder Burglary

1006

Rioting

213

Robbery

26

Dacoity 3

Fig 10.16 Incidences of Crimes in Kolkata (Year 2005).

As in the case of column charts we may also have multiple horizontal bar charts (Fig 10.17) drawn with reference to the Table 10.10 given below.

TABLE 10.10: NUMBER OF TEACHERS IN EDUCATIONAL ' INSTITUTIONS IN CERTAIN STATES OF INDIA (2004-05) States of India

Number of Teachers Women Men 96,740 10,886 1,23,155 15,312 64,517 12,996 78,176 5,679 1,11 ,686 29,233 1,00,503 42,644 31,718 7,240

West Bengal Utta.r Pradesh Andhra Pradesh Bihar Bombay Madras Punjab

Teachers in Educational Institutions (2004-05) Number of Teachers (in thousand)

0

30

60

90

120

150 ~ Men

c::J

Punjab

~~ljJ·~-

Fig 10.17 Number of Teachers in Educational Institutions in certain States of India (Data ·Table 10.10).

Women

/

236

BUSINESS MATHEMATICS AND STATISTICS

Ratio Charts or Logarithmic Charts The graphs previously described and illustrated give absolute changes in the values represented by the vertical axis. Take, for example, the line chart of Fig 10.3, where productions have been shown on the vertical axis. In this figure a change in the amount of production by, say, 10,000 tonnes, is represented by the same, length in the vertical scale, whether the production rises from 65,000 -tonnes to 75,000 tonnes, or from 89,000 tonnes to 99,000 tonnes. Such scales, therefore, give equal lengths in the graph for equal lengths in the values of a variable. They are called Arithmetic Scales (or Natural Scales). · But sometimes we are interested to measure relative changes in the values of the dependent variable (not Absolute Changes): We clarify the idea with the help ·of the following example: Suppose the production of Iron in four years are as shown below: (2)

(1)

l

l

2000 65,000 tonnes

2001 75,000 tonnes

.1 2002 89,000 tonnes

l 2003 99,000 tonnes

In both (1) and (2) absolute changes in production= 10,000 tonnes. But in (1) the percentage ofincrease = ~~ggg x 1003 = 15.43 and in (2) the percentage of increase = ~gggg x 1003 = 11.2%. If we draw graphs in the conventional Natural Scale, we cannot visualize such relative (or percentage) changes. In order to compare such relative changes over ll.'P..eriod of time we use a special form of graph called Semi-Logarithmic Graph. In this graph the horizontal axis (where usually time is represented) is marked off as in the case of our usual Arithmetic Scale. But the vertical axis should be scaled off in such a manner that equal lengths on this scale show equal ratios of change. Such a scale may be prepared by marking the divisions proportionar to the logarithms of natural numbers. _,.Suppose we Wl ld like to prepare a semi-logarithmic graph-paper and we wish to represent the following numbers: 1, 10, 20, 30, 40, 50, 100, 1000. Their logarithms are: 0, 1.00, 1.30, 1.47, 1.60, 1.69, 2, 3. Thus, we may take 1 at the origin, 10 at a height 1 (unit of length) above X-axis, 20 at a height 1.30 units of length above the X -axis, and so on. It will be observed that the lines parallel to the horizontal axis will be unevenly spaced, but that the spacing will be similar at intervals 10, 100, 1000, 10,000, etc. A complete group of such lines in each interval is known as a Cycle. Since, log la = YI log Y2 - log Y1, the distance between the lines y = Y1 and y = Y2 on such a chart will really indicate the ,value of the ratio ~ (not the difference Y2 - Y1). We usually take the. top and bottom limiting lines as the powers of ten when the semi-logarithmic chart is used. It should be noted that there is no zero point on the vertic~l semi-logarithmic scale since log 0 is not defined. Fig 10.18-represents the production of Gypsum in India (as given by the table below) by using a semi-logarithmic graph-paper. We have taken 10 (thousand tons) at the origin; hence, 79 (thousand tGns) should be 0.89 unit of length above the origin since log 79 = 1.89 and log 10 = 1. Similarly, other points have been plotted.

237

CHAP. 10: DIAGRAMMATIC REPRESENTATION OF STATISTICAL DATA

I TABLE 10.11: PRODUCTION OF GYPSUM ORE IN INDIA I Year

Quantity ('000 tons)

Year

Quantity ('000 tons)

Year

Quantity ('000 tons)

1997 1998 1999

79 140 206

2000 2001 2002

204 411 586

2003 2004 2005

612 618 850

'

Thousand tons 1000 800

/~

600 400 200

0

..... .....

-

1997 '98

v [.;"

!/

'99 2000 '01

.

~

'02

'03

'04

'05

Fig 10.18 Production of Gypsum ore in India during 1997-2005.

If both the scales (vertical and horizontal) are logarithmic, it i:s called Double Logarithmic Chart. They are seldom used in business concerns.

Advantages of Ratio Scale • Relative changes are best represented. The slope or steepness of the lines on a Logarithmic chart indicates the rates of change. The steeper the rise or fall of these lines, the greater is the ratio of increase or decrease. The relative rate of change of two sets of figures can then be easily compared by comparing the slopes of the lines. • Extrapolation is possible. • When a large range of values (say, from 1000 to 20,00,000) has to be accommodated within a comparatively small space the Ratio Chart is useful.

Disadvantages of Ratio Scale • Absolute changes cannot be easily compared on a ratio chart. • Zero and negative values cannot be represented. • It requires a little knowledge of logarithms. Instead of making a special graph-paper on which Y-axis is graduated according to the logarithms of numbers we may draw a logarithmic chart by plotting the values of. logarithms of the values of yon an ordinary graph-p[t.per (i.e., a graph-paper, where arithmetic scales are taken).

238

BUSINESS MATHEMATICS AND STATISTICS

30

,. ,

25

20

,/

~

15

10

5

Years 1997 '98 '99 2000 '01

'02 '03 '04 '05

\

Fig 10.19 Logarithmic Graph on a Natural Scale.

In Fig 10.19 we have taken an ordinary graph-paper. On the vertical axis we have represented the logarithms of the numbers. The data are the same as in t~e Fig 10.18.

Pie Charts The Pie Chart is a very useful pictorial device for visualizing the weight of different items in a composite quantity. In fact, like component bar charts, Pie Charts can effectively display the comparison between the various components or between a part and the whole. A Pie Chart consists of a circle subdivided into sectors by radii in such a way that the areas of the sectors are proportional to the values of the component items under investigation-the whole circle, of course,'representing the whole of the data under investigation. It should be remembered that the areas of sectors are proportional to the angles at the centre. So we may also consider a Pie Chart to consist of a circle, broken up radially into sectors, in such a way that the angles of the sectors at the centre would be proportional to the various components of the given aggregate. In order to draw a Pie Chart, we first express the different components of the given data as percentages of the ' whole. Now the total angle at the centre being 360°, it will represent the whole, i.e., 1003. Therefore, 3.6° will represent 13 of the whole. Consequently, if x be the perceµtage of a certain component, the angle of the corresponding sector at the centre = x x 3.6°. (Conversely, if a certain component is represented by a certain sector, we measure the angle subtended by it at the centre from which we can find the percentage of that component with respect to the whole.) In this way, the central angles of the sectors correspondi~to the different components are determined. A circle drawn with a convenient radius is then divided into different sectors with those central angles. The different sectors thus obtained are finally shaded differently.

239

CHAP. 10: DIAGRAMMATIC REPRESENTATION OF STATISTICAL DATA

''

Example 12. Draw a Pie Chart to represent the following data relating to the production cost · of ~anufacture: Cost of Materials: Rs. 38,400; Cost of Labour: Rs. 30,720; Direct Expenses of Manufacture: Rs. 11,520; Factory Overhead Expenses: Rs. 15,360.

"

Solution. We first express each item as a percentage of the total cost, viz. , Rs. 96,000. TABLE 10.12: CALCULATIONS FOR PIE CHART Percentage

Central Angles

( 1) Cost of Materials

~:~~ x 100%

40%

(2) Cost of Labour

~~~~g

32%

(3) Direct Expenses

!!~

(4) Factory Overhead

!~~g

= x 100% = x 100% = x 100% =

Total

12% 16%

= 144° 3.6° x 32 = 115.2° 3.6° x 12 = 43.2° 3.6° x 16 = 57.6°

3.6° x 40

360°

100%

A circle of-convenient radius is now drawn and the above angles are marked out at/ the centze of the circle. 4 radii will then divide the whole circle into four required sectors. \ jfhe different sectors are generally differently shaded. Fig 10.20 gives the Pie Chart required. 1

Cost of

. .

Materials 40%

·.. u

100 90

80 70 60

c::

so

~

40

OJ ::J

u..

~

-~

:;

E ::J

u

30 20 10 0

49.5 995 1495 199.S 249.S 2995 3495 Profits (Rs. Lak h)

x

-+

Fig. 10.29: "Less than " and "More than" type ogives.

Th~ cumulative frequencies (both "less than" and "more than" types) are plotted on a graph-paper against the corresponding class-boundaries, taking profit (in Rs. lakh), i.e., classboundaries along the horizontal axis OX and no. of companies (i.e., c.f.) along the vertical axis OY. The two ogives ("less than" and "more than" types) are drawn in Figl0.29.

Example 22. Draw a suitable diagram from the following data: Year

2002 2003 2004 2005

Sales

Gross Profit

Net Profit

('000 Rs.)

('000 Rs.)

('000 Rs.)

120 135 140 150

'10 45 55 60

20 30 35 40

'-,

Solution. To show Sales, Gross Profit and Net Profit in each of the years 2002- 2005, it will be suitable to draw multiple bar diagrams for each year. We represent years along the horizontal line OX and Sales, Gross Profit and Net Profit along the vertical line OY in the same scale. 160 140 120 100

80 60 40 20

0

2002

2003

2004

2 0 05

Fig 10.30 Multiple bar diagram showing Sales, Gross Profit and Net Profit in the years 2002-2005.

Example 23. The budget of two famili es are given below. Repre:;en t the data by percentage rectangular diagram:

250

BUSINESS MATHEMATICS AND STATISTICS

Items of Expenditure

Family A

Family B

1600 800 600 200 800

1200 320 480 160 240

Food Clothing Rent Light and Fuel_ Miscellaneous

[D.U. B.Com. 1991)

Solution. We first convert the given data (figures) into percentages of the total for both Family A and Family B , and then draw the rectangular diagram (the component bar charts). As total percentage is 100 for both the families A and B, we take widths of the two rectangles in the ratio 4000: 2400 = 5: 3. TABLE 10.19: PERCENTAGE CALCULATIONS Items. of Expenditure

Family A

Rs. 1600

Food

Family B

Percentage

Rs.

= 40

1200

!~og x 100

Percentage

= 50.00 x 100 = 13.33

~~gg x 100

Clothing

800

4 i0~ x 100 = 20

320

2342000

Rent

600

4 0~0 x 100

480

2~ ~ x 100

Light and Fuel

200

Miscellaneous Total

8

6

800

= 15 J0~0 x 100 = 5 8 4 : 0 x 100 = 20

240

4000

100

2400

\

8

6

= 20.00

2~ ~ x 100

160

2 4~ x 100

2

= 6.67

= 10.00 100

We represent Family A and Family B along the horizontal line OX and expenditure on different items along the vertical line OY. Percentage rectangular diagram showing the budgets . of two families A and B is drawn in Fig 10.31. 100

100

90 ~=========i 83.33

80

=====::::! 63.33

60

50 40 20

0

Family A (Rs. 4,000)

Family B (Rs. 2,400)

Fig 10.31 Percentage rectangular diagram showing the budgets of two families A and B.

'251

CHAP. 10: DIAGRAMMATIC REPRESENTATION OF STATISTICAL DATA

- - - - - - - - - 1 EXERCISE ON CHAPTER 10 Theory 1. .What is graphical representation? Discuss its advantages.

' '·)

2. How is a set of quantitative data presented by graphs and charts? Name at least four such graphical representations with examples. 3. Define and illustrate: Histogram, Pie Chart, Bar Chart, Line Chart. 4. What do you understand by Bar Chart? How is it drawn? 5. Define a Histogram and describe how it is constructed. 6.

[B.U. B.Com.(H) 1988)

(a) What is a Pie Chart? (b) Discuss the types of data which are usually represented by Pie Diagram: Stat~ how they are drawn.

7. Explain the following with illustrations: (a) Pie Chart, (b) Column Chart, (c) Logarithmic Graphs. 8. Explain briefly what is meant by Bar Charts, Pie Charts and Histograms, and describe briefly how they are useful in business analysis. Illustrate also any one of these methods of graphical representation. [C.U. B.Com.(H) 1991) 9. Write the advantages and disadvantages of diagrammatic representation of data. [C.U.B.Com. 2006}

Problems 1.

(a) The monthly productions of Hind Motor Cars for the first six months of the year 1977 are given below: January-150, February-200, March-180, April-230, May-210, June-140. Represent the production figures by a Line Chart. (b) The monthly productions of Maruti Udyog Limited for the first six months of the year 1985 are given below: January~250, February-300, March-340, April-320, May-270, June-240. Represent the production figures by a Bar Chart. [C.U.B.Com. 1986)

2. Draw a Bar Chart for the number of students of a college: B.Com. 1st year class-600, B.Com. 2nd year class-500, B.Com. 3rd year class-350. 3. Draw a Pie Chart to represent the following data relating to the production cost of a manufacturer: Cost of material-Rs. 20,000, Cost of labour-Rs. 18,500, Other expenses-Rs. 11,500.

252

BUSINESS MATHEMATICS AND STATISTICS

4. Draw a Histogram to represent the following frequency distribution: (a)

Daily Wages (in Rs.) 5-10 10-15 15-20 20-25 25-30

(b)

Income (in Rs.): Frequency:

Number of Workers 15 25 30 20 10

11-15 4

16-20 13

21-25 22

26-30 33

31-35 12

36-40 6 [B.U. B.Com. 1988]

5. (a) Construct a Frequency Polygon from the following table: Weight in kg 30-40 40-50 5o-60 60-70 70-80

No. of Students 400 500 700 300 100

(b) A frequency distribution of marks obtained by 200 students in a competitive examination is given below: Marks

20-30

3o-40

40-50

5o-60

60-70

70-80

Total

JO

40

75

20

25

30

200

Number of Students

Exhibit the distribution by a Histogram and hence, draw the Frequency Polygon. 6. The monthly production of scooters in India is as given below: January-550 May-650 September-425

February-500 June--c-600 October-400

March-625 July-475 November-700

April-525 August-575 December--625

Represent the above data by Bar Graph. 7. (a) The daily profits and losses of a business concern for the first 10 days of a month are given in the following table: Days of the Month Profit in Rs. Loss in Rs.

1 400

2

3

4

800

600

900

5 200

6 200

7

8

400

100

9 800

10

1000

Represent the above data by a Line Chart, choosing suitable scales. {b) The daily selling price of gold in India (each 10 gm) from the 10th to 19th October in the year 2007 were as follows:

253

CHAP. 10: DIAGRAMMATIC REPRESENTATION OF STATISTICAL DATA

Days of October Prices (in Rs.)

10

11

12

13

14

15

16

17

18

19

8600

8750

9200

9600

9600

10200

9500

8800

9750

10000

Draw a Line Chart to represent the above data. 8. The monthly production of cycle in India is as under: January-5720 May-6040 Septelllber-5605

February--4900 June-4610 October-,--3275

March--6110 July~3060

Novelllber-6850

April-5930 August-4700 Decelllber-6130

Represent the above data by (a) a Line Chart and (b) a Bar Graph. 9. Draw a Bar Chart for the number of students of a college: Pre-University Class B.Colll. 1st-year ClassB.Colll. 2nd-year Class B.Colll. 3r~-year Class

200 1,000 600 400

10. Prepare a Bar Chart from the following data: India's Foreign Debt as at 31st March, 2006: Source of Borrowing

Amount of Loan in Crore (Rs.)

Federal Republic of Gerlllany Japan United Kingdolll U.S.A. I.B.R.D. {International Bank) l.D.A.

350 160 480 1200 260 450

11. Draw the graph of the following: Year Yield (in lllillion tonnes)

1998 12.8

1999 13.9

(Hints: Draw a Line Chart.)

12. Draw a Pie Chart from the following data:

(a) Revenue of the Central Govt.

2000 12.8

2001 13.9

2002 13.4

2003 6.5

2004 2.9

2005 14.8

255

CHAP. IO: DIAGRAMMATIC REPRESENTATION OF STATISTICAL DATA

14. (a) The cost of production of a certain commodity in a factory is given below. Represent the data in a Pie Diagram: Items

(in thousands of.Rs.)

555 730 150

Raw-material Labour Transport

Items

(in thousands of Rs.)

Packing

30 56 29

Factory Overhead 1 Advertisement

[C.U. B.Com. 1987]

(b) Represent the, following data by a circular or Pie Chart: hem Expenses

Food 3400

Clothing

Fuel

Rent

Educat'on

Misc.

Total

960

200

780

1500

1080

7920

15. From the following table, draw a Ratio Chart on a graph-paper: Year Units Produced

1997 2

1998 4

1999 8

2000 16

2001 32

2002 64

2003 128

2004 256

I

I

16. (a) Draw a Pie Chart to represent the following data on the proposed outlay· of the Fifth Five-Year Plan: Fifth Five-Year Plan Agriculture Irrigation and Power Industries and Minerals Education Roads and Communication Total outlay

Rs. (in crore)

12,000 5,000 8,000 9,000 6,000 40,000 [C.U. B.Com. 1991, '97 Type]

(b) Draw a Pie Chart to represent the following data on the proposed outlay during a Five-Year Plan of a Government: Fiftb Five-Year Plan

"'

Agriculture Irrigation and· Power Industries and Minerals Education Roads and Communication

'.Rs. (in crore) 8,000 7,000 4,000 5,500 2,500

I

[V.U.B.Com. 1994

256

BUSINESS MATHEMATICS AND STATISTICS

(c) Represent the following data by a Pie Chart: Expenditure of a Govt, in 1995-96

Items

Rs. (in crore) Agriculture

8,000

Industries and Minerals

7,000

Irrigation and Powers

4,000

Communications

5,500

Miscellaneous

2,500 [C.U. B.Com. 2003]

17, (a) The total costs and its components in two different months are given below: Items

January

February

Direct Material

700

600

Direct Labour

800

700

Direct Expenses

100

70

Overhead

200

300

1,800

1,670

Total

Draw component Bar Charts showing the total cost and its components. (b) In an examination 70% passed in English, 75% passed in Mathematics and 50% passed in both the subjects while 40 students failed in both the subjects. Draw a Pie Chart to represent the number of students, (i) who passed in both, (ii) who passed in English only, (iii) who passed in Mathematics only, and (iv) who failed in both.

[N.B.U. B.Com. 1992]

[Hints: Calculations for Pie Chart are shown below:]

Percentage

Item

Central Angles

Passed in both subjects Passed only in English

5%

= 180° = 72° 25 X 3.6° = 90° 5 X 3.6° = 18°

100%

360°

50% 70-50=20%

Passed only in Mathematics Failed in both subjects Total

75-50 = 25% 100 - (50 + 20 + 25)

=

50 x 3.6°

20 X 3.6°

257

CHAP. 10: DIAGRAMMATIC REPRESENTATION OF STATISTICAL DATA

18. Draw a Histogram to represent graphically the following Frequency Distribution:

Frequency Distribution of Income of 2,775 workers Frequency of Workers

Income in Rs.

375 400

70-80 80-90 90-100

650 575 350 425

100-110 110-120 120-130

Total 19.

2,775

(a) Draw a Histogram to represent the followlng Frequency Distribution: Frequency Distribution of the weights of 2,000 students Weight in lb

Number of Students

90-100 100-110 110-120 120-130 130-140

500 700 300 400 100

Total

2,000

(b) Draw a Histogram of the following distribution: Age (in years)

20-24

25-29

30-34

35-39

40-44

No. of Workers

50

70

100

180

150

45-49 / 50-54

59,,_59

[C.U. B.Com. 2000]

20.

(a) Draw a Histogram to represent the following frequency distribution: Output (units per worker) Number of Workers

I

500-510 510-520 520-530 530-540 540-550 550-560 560-570 57(}-580 18

8

23

37

47

26

16

5

(b) The weekly wages of 430 workers are given below: Weekly Wages (Rs.)

20-30

30-40

40-50

50--60

60-70

70-80

80-90

Number of Workers

30

50

80

100

70

60

40

Draw the Histogl:l,tm and Frequency Polygon for the distribution. [B.U. B.Com. 1986]

21.

(a) Draw a Histogram representing the following frequency distribution of 1,000 students by weight: Frequency Distribution of the Weights of 1,000 Students. 17

258

BUSINESS MATHEMATICS AND STATISTICS

Weight in lb

Number of Students

80 and under 90 90 and under 100 100 and under 110 110 and under 120 120 and under 130 130 and under 140 Total

85 300 215 150 50 200 1,000

(b) Draw the Histogram of the following distribution and use it to find the total ntJ.mber of wage Earners in the age-group 19-32 years: Age-group No. of Wage Earners

14-15

16-17

18:20

21-24

25-29

3~34

35-39

60

140

150.

110

110

100

90

(Hints: Find frequency densities and then proceed as in worked-out Example 19.] 22~

Draw the Histogram to represent the following frequency distribution: (a)

Weekly Wages (in Rs.) No. of Men

30(}-350 350-400 340 115

400-450 450--500 220 175

500-550 80

55!Hl00 35

600-650. 650--700 20 15

[c.u._13.com; (b)

Income (in Rs.) No. of Workers

101-110 225

111-120 375

121-130 500

131-140 725

141-150 650

1994]

151-160 161-170 350 175 --(N.B.U. B.Com. 1996]

23. Draw (a) Histogram and (b) Cumulative Frequency Polygons (both Less than _and More than) from the foll~wing distribution:

I

Weekly Wages Distribution of 250 Workers Weekly Wages (in Rs.)

Number of Workers

3~31

2 9 25 30 49 62 39 20 11 3 250

32-33 34-35 36-37 38-39 4~1

42-43 44-45 46-47 48:49 Total

__,,,.

.259

CHAP. 10: DIAGRAMMATIC REPRESENTATION OF STATISTICAL DATA

24. Draw a G1'ouped Bar Chart with the following data: World Production of Raw Cotton during, 1996-98 Production ('000 bales of 478 lb each) 1997-98 1996-97

Country U.S.A. China U.S.S.R. India Pakistan Brazil Egypt

13,027 / 6,000 f

10,900 6,500 5,800 4,300 1,350 1,340 1,870

6,200 4,180 1,323 1,340 1,498

, 25. Draw Histogram and Frequency Polygon to present the following data: Income (in Rs.)

Number of Individuals

100-149 150-199 200-249 250-299 300-349 350-399 400-449 45D-499

21 32 52 105 62 43 18 9

Total

342

26. Construct a Frequency Polygon from the following table: Output of Workers Output (units per worker) Number 500-509 510-519 520-529 530-539 540-549 550-559 560-569 570-579

of Workers 8 18 23 37 47 26 16 5

27. Draw a Cumulative Frequency Diagram (less than type) from the following data: Monthly Wages (in Rs.) 125-175 175-225 225-275 275-325 325-375 375-425 425-475 . Number of Workers

6

10

25

35

12

10

4

260

BUSINESS MATHEMATICS AND STATISTICS

28. Draw Cumulative Frequency Diagram (less than type) of the following frequency distribution and hence determine the median: Monthly Wag«!S (in Rs.) 12.5-17.5 17.5-22.5 22.5-27.5 27.5-32.5 32.5-37.5 37.5-42.5 42.5-47.5 47.5-52.5 52.5-57.5

Number of Workers 2 22 10

14 3 4 6 1 1

Total

29.

63

(a) Draw Ogives (both less than and more than types) from the following frequency distribution: Wages (in Rs.) Number of Labours

200-209

210-219

220-229

230-239

240-249

250-259

10

20

22

20

25

3

[C.U. B.Com. 1,988, '98]

(b) Draw Ogives (both "less than" and "more than") for the following distribution: Wages (in Rs.)

50-59

60-69

70-79'

80-89

90-99

100-109

110-119

5

10

15

8

6

4

2

Number of Employees

[C.U. B.Com. 2006]

[Hints: See worked-out Ex. 21.J

30. Draw the Cumulative Frequency Diagram of the following frequency distribution and hence determine the median weight of an apple: Weight}n gm 110-119 120-129 130-139 140-149 150-159 160-169 170-179 180-189 Total

Frequency 5 7 12 20 16

10 7 3 80

31. The trends in Revenue and Expenditure of the Central Govt. for the years 1995-2001 are given in the table below:

261

CHAP. 10: DIAGRAMMATIC REPRESENTATION OF STATISTICAL DATA

Revenue (Rs. in crore) Expenditure (Rs. in crore)

1995-96 4800 4400

1996-97 5600 4700

1997-98 6800 6300

1998-99 6700 6800

1999-2000 7400 7600

2000-01 8200 8300

Draw Multiple Bar Charts to represent the Revenue and Expenditure of the Government since 1995. Read off the surplus or deficit from the figures so drawn. [Hints: For ea.ch financial year draw a pair of column charts-one representing Expenditure and the other representing Revenue.]

32. Draw a Pie Chart representing the following data on the proposed outlay of a Five-Year Plan of a Government (Rs. in crore): Agriculture-12,000, Industries and Minerals-9,000, Irrigation and Power-·5,000, Education-8,000, Communications-4,000, Miscellaneous-2,000. [C.U. B.Com. 1991]

33. Draw a Histogram of the following frequency distribution and find the proportion of firms with annual sales greater than Rs. 70,000: Annual Sales (Rs. '000) No. of Firms

0-20

20-50

50-100

100-250

250-500

500-1000

20

50

69

30

25

19 [C.U. B.Com. 1998]

34. Draw a Pie Chart of the following data: Type of Commodity

Family Expenditure (Rs.)

Food Clothes House Rent Education Savings Misc.

3,000 1,250 2,000 1,100 750 900 [C.U. B.Com. 2008]

(Hints: Total family expenditure = Rs. 9,000. We express ea.ch type of family expenditure as percentage of total family expenditure as shown below:

Type of Commodity

Percentage Expenditure

Food

~ggg x 100 = ~ 3

l~O

~~~g x 100 = ~ 3

1

House Rent

~ggg x 100 = 2 ~0 3

2~0 X

Education I

1

1

;gg

Misc.

:g0~ x 100

Now draw the Pie Chart of the given data.]

5

0

~~gg x 100 = ~ 3

Savings

Total

0

Central Angles

Clothes

1

0

x 100 =

¥3

= 103 1003

X

= 120° = 50° 3.6o = 800 3.6o = 440

3.6°

~ 5 x 3.6°

l~O X

¥ X 3.6° = 10 x 3.6°

30°

= 36° 360°

Chapter 11

Measures of Central Tendency: Mean, Median and Mode

11.1

Introduction

We may condense statistical data to a large extent by classification and tabulation. For in..-_ terpretation of statistical data mere tabulation is not enough, especially when comparison of two or more series of data is needed. In the present chapter our object will be to have some ,_,,,_ mathematical measures which will characterize the whole series of tabulated data. --~

11.2

Central Tendency of Data

In many frequency distributions, the tabulated values show small frequencies at the beginning and at the end, and very high frequency at the middle of the distribution. ['his indicates that the typical values of the variable lie near the central part of the distribution and other values cluster around these central values. This behaviour of the data about the 'concentration of the values in the central part of the distribution is called central tendency of the data. We shall measure this central tendency with the help of mathematical quantities. A central value which 'enables us to comprehend in a single effort the significance of the whole' is known as Statistical Average or simply Average. In fact, an average of a statistical series is the value of the variable which is representative of the entire distriqution and, therefore, gives a measure of central tendency.

Measures of Central Tendency There are three common measures of central tendency: (1) Mean, (2) Median, (3) Mode. 262

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

263

The most common and useful measure of central tendency is, however, the Mean. In the following articles the method of calculation of various measures of central tendency will be discussed. In all such discussions we need a very useful notation known as Summation Notation which we explain before proceeding further. 1

11.3

Summation Notation (I;)

The symbol E (read: Sigma) means summation. If xi, x2, X3, · · · , Xn be then values of a variable x, then their sum =xi + x2 + · · · + Xn is shortly written as n

LXi or Exi or simply Ex,(sometimes, Ex1). i=l

Similarly, the sum W1X1 + W2X2 + · · · + WnXn is denoted by n

L WiXi or EwiXi or simply Ewx (or sometimes, Ew1x1). i=l

Some Important Results n n n L(Xi ±yi) = LXi ± LYi· i=l

i=l

i=l

n

LA

= A + A + · · · + A = nA (A

i=l

is a constant).

n terms

n

- L

n

Axi

= Ax1 + Ax2 + · · · + Axn = A(xi + X2 + · · · + Xn) = AL Xi·

~1

11.4

~l

Mean: Arithmetic Mean (x)

There are three types of mean: (1) Arithmetic Mean (A.M.), (2) Geometric Mean (G.M.), and (3) Harmonic Mean (H.M). Of the three means, Arithmetic Mean is most commonly used. In fact, if no specific mention be made, by Mean we shall always refer to Arithmetic Mean (A.M.) and calculate accordingly.

Simple Arithmetic Mean Definition 1. The Arithmetic Mean (x) of a given series of n values, say, xi, x2, · · ·, Xn is defined as the sum of these values divided by their total number; thus _ , x1

x

=

+ X2 + X3 + · · · + Xn n

.

Ex

= n'

1

i.e.,

Ex n

3:=-.

264

BUSINESS MATHEMATICS AND STATISTICS

:Ex

Thus, Arithmetic Mean (or simply Mean)=-.

n

Example 1. Find the Arithmetic Mean of 3, 6, 24 and 48. Solution.

. 3 + 6 + 24 + 48 81 The reqmred A.M. (x) = = 4 = 20.25. 4

Weighted Arithmetic Mean X1, x2, · · · , Xn be n values of a variable x and if Ji, h, · · · , f n be their respective weights or their respective frequencies, then the weighted arithmetic mean x is defined by _ fix1 + hx2 + · · · + fnXn Ef X Ef X

Definition 2. If

x-

-

where N

=

Ji + h + .. · + f n

------

N '

L.f -

Ef = total frequency.

Thus, Weighted Arithmetic Mean or simply Mean (x)

= :E~x,

where N

= :Ef.

We see that in this case along with each value of the variable we have given due weight or importance to the number of times it occurs. If the weights Ji, f2, · · · , fn are all equal, then the weighted arithmetic mean becomes a simple arithmetic mean. In this case equal importance is given to all the items. Example 2. Find the Arithmetic Mean from the following frequency distribution: Weight in kg

50

55

60

65

70

Total

No. of men

15

20

25

30

10

100

Solution. Here we shall give due importance to the frequency of each value of the variable. Weighted Arithmetic Mean =

¥/-, where N = Ef =

total frequency.

I TABLE 11.1: CALCULATION OF WEIGHTED A.M. I Weight in kg x

No. of men

50 55 60 65 70 Total

:. Weighted Arithmetic Mean (x) =

f

15 20 25 30 10 100= N

¥/- =

°: = 60 kg.

6 1

fx 750 1100 1500 1950 700 6000 = 'Efx

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIANc-4ND MODE

11.4.1

265

Important Properties of A.M.

1. The sum or total of the values is equal to the product of the number of values and their A.M. Proof. (i) Simple A.M.

If x1, x2, · · · , Xn be n values of a variable x, then mean xis Ex x = - , or, nx =Ex, i.e., Ex= nx.

(1)

n

Proof. (ii) Weighted A.M.

If X1, x2, · · · , Xn be n values of a variable and if fl, then the weighted A.M. x is given by

f2, · · · , f n be their respective weights,·

2. The algebraic sum of the deviations of the values from their A.M. is zero. Proof.

Case I. Simple A. M. If x1, X2, · · · , Xn are then values of a variable x and x, their A.M., then x 1 - x, x 2 - x, · · · , Xn - x are called the deviations of x 1, x2, · · · , Xn respectively from x. n

Algebraic sum of the deviations

=

L (xi -

x)

i=l

+ (x2 - x) + ... + (xn - x) = (x1 + X2 + · · · + Xn) - (x + x + · · · to n terms) = Ex - nx = nx - nx [·: Ex = nx by (1)] = o. (x1 - x)

Case II. Weighted A.M. n L f i (xi -x) i=l

ft

(x1 - x)

+h

(x2 - x)

+ ... + f n (xn

- x)

=

fix1 - fix+ hx2 - hx + · · · + f nXn - f nX (fix1 + hx2 + · · · + f nXn) - (fix+ hx + .. · + f nX) Ef x - (/1 + h + · · · + f n)x

=

NX-NX

=

0.

= =

_ Efx ] [·: X=Nandf1+!2+···+fn=N

Example 3. Verify the truth of the previous statement (No. 2 above) taking the values of a variable as 3, 6, 12, 16, 18.

. 266

BUSINESS MATHEMATICS AND STATISTICS

Solution. A.M. (x) = 3 + 6 + 12 + 16 + 18 = 55 = ll. 5 5 The algebraic sum of the deviations of 3, 6, 12, -16, 18 from their A.M. is (3 - 11) + (6 - 11) + (12 - 11) + (16 - 11) + (18 -11)

-8-5+1+5 + 7 0.

=

Example 4. Verify the truth of the statement (No. 2} taking the values of a variable as 2, 5, 9, 11 with weights 8, 7, 3, 2 respectively. . . _ 8x2+7x5+3x9+2xll 100 · Solution. Weighted A.M. (x) = + + + = 2o = 5. 8 7 3 2 :. algebraic sum: of the deviations from x = = =

Ef (x - x)

8(2 - 5) + 7(5 - 5) + 3(9 - 5) +2(11- 5) 8x-3+7x0+3x4+2x6 -24 + 0 + 12 + 12 = o.

Note: If the deviations are taken from any arbitrary number A (not x), then the sum of the deviations will not be zero, as can be easily verified. -

Example 5. If the algebraic sum of the deviations of 12 observations measured from 40 be 60, find the A.M. [V.U. B.Com. 1995) .Solution. By the question, 12

12

~)xi

- 40) =

12

L:xi- L40=60

60 or,

i=l

i=l

i=l

Exi - 40 x 12 = 60 Exi = 60 + 480 = 540.

or, or, Hence

= 540 = 45 A .M. = Exi 12 12 . 3. The sum of the squares of the deviations of the values is the minimum, when deviations [B.U. B.Com.(H) 1990] are taken from the arithmetic mean. Proof. Let xi, x2, • · · , Xn be n values of the variable x and let that A is any arbitrary number.

x be their A.M.

Suppose

n

The deviations from A are x1 -A, x2 -A, · · · , Xn -A. We have to show that is minimum when A= Now

L i=l

x. '.

xi -

.

A = (z1 - ~) + (x - A). \

(Xi -

A)

2

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

n

:. L (Xi - A)

267

n

2

=

i=l

L {(xi - x) + (x -

A)}

2

i=l

= =

=

E {(xi - x) E(xi - x)

2

E(xi .:_ x)

2

2

+ (x -

+ E(x + n(x -

2

+ 2 (x, - x) (x _: A)} A) + 2E {(xi - x) (x - A)} A) 2 + 2 (x - A) E (xi _::. x) A)

2

n

=

L (xi - x)

2

+ n(x- A) 2

E (xi - x) = O]

[·:

i=l

We see that the right-hand side is the sum of two quantities of which the first term is positive and does not depend on the value of A we choose, and the second term~ 0. n

n 2

:. L(Xi-Af~L(xi-x) • i=l

i=l

Equality occurs when the second term is zero, i.e., when A= x. ,---·

n

:. L

2

(:z:i - A) is minimum only when A=

:z:.

i=l

Similarly, we can prove the result for a frequency distribution.

11.4.2

Short-~ut Method of ·Calculating A.M. for Discrete Series: Method of Assumed Mean

Simple A.M. Let x1, x2, · · · , Xn be then values of a variable x and di, d2, · · · , dn be the deviations ohhe. n values from any arbitr~ry value A (called assumed mean). Then di= xi -A or, Xi= A+ di.

x = Exi = E(A +di) ,,,; nA +Edi = nA n n n n Thus, :z:- =

+ Edi

= A+ Edi.

n

n

A+~. n , i.e.,

. A d A .M . = ssume mean

+

Sum of the deviations from A . . No. of values of the variable

Weighted A.,M. x

= =

Thus, x = A

+ E{V"'.

E/ixi = E/i(A +di) = EAfi + Efidi N N N Efi Efidi A Efidi [ A N +"fr = + "fr ·: Efi = N J.

268

BUSINESS MATHEMATICS AND STATISTICS

Example 6. The monthly incomes of 5 persons are Rs. 250, Rs. 360, Rs. 280 1 Rs. 480 and Rs. 410. Find the Arithmetic Mean .of the incomes of 5 persons. Solution. We take the assumed mean A== 360. Here n = 5, 'Ed= -20. 'Ed

:. x = A + -;- =

-20 360 + - 5

= 360 -

4 = Rs. 356.

TABLE 11.21 CALCULATION OF DEVIATIONS Income (Rs.) :i: 250 360 280 480 410

Deviation from 360(d = :i: - 360) -110 0 -80 120 50

Total

-20 =Ed

Note: The students can try with another assumed mean and arrive at the same result.

Example 7. Calculate the Arithmetic Mean of the following data: Value (x) Frequency (!)

1 7

2 11

3 16

4 17

5 26

6 31

7 11

8 1

9 1

Total 121

Solution. TABLE 11.3: CALCULATION OF A.M.

Va~ue 1 2 3 4 5 6 7 8

I

Frequency

Deviations froni

fd

I

A(=5)[d=:i:-5] -4 -3 -2 -1

-28 -33 -32 -17

9

7 11 16 17 26 31 11 1 1

Total

121

0 1 2 3 4

=N

-

0 31 22 3 4 -110 + 60

= -50 = Efd

Let us take the assumed mean A= 5. Here N = 121, "Efd = -50 . . . Arithmetic Mean= A+ 'Eid = 5 +

~~~· =

5 - 0.41 = 4.59.

269

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

11.4.3

Calculation of Arithmetic Mean (or simply, Mean) from a Grouped Frequency Distribution-Continuous Series

In a grouped frequency distribution, the individual values of the variable and their frequencies are not known. Here, the mid-value of a class-interval is taken to represent all the values of the variable falling within that class with a frequency equal to that of the class-interval. From the_ given data the mid-values of all the class-intervals are found by taking the average of the two ·· class-limits (or class-boundaries). Then Arithmetic Mean may be calculated by any one of the ·methods discussed below.

Ordinary Method (or Direct Method) This method is already discussed before. In this method, the mid-values of the class-~ervals (or values) are multiplied by the corresponding class-frequencies. The sum of the products thus obtained is divided by the total frequency to get the Mean. The mean x is given by the formula

x=

L.£x, where x =mid-value of a class and N =total frequency.

Example 8. Calculate the mean of daily wages from the following table: Wages (Rs.) No. of workers

4-6

6-8

8-10

10-12

12-14

6

12

17

10

5

Solution. TABLE 11.4: CALCULATION OF MEAN OF DAILY WAGES Class-intervals

Mid-values (Rs.) x

4-6 6-8 8-10 10-12 12-14

5 7 9 11 13

Total

-

Frequency

f

fx

6 12

30 84 153 110 65

17 10

5 50=N

44f?

= Efx

'

. L.fx 442 :. Mean of d aily wages= - - = =Rs. 8.84. N 50

Short-cut Method (or Method of Assumed Mean) This is similar to the method described earlier. In this method, the mid-value of one classinterval (preferably corresponding to the maximum frequency lying near the middle of the distribution) is taken as the assumed mean (or the arbitrary origin) A and the 'cleviations from A are calculated. The mean is then given by the formula: '°L.fd

x =A+ N' where d = x -A= (Mid-value)-(Assumed mean).

270

BUSINESS MATHEMATICS AND STATISTICS

Example 9. Compute the Arithmetic Mean of the following frequency distribution:

I Marks I No. of students

20--29

30--39

5

11

I 40--49 I 18

50--59

6(}--69

22

16

I 10--19

I

8

Solution. Efd

A+ N

. . Arithmetic Mean =

= 54.5 +

-230

80 =

54.5 - 2.875

51.625 = 51.6 (approx.).

=

TABLE 11.5: CALCULATION OF A.M. OF MARKS Class-intervals

Mid-values

20--29

24.5

..,. 30

30--39

34.5

- 20

40--49

44.5

-10

18

50--59 6(}--69

54.5 =A

0

22

64.5

10

70--79

74.5

20

5 11

/

-

-

\otal

/d.

Deviations from 54.5 Frequency ,, ' (d. = :r; - 54.5) I

:r;

-150 - 220 - 180

16

0 160

8

160

80=N

\

-550 + 320 = -230 = Efd

Method of Assumed Mean Using Step Deviation In the preceding method, the deviations of the mid-values of the class-intervals from the assumed mean were used. The deviations are often divided by a common factor i, usually the common width of the class-intervals if the classes are of equal width or by the H.C.F. of the deviations, if the cl&ases are not of equal width. The arithmetic mean in this case is given by Mean (x) -where d

= xiA

Examp~e

'Ef d

= A+ N

x

. i,

and i = the common width of the classes or H. C.F. of the mid-values.

10. Find the Arithmetic Mean of the weekly income from the following frequency

distribution:

I Weekly income in Rs. j

No. of workers

10--15 200

I 15-20

I

20--25

700

900

I 25-30 I 800

30--35 600

Solution. A.M. (x)

Efd = A+ N

x

. i,

x-A

where d = - i - .

I 35-40 I 400

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

271

TABLE 11.6: CALCULATION .OF A.M. OF WEEKLY INCOME , d . a-A Class-intervals Mid-values Frequency fd • :r: (A= 22.5,i = 5) I 10-15 12.5 -2 -400 '200 1~20 17.5 -1 700 -700 20-25 22.5 =A 0 900 0 2~30 27.5 1 800 800 30-35 32.5 2 600 1200 ' 37.5 3~0 3 400 1200 Total -1100 + 3200 3600 = N

=--.--

= 2100 = E/d .·.

11.4.4

A.~.

.

(x)

2100

.

= 2~.5 + 3600 x 5 = 22.5 + 2.92 = Rs. 25.42.

Calculation of A.M. from Grouped Frequency Distribution with Open' Ends

If in a grouped frequency distribution, the lower limit of the first class and the upper limit gf the last class are not known, it is difficult to find the A.M. When the closed classes (other than the first and last classes) are of equal widths, we take the widths of the open classes equal to the common width of closed classes and determine the A.M. by any of the preceding metho~. For example, consider the following data: Marks Below 10 10-20 20-30 3o-40 Above 40

No. of students

6I 9 15

12 8

Here, since the closed classes are of equal width (i.e., 10), we take the lower limit of the first class as 'O' and the upper limit of the last class as 50. The first and the last class-intervals would be 0-10 and 40-50 respectively. Let us con.sider another example: Marks Below 5 ~15 1~30

30-50 Above 50

No. of students 3 6 11

8 2

Here, since the widths of 2nd, 3rd, 4th classes are respectively 10, 15, 20 which increase by 5, we take the lower limit of the 1st class as 'O' and the upper limit of the last class as '75'.·

272

BUSINESS MATHEMATICS AND STATISTICS

The first and the last classes would be 0-5 and 50-75 respectively. The A.M. can be obtained by any of the previous methods.

11.4.5

Mean of Compositive Group

If a group of ni values has A.M. xi, and another group of ni values has A.M. x2, then the A.M. (x) of the composite group (i.e., the two groups combined) of ni + n2 values is given by

nix1 + n2x2 n1 +n2

_

x=----Proof. Let x1 be the A.M. of n1 values x1, x2, · · · , • · · , Yn 2 • Then X1

or,

"'---

_ X1 + X2 + · · · + . x-

.·

Ex·

= -' ni

Xn 1

and x2

= Exi and

nix1

and x2 be the A.M. of n2 values yi, Y2, Ey·3

=-

n2 n2x2 = EY1·

+ Yl + Y2 + · · · + Yn 2 - Exi + Eyj - n1x1 + n2X2 ---ni + n2 ni + n2 n1 + n2 ·

Xn 1

-

For r groups the A.M. (x) is given by _

nix+ n2.X2

+ · · · + nrXr

x = --------n1 + n2 + · · · + nr Example 11. The means of two samples of sizes 50 and 100 respectively are 54.1 and 50.3. Obtain the mean of the sample of size 150 obtained by combining the two samples. Solution. Here ni

= 50, n2 = 100, x1 = 54.1, x2 = 50.3. =

= Example 12. The A.M. of 7, x - 2, 10, x

50 x 54.1+100 x 50.3 2705 + 5030 = 150 50+100 7735 = 51.57 (approx.). 150

+ 3 is 9;

find the value of x. [C.U. B.Com. 1998, 2000, '0'7]

Solution. A.M. of 7, x - 2, 10, x

+ 3 is

7 + (x - 2) + 10 + (x + 3) 18 + 2x 9+ x 4 = 4 = -2-· ..

9

; x

= 9 [·:

A.M. = 9] or, 9 + x

= 18 or, x =

18 - 9 = 9.

273

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

11.5

Mean: Geometric Mean (G.M.)

Besides A.M. the geometric mean of a set of values of a variable can also be taken as one of the measures of central tendency of the values. --Definition 1. The geometric mean G of n positive values, say x1, . the positive nth root of their product; thus G

=

\lx1 X X2 X • • • X Xn

= (x1

x2, · · ·, Xn,

is defined by

X X2 X • • • X Xn)l/n.

In order to calculate G we often use logG

1

1

= -(logx1 + logx2 + · · · + logxn) = -E logx. n n

(Now obtain G by taking antilog of the right-hand side.) Illustration 1. If x is the G.M. of 28 and 7, then x

= ±v'28 x 7 = ±2 x 7 = ±14 . .\\

[C.U.B.Com. 1997]

Example 13. Find the G.M. of the following: l, 3, 9, 3.

[C.U. B.Com. 2001)

14 Solution. G.M. of 1, 3, 9, 3 = (1 x 3 x 9 x 3) 1/ 4 = (3 4) / = 3. Exa!"ple 14. Find the G.M. of 4, 6, 9 with weights l, 2, 1 respectively.

[C.U. B.Com. 1996]

Solution. Weighted G.M. = ( 41 x 62 x 91) l/(1+ 2+1> = (22 x 22 x 32 x 32) 1/ 4 = (~'1 x 34) 114 = 2 x 3 = 6.

Generalized Definition If Xi, x2, · · · is given by

, Xn

G

occur with frequencies (or weights) J

J

~

= ( X1 1 x X2 2 x ... x x~n

)

1/N

'

Ji, /2, · · · , fn respectively, then the G.M.

where N

=Ji+ h + ... +In·

Here

Hence,

G = antilog {~

L Flogx}.

Example 15. (i) Calculate the Geometric Mean of 3, 6, 24 and 48. (ii) If the G.M. of x, 9, 12 be 6, find the value of x.

[C.U. B.Com. 1991, '97] [V.U. B.Com. 1997)

Solution. (i) G.M. = (3 x 6 x 24 x 48) 1/ 4 = (3 x 3 x 2 x 3 x 23 x 3 x 24) 1/ 4 = (3 4 x 28 ) 114 = 3 x 22 = 12.

(ii) Bjr definition, (x x 9 x 12) 1/ 3 = 6; or, l08x = 63 = 216 or, x DuB

l.1111'!1.t...

•- ~t rl"' I I 1

4 0

= 2.

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BUSINESS MATHEMATICS AND STATISTICS I

I

Example 16. Find the G.M. of 12, 18, 48, 61 whose weights are 5, 3, 2, 8 respectively.

Solution.

I TABLE ii. 7: CALCULATIONS FOR G.M. log a:

a: 12 18 48 61 Total

logG

=

1

NEflogx =

f logx

f

1.0792

5

5.3960

1.2553 1.6812

3 2

3.7659 3.3624

1.7853

8

14.2824

-

1

18

18

=N

26.8067

=

~!log

x

x 26.8067 = 1.4893; :. G = antilog (1.4893)

= 30.85.

Uses of G.M. The G.M. is not as widely used as the A.M. But as a measure of central tendency it is sometimes more significant than the A.M. It is useful in averaging rates of changes when ratio changes are more important than the absolute changes. It is used in the construction of index numbers.

11.6

Mean: Harmonic Mean (H.M.)

Definition 1. The Harmonic Mean H of then values x1, x2, · · · ,

Xn

is defined by

1

H

i.e., H.M. is the number of values divided by the sum of the reciprocals of the values,,y,nd it cannot be defined if some of the values is zero.

Generalized Definition If x1, by

X2, · · · , Xn

have the frequencies Ji,

f2, · · · , fn

respectively, then the H.M. (H) is given

Uses of H.M. H.M. is of very limited use. It is useful in finding averages involving rate, time, price and ratio.

Example 17. Find the harmonic mean of the following numbers: l, ~'

!, ~· [V.U. B.Com. 1995]

CHAP. ll! MEASURES OF CEJTR;AL TENDENCY: MEAN, MEDIAN AND MODE I

Solution. H.M.

=

4 1

275

I

1

1

1

1+172+173+174

4

4

2

1+2+3+4

10

5

= -----

Example 18. An aeroplane flies around a square the sides of which measure 100 km each. The aeroplane covers at a speed of 100 km per hour the first side, at 200 km per hour the second side, at 300 km per hour the third side and at 400 km per hour the fourth side. Use the correct mean to find the average speed round the square.

Solution. Here H.M. is the appropriate mean. Let the required average speed H km per hcur. Then H

=

4 1,

IM +

_ 1_ 200

+

_ 1_ 300

+

_ 1_ 400

=

4

12 +6 +4+ 3 1200

=

4 x 1200 = 4 x 48 = 192 km/hour. 25

_Example 19. Find the H.M. of 3, 6, 12, 24 whose weights are 6, 2, 4, 8 respectively. Solution.

11. 7

Advantages and Disadvantages of A.M., G.M. and H.M.

Arithmetic Mean Advantages • The A.M. is the most familiar and widely used measure of central tendency. It is simple to understand and easy to calculate. • It is rigidly defined. • Its calculation depends upon all the values in the series. • It is suitable for algebraic treatment. • It is least affected by sampling fluctuations.

Disadvantages • It cannot be determined by inspection. • It is very much affected by the presence of a few extremely large or small values of the variable. • Mean cannot be calculated if a single item is missing. • Calculation of A.M. from a grouped frequency distribution with open-end Classes is impossible, unless some assumptions are made regarding the sizes of these classes.

276

BUSINESS MATHEMATICS :\ND STATISTICS

Geometric Mean Advantages • G.M. is not as widely used as the A.M. It is particularly suitable for averaging rates of changes. • It is rigidly defined and it depends upon all the values in the series. '• It is suitable for algebraic treatment.

• G.M. is not much affected by the presence of extremely large or small values of the variable.

Disadvantages • Unlike A.M., G.M. is neither simple to understand nor easy to calculate.

• If any value in the series is zero, G.M. cannot be calculated. • Calculation of G.M. is impossible unless all the values are positive.

Harmonic Mean Advantages • It is useful in averaging rates, ratios and prices. • It is suitable for algebraic treatment. • Its calculation is based on all the values in the series.

Disadvantages • It is of very limited use and it is not easy to understand. • H.M. cannot be calculated if any one of the values is zero.

Relation between A.M., G.M. and H.M. For any set of positive .values of a variable, we can write

A.M. :;::: G.M. :;::: H.M. (equality occurrfog only when the values are all equal). It can be proved that

A.M. G.M.

G.M.

= H.M.,

( )2 or, A.M.xH.M.= G.M.

277

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

We prove these results for two positive values, say x1 and x2 of a variable x. We have x1 A.M. =

+ x2 2

' G.M. =~and H.M. = (1/x1)

2

2x1x2 x1 + x2

+ (1/x2)

Since the square of any real quantity ? 0, , (Fi. -VXi,) 2 ? 0, or, x1 +x2 -2~? 0, or, x1 +x2? 2~ , x1+x2 ~ . or, ? yX1 • x2, i.e., A.M. ? G.M. 2

(1)

[C.U. B.Com. 2007]

Equality occurs only when x1 = x2. Now, A.M. x H.M. = i.e.,

A.M. x H.M.

x1

=

+ x2 x

2x1x2 = 2 X1 + X2 (G.M.) 2 .

X1. X2

= (~)

2

2 = (G.M.) '

(2) (3)

From (2), A.M. x H.M. = G.M. x G.M. Since by (1), A.M. ? G.M., :. H.M.

~

G.M., i.e., G.M. ? H.M.

(4)

Combining (l) and (3) we get, A.M. ? G.M. ? H.M., equality occurring only when X1

=

X2.

Illustration 1. (i) Take three numbers 2, 4, 8: A.M. = \ .

-

\j.M. - (2.4. 8 )

14 ; 3

. 3 - 24 - 4 ' H.M. - 1/2 + 1/4 + 1/8 - 7·

1/3 -

Now see that A.M. x H.M. = 134 x 274 = 16 = (G.M.) 2 • (ii) If A.M. between two numbers is 6.5 and their G.M. is 6, then 2 2 36 A.M. x H.M. = ( G.M.) , or, 6.5 x H.M. = 6 = 36 or, H.M. = - = 5.54. 6.5

11.8

Median

Median is another measure of central tendency. If a series of values of a variable is arranged in ascending or descending order of magnitudes, then the value of the middle term or the mean of the two middle terms according as the number of values is odd or even is called the median. Median divides the series into two equal- parts. It is a positional average and it is unaffected by the presence of an extremely large or 'small value. It can be calculated from a grouped frequency distribution with open-end classes.

Example 20. (i) Find the median of 33, 86, 68, 32, 80, 48, 70, 64. (ii) Find the median of 88, 72, 33, 29, 70, 86, 54, 91, 61, 57.

[C.U. B.Com. 1997]

278

BUSINESS MATHEMATICS AND STATISTICS

Solution. (i) Arranging the given numbers in ascending order of magnitudes, we get 32_,..33, 48, 64, 68, 70, 80, 86. There are 8 numbers and, therefore, there are two middle terms which are the 4th and 5th terms. Thus the two middle terms, are 64 and 68. Hence, Median·± 64 68 = 1 ~ 2 = 66. (ii) Arranging the given numbers in ascending order of magnitudes, we get 29, 33, 54, 57, 61, 70, 72, 86, 88, 91. There are 10 numbers and, therefore, there are two middle terms which are 5th and 6th terms. :. middle terms are 61 and 70. Hence, Median = 61 0 = 1 ~ 1 = 65.5.

!

f

11.8.1

Calculation of Median

Simple Series The given values are arranged in order of magnitudes. If the number of values is odd, there is one middle term which is the median. If the number of values is even, then there are two middle terms and the A.M. of these two middle terms is the median. The formula is Median = value of the ("i 1 )th term, where n = total number of terms in the series.

Simple Frequency Distribution In this case, we first calculate cumulative frequency (less than type) corresponding to each value of the variable. Then the value of the variable corresponding to the cumulative frequency is the median, where N = total frequency.

Nil

Grouped Frequency Distribution (i) Median by Formula: The cumulative frequency (less than type) corresponding to each class-boundary is first calculated. And then Median is that value of the variable which corresponding to the cumulative frequency J¥- and the class in which the cumulative frequency J¥lies is called the median-class. Median is given by the formula: Median= li where Li N F

+ (N/2) f rn

F

x c,

lower boundary of the median-class, total frequency, cumulative frequency below L1 (or sum of the frequencies of all classes lower than the median-class), frequency of the median-class, fm width of the median class. c (ii) Median by Simple Interpolation: We first define simple interpolation. Simple interpolation is the method of computing an intermediate value of the function y for a specified value of x on the basis of given values of x ·and y, assuming that y change uniformly with x (i.e.,

279

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

assuming a linear relationship between x and given values of x and y: x 12 15----+ 20

y). For example, consider the following pairs of y 60

92

Suppose, we have to compute the value of y, say y1 , when x uniformly with those of x, then we can write,

= 15. If the values of y changes

15 - 12 YI - 60 . . [·: y changes umformly with x] 20 -12 92 - 60 - 60 3 83 = YI 32 , or, 8 x 32 =YI - 60, or, 12 = y 1 - 60 YI = 12 + 60 = 72. --- =

or, or,

Median can be determined by applying simple interpolation in a cumulative frequency distribution. If the cumulative frequency J¥- lies between the cumulative frequencies Fi and F2, then F1 and F2 correspond to the lower class-boundary li ano upper class-boundary l2 of the median class. In this case, Median (M) is given by

M-li (iii) Median by Graphical Method: First draw the ogive (less than or more than type) or cumulative frequency polygon taking the variable (or class-boundaries) on the horizontal axis (i.e., x-axis) and the cumulative frequency on the yertica~ axis (i.e'., y-axis). Locate J¥on the y-axis and from It draw a horizontal line which meets the ogive, say at P. From P draw a perpendicular on the x-axis and then read the point where it meets-the x-axis from the horizontal scale. This gives the median. [See Example 24 given below.] Example 21. Find the median from the following data: Wages (Rs.) No. of persons

2.30

2.40

3.20

3.40

3.50

4.50

5.10

5.50

6

4

3

4'

4

4

4

1

Solution. TABLE 11.8: CALCULATION OF CUMULATIVE FREQUENCY

-

:z:

f

Cumulative Frequency

2.30

6

2.40

6 4

3.20

3

3.40 3.50 4.50

4

17+-

4 4

21

5.10 5.50

4 1

10 13

25 29 30

=N

280

BUSINESS MATHEMATICS AND STATISTICS

Here NtI = 3Q{ 1 = 15.5 and Median = the value corresponding to cumulative frequency 15.5, i.e., when the values are arranged in order of magnitudes, median is the A.M. of the 15th and the 16th values. From the last column we see that 15.5 lies between 13 and 17, and 14th to 17th values are each 3.40. 3.40 + 3.40 . 15th value + 16th value Hence, Median = = = 3 .4 0 · 2 2 Example 22. Calculate the median from the following:

Class-intervals Frequency

2-4

4-6

6-8

8-10

3

4

2

1

Solution. First method (Median by the application of formula):

I TABLE 11.9:

CALCULATION OF CUMULATIVE FREQUENCY

Class-boundary 2-4 4-6 6-8 8-10

Frequency

Cumulative Frequency

3

3

4 2 1

9

7

I

+--

lO=N

f

Median= the value corresponding to cumulative frequency (= 5), and 5 is greater than cumulative frequency 3, but less than the next cumulative frequency 7. Therefore, medial-class is 4-6 and median is given by Median = li

+

(N/2) - F

fm

x c.

f

Here li = 4, = 5, F = 3, Im = 4, c = 2. Hence, Median = 4 + 5 43 x 2 = 4 + 1 = 5. Second method {Median by the application of simple interpolation): Median= the value corresponding to the cumulative frequency

f.

TABLE 11.10: CALCULATION OF CUMULATIVE FREQUENCY Class-boundary

Cumulative Frequency (less than)

2 4

0

6

7

3

Median (M)----+

+--

8 10

Since 4 and 6.

f

~=5

9

lO=N

= 5 lies between the cumulative frequencies 3 and 7, Median (M) must lie between

281

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

Now by simple interpolation, M-4 _ 6 4

=

5-3 M-4 _ , or, - 2 7 3

2

1

1

= 4 = 2, or, M = 4 + '2 x 2 = 4 + 1 = 5.

Hence median= 5. Example 23. Prom the following distribution determine the median:

I Marks

I 20-29

I No. of students I

7

40-49 34

30-39 16

50-59 55

60-69 28

70-79 20

Total

160 [C.U. B.Com. 1996)

Solution. Median = the value corresponding to cumulative frequency (less than type) ~· TABLE 11.11: CALCULATIONS FOR MEDIAN Marks (Class-intervals) ·'

'''

20-29 30-39 4Q-49 50-59 60-69 70-79

Frequency

Cumulative Frequency

f

(less than type)

7 16 34 55 (= /m) 28 20 160 = N

7 23 57 (= F) 112+140 160= N

N = 160 = 80 2 2 .

Median = l 1 + (Nl'j-F x c. From the last column, we see that the cumulative frequency just greater than 80( = ~) is 112 and the class-interval corresponding to cumulative frequency 112 is 50-59. Hence, the median class is 50-59. Here l1 = lower 'class-boundary of the median class = 50 - 0.5 = 49.5, F = 57, f m = 55, c = 10. Hence, Median = 49.5 + 805557 x 10 = 49.5 + ~35° = 49.5 + 4.18 = 53.68. Example 24. The table below gives the frequency distribution of weights of 80 apples at random from a big consignment: Weight in gm Frequency

110-119 5

120-129 7

130-139 12

140-149 20

150-159 16

160-169 10

170-179

180-189

7

3

Draw the cumulative frequency diagram and hence determine the median weight of an apple. [C.U. B.Com. 2002 Type]

Solution. Here class-limits are given. For the cumulative frequency distribution, first we have to find the class-boundaries corresponding to the class-limits. The class-boundaries are 109.5119.5, 119.5-129.5, etc.

282

BUSINESS MATHEMATICS AND STATISTICS

I TABLE 11.12:

CALCULATION OF CUMULATIVE FREQUENCY

Class-boundary (Weight in gm)

I

Cumulative Frequency Less than More than

109.5 119.5 129.5 139.5 149.5 159.5 169.5

80 75 68 56 36 20

0 5 12' 24 44 60

10

70 77

179.5 189.5

3 0

80

The cumulative frequencies (both less than and more than types) are plotted on a graphpaper (see Figll.1) against the class-boundaries taking the former along the vertical axis OY and the latter along the horizontal axis OX. The two cumulative frequency diagrams (or Ogives) are drawn. They meet at P. Locate 1¥- = 40 on the Y-axis and from it draw a horizontal line which meets the ogive at P. From P draw a perpendicular on OX. If M be the foot of this perpendicular, then Median =OM= 139.5 + 8 = 147.5 gm.

.

y

80

[']

I

More than type ogive

LI

Less than type ogive

60

,... ,...

._

p

20

...

,,:

'"""' 0

. It)

...oi 0

ltl

....oi...

.

It)

oi

...

C\I

It)

~

...

t~

c:..,. as ...

'13.:l x

It)

~

~

:::!: Weight (in,gm)Fig 11.1 Cumulative Frequency Diagram.

CHAP. 11: MEASURES OF CENTRAL TEN.DENCY: MEAN, MEDIAN AND MODE

Note: Quartiles ca.n also be determined from the cumulative frequency polygon \or ogive) by using the following definitions: the value corresponding to cumulative frequency and

From the ogive, Qi

11.8.2

. . fr equency the value correspond mg to cumu1at1ve

Q3

N N 4 ; 4 = 20 3N ; 3N = 60. 4 4

= 136.5, Q3 = 159.5. [For more details, see Section 11.9.1.]

Advantages and Disadvantages of Median

Advantages • It is simple to understand and easy to calculate. • It is rigidly defined. • It is unaffected by the extreme values.

Disadvantages • It is not based on all the values. • It is much affected by sampling fluctuations in comparison to mean. • It is not suitable for algebraic treatment. • Its calculation depends on the arrangement of the data in order of magnitudes.

11.9

Quartiles, Deciles and Percentiles

If a series of values of a variable is arranged in ascending order of magnitude, the midd.lemost value which divides the series into two equal parts is the median. By extending the idea, we can find three values Qi, Q2 and Q3 which would divide the series into four equal parts. These three values Qi, Q2 and Q3 are called the first (or lower) Quartile, second Quartile and third (or upper) Quartile respectively. Obviously, the second quartile Q 2 is the median. Similarly, the nine values which divide the series into ten equal parts are called Deciles which are denoted by Di, D2, · · · , Dg and the 99 values dividing the data into 100 equal parts are called the Percentiles denoted by Pi. P2, · · · , P99 respectively. Quartiles, Deciles and Percentiles are also called partition values.

11.9.1

Calculation of Quartiles, Deciles and Percentiles

The calculation of Quartiles, Deciles and Percentiles is similar to that of Median.

284

BUSINESS MATHEMATICS AND STATISTICS

For a simple series (or a simple frequency distribution) the data are first arranged in ascending order of magnitude. Then

N+l

Qi

=

Qa

=

D7

=

D9

=

the. value of the - -th item, 4 3(N + 1) . the value of the th item, 4 . 7(N + 1) . 7th decile = the value of the th item, 10 1 9 9th decile = the value of the (~; ) th item,

. Pss

-

58th percentile = the value of the

58

(

%;

1 ) th item·.

For a grouped frequency distribution, cumulative frequencies (less than type) are first calculated. Then

~,

Qi

=

the value corresponding to cumulative frequency

Qa

=

the value corresponding to cumulative frequency

D6

=

6th decile = the value corresponding to cumulative frequency

Pas

=

35th percentile = the value corresponding to cumulative frequency

+

li 4 - Fi

fi

Qi

=

li

l1

=

lower boundary of the Qi-class

where

3

~,

=

Ji =

~~~.

X Ci,

(i.e., the class in which cumul-ative frequency Fi

~~,

~

falls),

cumulative frequency below li, frequency of the Qi-class, ci =width of Qi-class.

Similarly, Qa =la+

aN 4 - Fa

fa

x ca,

D 6

=

l 6

+

~~ - F5

!

6

l . ait~ - Fas x C6, Pas= as+ fas x cas.

For a symmetrical distribution, Q2 - Qi = Qa - Q2, or, Qi+ Qa = 2Q2 . ./

Illustration 1. If for a symmetrical distribution, Qi = 24 and Qa = 42, then 2Q2 = 24 + 42 = 66, or, Q 2 = 33.

Uses of Quartiles Quartiles are used for measuring central tendency, dispersion and skewness. Second Quartile Q2 is the median which is a measure of central tendency. Quartiles Qi and Qa are used to

285

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

define Quartile Deviation which is a measure of dispersion. Skewness based on· Quartiles Qi. Q2 and Q3 is given by Skewness= (Q3 - Q2) - (Q2 - Qi) = Q3 - 2Q2 +Qi. (Q3 - Q2) + (Q2 - Qi) Q3 - Qi Example 25. (i) Find the quartiles of the following numbers: 29, 12, 26, 19, 24, 36, 21, 33, 35. Find also the 3rd and 7th deciles.

(ii) Find the 23rd and 57th percentiles of the following values: l, 3, 5, 7, · · · , 197. Solution. (i) Arranging the given numbers in ascending order of magnitude, we get, 12, 19, 21, 24, 26, 29, 33, 35, 36. Here n = 9, n+l 9+1 n+l 9+1 3(n+l) 3(9+1) -4- = -4-, -2- = -2-, 4 = 4 :. 1st Quartile (Qi)= value o£the 2:.t1th, i.e., 2!th item =value of the 2nd item+ ! (3rd item - 2nd item) = 19 + x 2 = 20; 2nd Quartile (Q2) =value of the ~th, i.e., 5th item= 26; and 3rd Quartile (Q3) =value of the 3

Here ~ =

149.5

72

154.5

89

159.5

99

164.5

100

=

.q:

=

N

g = 50. Median (M) =the value corresponding to cumulative frequency 50.

1 0

:. Median class is 145...'.149 and Median= l

1+ lf1:F x c.

Here l1 = 144.5, ~ = 50, F = 48, frn = 24, c = 5. :. Median

144.5 +

=

50 - 48 24 x 5

144.5 + 0.42

5

= 144.5 + 12

= 144.92.

The empirical relation between Mean, Median and Mode is Mean - Mode= 3 (Mean - Median) or, 145.35 - Mode= 3 (145.35 - 144.92) or, 145.35 - 1.29 = Mode, i.e., Mode = 144.06.

=3 x

0.43

Example 10. The A.M. of the following distribution is 28.8. Find the missing frequency: Marks No. of students

0-10

10-20

20-30

30-40

40-50

50-60

4

6

20

?

7

3 [C.U. B.Com. 1996)

Solution. Let z be the missing frequency. Then we have TABLE 11.21: CALCULATIONS FOR ARITHMETIC MEAN

d=x-A = x - 35

fd

4

-30

6

-20

-120 -120

20

-10

-200

0

0 70

Class-

Mid-

Frequency

interval

value x

f

0-10 10-20

5 15

20-30

25

30-40

35 45

z 7 3

40-50 50-60 Total

55 -

Mean

10 20 -440 + 130

40 +z= N

Efd

= A + -N = 35 -

iO

310

=

-310

- - . Also Mean = 28.8. 40+z

= 'Efd

296

BUSINESS MATHEMATICS AND STATISTICS

28.8=35-

~

or,

40+z

or, or, or,

310 + z = 35 - 28.8 = 6.2 40 310 = 248 + 6.2z 6.2z = 62 62 z = . = 10. 62

Hence, the missing frequency is 10. Example 11. The Median and Mode of the following frequency distribution are known to be 27 and 26 respectively. Find the values of a and b: Values Frequency

0-10 3

10-20 a

20-30 20

30-40 12

40-50 b [C.U. B.Com. 1998, 2001]

Solution. Median = 27 and Mode = 26; Median and Modal classes are the same which is 20-30. :. L = 20, fm = 20, Ji= a and h = 12. Also i = 10. Now,

Mode::;:: L or, or,

+ ( 2mf fm

f Ji J ) xi. Also Mode= 26. i-

2

20 - a 26 = 20 + ( ) x 10, or, 6 = (20 - a)lO 2 x 20 - a - 12 28 - a 168 - 6a = 200 - lOa, or, 4a = 200 - 168 = 32, or, a= 8.

Again,

fY.._p

Median=

Li + 2 f m

x c.

Here l1 = 20, N = 3 +a+ 20 + 12 + b = 35 +a+ b, F = 3 +a, Median= 27.

Hence, a

= 20,

c=

10 and

~ - (3 + a)

x 10 = 20 + 35 + b - a - 6 20 4 7 x 4 = 29 + b - a, or, a - b = 1 8 - b = 1, or, b = 8 - 1 = 7. 27 = 20 +

or, or,

fm

= 8 and b =

7.

Example 12. In the following frequency distribution, two class-frequencies are missing: Intelligence Quotient No. of Students

55-{i4 2

65--74 19

75-84 78

85-94 ?

95-104 301

105-114 ?

115-124 92

125-134 14

135-144 4

Is it however known that the total frequency is 900 and the Median 100.048. Find the two missing frequencies. Solution. Let the two missing frequencies be Ji and f2.

297

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

Median = the value corresponding to cumulative frequency N /2. The class-boundaries of the given classes are 54.5-64.5, 64.5-74.5, · · · , 134.5-144.5. Here Median = 100.048 and N /2 = 900/2 = 450.

I TABLE 11.22: CALCULATION OF CUMULATIVE FREQUENCY I Class-boundary

Cumulative Frequency (less than)

54.5

0

64.5

2

74.5

21

84.5

99

94.5

99+ Ji

104.5

400 +Ji

114.5

400 +Ji+ h

124.5

h 506 +Ji+ h 510 +Ji + /2 = N = 900

100.048--+

+-

134.5 144.5

450

= £¥-

492 +Ji+

. By simple interpolation, we get 450 - (99 +Ji)

100.048 - 94.5

-----=-----..,..--104.5 - 94.5 (400 +Ji) - (99 +Ji) or,

5.548 _ 351 - Ji 5.548 x 301 _ 10 301 ' or' 10 - 35 1 -

or,

167 = 351 -

or,

Ji = 351 -

f

1

f I , [· . . 5.548 x 301 = 1669.948 = 166 .9948 = 167] 10

10

167 = 184.

Again, 510 +Ji+ h = 900 or, Ji+ h = 390 or, 184 + h = 390 or, Hence, the two required missing frequencies are 184 and 206. Otherwise: Median= li

+

!::!__p 2 fm x c,

or, or, or,

h

= 390 -184 = 206.

100.048 = 94.5 + 450 - (99 +Ii) x 10 301 5.548 = (35 l - Ji) x 10 301 5.54 x 301 = 351 - Ji, etc. 10

Example 13. Calculate the median and mode of the following: Annual Sales (Rs. '000)

Frequency

Annual Sales (Rs. '000)

Frequency

less than 10 less than 20 less than 30

4 20 35

less than 40 less than 50 less than 60

55 62 67

298

BUSINESS MATHEMATICS AND STATISTICS

Is it possible to calculate the arithmetic mean? If possible, calculate it. [Utkal U. B.Corn. 2000 Type)

Solution. Here cumulative frequency distribution (less than type) is given. Median = the"value corresponding to cumulative frequency N /2.

I TABLE 11.23:

CALCULATION OF CUMULATIVE FREQUENCY

Yalu~' (Annual Sales)

Cumulative Frequency (less than)

in Rs. '000

4 20

10

20 Median (M)-->

+--

30 40 50 60

33.5 = ~

35 55 62 67=N

N = 67; :. J¥- = 33.5. By Simple Interpolation, we get

M-20 30-20

33.5-20 35 - 20

or' or, or, or,

M - 20 13.5 10 = 15 13 5 M - 20 = · x 10 15 135 M - 20 = = 9 15 M = 9 + 20 = 29.

Hence, Median = Rs. 29,000. Using formula, Median= li

+

!:!. 2 - F

fm

x c = 20 +

33 5 - 20 · 15 x 10

135

= 20 + l5 = 29.

' To calculate Mode, we construct grouped frequency distribution.

I TABLE 11.24:

ORDINARY FREQUENCY DISTRIBUTION

Class-intervals

Frequency

0-10 10-20 20-30 30-40 40-50 50-60

4 16 15 20 7 5 67

Total

I

I

299

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

Maximum frequency is 20 and, therefore, the modal class is 30-40.

f m - !1 20 - 15 ) x 10 x i = 30 + ( 2fm - Ji - f2 2 X 20 - 15 - 7 50 30 + = 30 + 2.778 = 32.778 =Rs. 32,778. 18 L

Mode

+

Yes. It is possible to calculate the arithmetic mean from the ordinary frequency distribution shown in Table 11.25.

TABLE 11.25: CALCULATIONS FOR A.M. Frequency d- ~ - o:-25 Mid-

Class-

-

i

-

value x

f

(A= 25)

0-10 10-20 20-30

5 15

4 16 15 20

-2

25 =A 35 45

30-40 40-50 50-60

-1

-8 -16

0 1 2

0 20 14

3

15

7 5

55

Total

67=N

A.M. (x)

=

A+

-24

Efd

N

fd

10

interval

25 x i = 25 + 67 x 10

+ 49 =

25 = L,f d

250

= 25 + 67

25 + 3.731 = 28.731=Rs.28,731.

Example 14. The table below gives the number (F) of candidates obtaining marks x or higher in a certain examination:

x

10

20

30

40

50

60

70

80

90

100

F

140

133

118

100

75

45

25

9

2

0

Calculate the mean and the medianJ1],axks obtained by the candidates. Solution. Here cumulative frequency distribution (more than type) is given. We first construct ordinary (grouped) frequency distribution and then calculate mean.

300

BUSINESS MATHEMATICS AND STATISTICS

TABLE 11.26: CALCULATIONS FOR MEAN Classintervals

Midvalue x

Frequency

d = '"-:A

fd

I

10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

15 25 35 45

7 15 18 25 30 20 16 7 2

(i = 10) -4 -3 -2 -1 0

-28 -45 -36 -25

55=A 65 75 85 95 ...

Total

140

•

0 20 32 21

1

2 3 4

8

...

=N

-134

+ 81 = -53 = L,fd

A+ -'2:.f d x t. = 55 + (-53) x 10 = 55 - -53

Mean

N 55 - 3.79

140

= 51.21

14

(approx.).

Example 15. Find the missing frequencies in the following distribution when the mean is 11.09: Class-limits

9.3-9.7

9.8-10.2

10.3-10.7

10.8-11.2

11.3-11.7

11.8-12.2

12.3-12.7

Frequency

2

5

h

f4

14

6

3

Solution. Mean (x)

=

11.09, N

12.8--13.2

60

= '2:.f = 60.

TABLE 11.27: CALCULATIONS FOR MEAN Classlimits

I

Mid-value x

9.3-9.7 9.8-10.2 10.3-10.7 10.8-11.2 11.3-11. 7 11.8-12.2 12.3-12.7 12.8-13.2

Frequency

I

9.5 10.0 10.5 11.0 =A 11.5 12.0 12.5 13.0

Total

d

2 5

h f4 14

6 3 1 31

+ h + f4 = N

= 60

= '"--:11 • i = 0.5

fd

-3 -2 -1 0 1 2 3 4

-6 -10

-

-h 0 14 12 9 4 23 -

'ff/: x i, where d = xi A. Here A = 11, i = 0;5. or, 11.09 = 11.0 + 6c/3 x 0.5, or, 11.09 - 11.0 = 6c/3 x ~ Mean (x) =A+

23

23

or, 0.09 x 120 = 23 - f3, or, h = 23 - 10.8 = 12.2 [·: 0.09 x 120 But the value of h cannot be a fraction;:. h = 12.

= 10.8)

Total

h = L,fd

301

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

Again, 31+h+f4=60, or, 31+12 + f4 = 60, or, f4 = 60 - 43 = 17. Hence, the required missing frequencies are fa = 12 and f4 = 17. Example 16. Find the third and the 8th deciles and also the 35th and 75th percentiles from the following frequency distribution: Class-interval Frequency

0-10 10

10-20 15

20-30 25

30-40 40

40-50 50

50-60 30

60-70 20

70-80 10

Solution. TABLE 11.28: CALCULATIONS OF CUMULATIVE FREQUENCY

N= 200

Class-boundary

Cumulative Frequency ("less than")

For

0 10 25 50

For Ds, ~ IO ai 160.

0 10 20 30 D3-->

+--

?35-->

+--

40 50

90 140

?75---+

+--

Ds

---+

60 70 80

+--

?35,

For P1s,

3 N 10

= 60

35N 100

= 70

75

= 150

N

100

150 = 75N/100 160 = ~1;;'

170 190 200 =N

Da = 3rd Decile = la +

•

Ds =8th Decile= ls+

?35

For

60 = 311;;' 70 = 35N/100

D3,

= 35th Percentile =

-3N 10 -Fa

f

3

BN 10 -

f

l35

8

+

F8

x ca = 30 +

x cs = 50 +

35 N - F35 100

f

35

60-50 x 10 = 30 + 2.5 = 32.5. 40 [·. · Da-class is 30-40] 160 - 140 x 10 = 50 + 6.67 = 56.67. 30 [·: Ds-class is 50-60)

x c35 = 30 +

70 - 50 x 10 = 35 40

x c75 = 50 +

150 - 140 x 10 = 50 + 3.33 = 53.33. 30

and P15

= 75th Percentile = l15

+

75 N - F15 100

h

5

302

BUSINESS MATHEMATICS AND STATISTICS

EXERCISE ON CHAPTER

11(1)

Theory 1. Explain what is meant by Central Tendency of data. What are the common measures of

central tendency? 2. What do you understand by 'Average'? Define at least 4 types of such averages, and then state their advantages and disadvantages. Give example of each type. 3.

(a) Define Arithmetic Mean, Geometric Mean and Harmonic Mean, and compare their relative advantages and disadvantages. [C. u. B.Com. 1986]

(b) Define Median of a distribution. Give one: example. (c) What are the requisites of a good average? 4.

[Utkal U. B.Com. 2000]

(a) Write short notes on Mean, Median and Mode. (b) Point out the merits and demerits of the mean, the median and the mode as measures of central tendency as well as the situations where to use each. (c) What are quartiles of a distribution? How do you use them in measuriJ1g dispersion? (d) What do you understand by Weighted Mean?

5. Define and discuss the 'quartiles' of a distribution. Explain their uses. 6.

(a) Explain the terms, Simple Mean and Weighted Mean.

[B.U. B.Com. 1988]

(b) Distinguish between simple and weighted average, and state the circumstances under which the latter should be employed. 7. Give a critical review of the different measures of central tendency with examples. 8. Prove that the Arithmetic Mean of two positive numbers x 1 , Geometric Mean. [Hints: Since square of any real number

> 0,

y'x2) 2 > 0, where XI i.e., A.M. > G.M.]

XI

:. (JXl -

#

x2 or,

x2

is greater than their [C.U. B.Com. 1988]

if the number# O;

+x2-2JXIX2

> 0 or, XI +x2 > 2JXIX2 or,

:i:i

!"' 2 > JxIX2,

9. Explain mean, median and mode as measures of central tendency. Give a comparative study of their relative advantages and disadvantages as measures of central tendency. [C.U. B.Com.(H) 1991]

.Problems (A) 1.

(a) The weekly wages of 5 labourers are (in Rs.) 40, 60, 36, 45, 25. Calculate their A.M. (b) Find the arithmetic mean of 14, 16, 19, 25, 21. (c) If the mean of 7, x - 3, 10, x

2.

+ 3 and

x - 5 is 15, find x.

[C.U. B.Com. 1988] [C.U. B.Com. 2002]

(a) Find the Geometric Mean and the Harmonic Mean of the numbers 1, 9, 81. (b) Find the Geometric Mean of:

303

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

(i) 1, 3, 9. (ii) 2, 9, 12. (iii) 3, 8, 9. (iv) 6, 24, 12. (v) 2, 8, 9 with weights 1, 1, 2.

[C.U. B.Com. 1986] [C.U. B.Com. 1989] [C.U. B.Com. 1987]

(c) If the G.M. of a, 4, 8 be 6, find a.

[C.U. B.Com. 2006]

3. Calculate -the Arithmetic Mean, Geometric Mean and Harmonic Mean of four numbers: (a) 3, 6, 24 and 48.

[C.U. B.Com. 1991, '97)

(b) 4, 6, 12 and 72. (c) 24, 72, 108 and 144. 4. Find the Harmonic Mean of the following numbers:

k· .. ·, ~· l, k, t• ... '2n~1·

(a) 1, ~' (b) 5.

(a) Find the Median of the following numbers:

(i) 7, 2, 5, 9, 6. (ii) 8, 3, 11, 7, 12, 6, 9. (iii) 33, 86, 68, 32, 80, 48, 70.

[C.U. B.Com. 1987] [C.U. B.Com. 1989]

(b) Find the Mode of the numbers: (i) 5, 3, 27, 5, 9, 3, 8, 5. (ii) 4, 3, 2, 5, 3, 4, 5, 3, 7, 3, 2, 6. (iii) 3, 2, 5, 4, 4, 2, 4, 3, 3, 4, 4, 5, 4, 2, 4, 4, 2, 4, 5, 4, 4.

[C.U.B.Com. 1987] [C.U. B.Com. 1989) [C~U.

B.Com. 2000]

(c) Find the Mode and Median of the following numbers: 25, 1275, 748, 162, 967, 162. 6. Find the Median of the following numbers: (a) 3, 9, 7, 4, 8, 6.

[C.U. B.Com. 1995]

(b) 94, 33, 86, 68, 32, 80, 48, 70.

[C.U. B.Com. 2007]

(c) 79, 82, 36, 38, 51, 72, 68, 70, 64, 63.

7.

(a) Find the Mean and Mode of the numbers: 4, 3, 2, 5, 3, 4, 5, 1, 7, 3, 2, 1. (b) Find the Mean and the Mode of the set of numbers: 7, 4, 10, 15, 7, 3, 5, 2, 9, 12. (c) Find the Median and Mode of the numbers: 3, 2, 5, 4, 4, 2, 4, 5, 4, 4.

4,

2, 4, 3, 3, 4, 4, 5, 4, 2, 4, [C.U. B.Com. 2000)

(d) Find the Median and Mode of the numbers: 4, 10, 7, 15, 7, 3, 5, 3, 7. [C.U.B.Com. 1990]

(e) Find the Mean, Median and Mode of the following numbers: 7, 4, 3, 5, 6, 3, 3, 2, 4, 3, 4, 3, 3, 4, 4, 3, 2, 2, 4, 3, 5, 4, 3, 4, 3, 4, 3, 1, 2, 3.

304

BUSINESS MATHEMATICS AND STATISTICS

8. Find the A.M. of the following distribution:

Weights (in pounds) No. of men

100

110

120

130

140

15

20

25

30

10

9. Calculate the Arithmetic Mean and Mode from the following data: Value Frequency

1

2 11

7

3

4

5.

6

16

17

26

31

7 11

8

9

Total

1

1

121

10. Find the mean of weekly wages from the following frequency distribution:

Wages (in Rs.) No. of workers 11.

30-40 10

40-50 20

50-60 40

60-70 16

70-80 8

80-90 6

(a) Find the arithmetic mean of the weekly income from the following frequency distribution: Weekly income (in Rs.) No. of workers

20-25 200

25-30 700

30-35 900

35-40 800

40-&

45-50.

600

400

(b) Calculate the mean from the following table: Monthly wages (in Rs.) No. of workers

10-20 44

!HO

15

20-30 104

3D--40

40-50

50-00

225

310

355

60-70 108

7Q-80

80-90

72

17

[B.U. B.Com. 1998]

12. Find the mode of the following distributions:

(a) (b) 13.

Marks No. of students

5

10

2

6

Marks No. of students

0-10 4

15 10

25 12

15

20-30 18

12

30

35

8

4

30-40 14

40-50 8

(a) Find the median from the following frequency distribution: Marks No. of students

0-10 10

10-20 20

20-30 35

30-40 25

40-50 10

(b) Find the median of the following distribution: Weight in lb No. of men

160-162 15

163-165 54

166-168 126

169-171 81

172-174 24 [C.U. B.Com. 2007]

14.

(a) A sample of size 50 has mean 54.4 and another sample of size 100 has mean 50.3. If the two samples are pooled together, find the mean of the combined sample. [B.U. B.Com. 1990]

305

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

(b) The mean annual salary of all employees in a company is Rs. 25,000. The mcau salaries of male and female employees are Rs. 27,000 and Rs. 17,000 respectively. Find the percentage of males and females employed in the company. [CA Foun. Nov. 1995]

15.

(a) If the Mean and Mode of a certain set of numbers be 60.4 and 50.2 respectively, find approximately the value of the Median. (b) For a moderately asymmetric distribution, Median= 27, Mean =26. Find the mode. [C.U. B.Com. 2000]

(Use empirical relation between Mean, Median and Mode.) 16.

(a) The table below gives the marks obtained in Statistics by 60 students: Class-interval Frequency

0-10 5

10-20 8

20-30 11

30-40 15

40-50 13

50-60 6

60-70 2

With the help of the cumulative frequency diagram or otherwise, determine the median mark of a student. [C.U. B.Com. 1990, 2004] (b) Obtain the quartiles of the following distribution: 50 4

Ages (in years) Frequency

52 12

54 18

58 23

60 30

62 26

64 22

66 16

68 5

70 4

[C.U.B.Com. 2002]

17. Below is ~iven the frequency distribution of weights of a group of 60 students in a class in a school: Weight (in kg)

30-34

35-39

4D-44

No. of students

3

5

12

50-54 14

45-49 18

55-59

60-64

6

2

Draw the cumulative frequency diagrams and hence determine the median weight. Find also the quartiles of the distribution. 18. The table below gives the frequency distribution of weights of 85 apples: Weight in gm

110--119

120--129

130--139

141H49

150--159

160--169

170--179

180--189

5

8

12

18

22

9

7

4

Frequency

Determine the median weight of an apple. 19.

[C.U.B.Com. 1989]

(a) The frequency distribution of wages of 100 workers is as follows: Wages (Rs.) No. of workers

250-259

260-269

270-279

280-289

290-299

300-309

8

16

30

34

10

2

Calculate the value of the median from the above data. (b) Find the mode of the following distribution: Bus. Math. & Stat. [C.U.}-20

[C.U. B.Com. 1988]

306

BUSINESS MATHEMATICS AND STATISTICS

Weight (in gm)

41D--419

420-429

43D--439

44D--449

45D--459

46D--469

47D--479

14

20

39

54

45

18

10

Frequency

20.

(a) Marks obtained by 22 students are given below: Marks obtained No. of students

0-10

10-20

20-30

30-40

2

4

9

7

Find the mode of the above distribution.

[C.U. B.Com. 1987)

(b) Find the mode of the following data:

I Monthly wages (Rs.) I No. of workers

1.25--175

175-225

225-215

8

10

25

I 215-325 I 35

325-375

375-425

425-475

12

10

4

(c) Find the mode of the following frequency distribution: Age (in years)

16-20

21-25

26-30

31-35

36-40

41-45

46-50

4

6

11

5

7

8

3

Population (in thousands)

[C.U.B.Com. 2007)

21. The A.M. of the following frequency distribution is 36.3. Find the missing frequency: Marks No. of students

10-20 12

20-30 18

30-40

40-50 25

?

50-60 15

22. The median of the following distribution is 33. Find the missing frequency: Marks No. of students

0-10

10-20

20-30

30-40

40-50

50-60

8

12

21

?

20

9

23. The mode of the following frequency distribution is Rs. 66. Find the missing frequency: Daily wages (Rs.) No. of workers'

30-40

40-50

50-60

60-70

70-80

80-90

8

16

22

28

?

12 [C.U.B.Com. 2005)

24.

(a) Construct the grouped frequency distribution from the following data and hence find its Arithmetic Mean: Marks Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60

No. of students 175 360 680 790 900 1000 [C.U. B.Com. 1995; V.U. B.Com. 1998)

307

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

{b) Construct a grouped frequency distribution from the following data and hence find the median and mode: Marks obtained No. of students ·175 Below 10 360 Below 20 Below 30 680 790 Below 40 Below 50 900 1000 Below 60

[c.u. B.Com.

2008]

25. Construct the grouped frequency distribution from the following data and hence find its A.M. and Median: ~

Frequency

~w~w~~~~~w~oo~m~oo

4

16

~

76

96

112

120

ft5

[N.B.l)..B.Com. 1994]

26. The A.M. and G.M. of two numbers are 25 and 15 respectively, find tJte two numbers. [C.U. B.Com. 2008] (Hints: Let the two numbers be x and y. Then ~ = 25 and ,/Xii= 15 or,·~+ y = 50 and xy = (x + y) 2 - 4xy = (50) 2 - 4 x 225 = 2500 - 900 = 1600; :. x +- y = ±40.

= 225.

:. (x - y)2

Solving x + y = 50 and x - y = 40, x the two numbers are 5 and 45.J

= 45, y = 5. If x + y = 50, x - y = ·-40, then x = 5, y = 45.

Hence,

Problems (B) 1.

(a) The monthly incomes of 5 labourers are Rs.150, Rs.140, Rs.165, Rs:170 and Rs.180. Calculate the Arithmetic Mean and the Geometric Mean. (b) The monthly incomes of 6 labourers are (in Rs.) 70, 42, 85, 75, 68, 55. Calculate the A.M. and G.M. (c) For a moderately asymmetrical distribution, median= 27, mean= 26. Find mode. [Hints: Use the formula: Mean - Mode= 3 (Mean - Median).

[C.U. B.Com. 2000]]

2. Find the Geometric Mean of the following numbers, correct to two decimal places:

(a) 90, 25, 1 ~5 , 81. (b) 126, 184, 267, 375, 458.

[N.B.U.B.Com. 1997)

3. Dealing in a certain security at the following prices took place on the Bombay Exchange. Calculate the Median price: 1001\,

IOOi,

1001\,

lOOft,

IOOt,

IOOi,

100-f2,

100M,

IOOft,

100,

991,

99f2,

99~~.

99i,

99t.

308 4.

BUSINESS MATHEMATICS AND STATISTICS

(a) The average weight of the following frequency distribution is 117 lb: Weight in lb

100

110

120

1

4

2

No. of persons

x

+ 25

140

Total

1

10

2

Find the value of x. (b) The mean of 20 observations is 85; but it was later found that two of the observations were wrongly read as 75 and 70 instead of 57 and 60. Find the actual mean. (B.U. B.Com.(H) 1988)

Solution. By definition, ~~ = mean = 85 or, Ex= 85 x 20 = 1700. Correct value of Ex = 1700 - (75 + 70) + (57 + 60) = 1672 . .". actual mean = Correct ;~ue of Ex = 1~~2 = 83 , 6 . (c) The A.M. of 25 observations is 44; later on it was reported that two of the observations 34 and 46 were copied as 28 and 42. Find the actual A.M. 5.

(a) Form a frequency distribution with 8 classes from the following data and work out the mean, the median and the mode from it: Data: In a workshop employing 30 persons the daily wages paid are as follows: (in Rs.)

2.30, 2.30, 4.50,

3.50, 4.50, 2.30,

2.30, 5.10, 3.50,

2.40, 4.50, 3.20,

3.20, 5.50, 2.40,

5.10, 2.40, 5.10,

4.50, 3.50, 3.40,

3.50, 3.20, 2.40,

2.30, 2.30, 5.10,

3.40 3.40 3.40

[V.U. B.Com. 1997)

(b) The heights of students of a college are given below. Find the median height: Height in cm

No. of students

160-163 22

164-167 80

168-171 98

172-175 148

176-179 104

180-183 43

184-187 5

[C. U. B.Com. 1994)

6. Find the Mean and the Mode from the following frequency distributions: (a)

Marks obtained

No. of candidates

0-9 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99

6 29 87 181 247 263 133 43 9 2 1000

Total

(b)

Output in units

No. of workers

300 to 309 310 to 319 320 to 329 330 to 339 340 to 349 350 to 359 360 to 369 370 to 379

9 20 24 38 48 27 17 6

309

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

7. (a) Find the Mode from the following distribution: Marks No. of candidates

1-5

6-10

11-15

16-20

21-25

26-30

31-35

36-40

7

10

16

32

24

18

10

5

41-45

(b) Find the mode of the following frequency distribution. The monthly incomes of 300 workers of a factory are as follows: Monthly incomes (Rs.)

10001100

No. of workers

16

11001200 24

12001300

13001400

14001500

15001600

16001700

17001800

59

100

41

31

19

10

[C.U.B.Com. 1997]

8.

(a) Calculate the A.M. and the Median of the frequency distribution given below. Hence calculate the mode using the empirical relation between the three: --

I Class-limits I 130-134 I 135-139 I Frequency I 5 I 15

140-144

145-149

28

24

155-159

150-154 17

160-164 1

10

I I

[B.U.B.Com. 1990]

{b) Find the Mean and Median of height from the following table: Height (x cm)

158-161

162-165

166-169

110-113

No. of men (/)

11

23

31

18

I 114-111 I

178-181

Total

5

100

12

I I

[V.U. B.Com. 1995]

9. In a small town, a survey was conducted in respect of profit made by retail shops. The following results were obtained: Profit or Loss in '000 Rs. -4 to -3 -3 to -2 -2 to -1 -1to0 0 to 1

No. of shops

Profit or Loss in '000 Rs.

No. of shops

4

1 to 2 2 to 3 3 to 4 4 to 5 5 to 6

56 40 24 18 10

10 22

28 38

Calculate: (a) th.e average profit made by a retail shop; (b) total profit by all shops. 10;

(a) Find out the missing frequencies of the following data. A.M. is 67.45 inches: Height (inches)

60-62

63-65

66-68

69-71

72-74

Total

No. of students

5

18

f3

f4

8

100 [C.U. B.Com. 2007]

(b) An incomplete frequency distribution is given below: Variable Frequency

10-20 15

20-30 12

30-4~

?

40-50 56

50-60 ?

60-70 32

70-80 18

Total 200

310

BUSINESS MATHEMATICS AND STATISTICS

You are given that the median value is 48.93. Using the median formula fill in the missing frequencies. [V.U. B.Com. 1996] (c) Find the missing frequencies in the following frequency distribution when the mean is 11.09: Class-limits 9.3-9.7 9.8-10.2 Frequency

2

10.3-10.7 10.8-11.2

5

11.3-11.7

11.8-12.2 12.3-12.7 12.8-13.2 Total

14

f3'

6

3

60 [C.U. B.Com. 2002)

11.

(a) Form an ordinary frequency table from the following cumulative distribution of marks obtained by 22 students and calculate(i) Arithmetic Mean, (ii) Median,

(iii) Mode. Marks

Below Below Below Below Below

No. of students

10 20 30 40 50

3 8 17 20 22 [B.U. B".Com.(H) 1987; C.U. B.Com.(H} 1990)

(b) Calculate the values of median from the following frequency distribution: Marks

less less less less less less

No. of students

than 10 than 20 than 30 than 40 than 50 than 60

160 210 350 480 730 1000 [C.U. B.Com. 2005, '08 Type)

12.

(a) Calculate the quartiles of the following data: Class-limits

Frequency

10-19 20-29 30-39 40-49

5 9 14 20

Class-limits

Frequency

50-59 60-69 70-79 80-89

25 15 8 4

...

Find also skewness based on quartiles. (b) Calculate the quartiles from the following data:

311

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

8-12 8

4-8 5

x

f

I 16-20 I 20-24 I 24-28 I 25 I 14 I 10

12-16 18

[C.U.B.Com. 2003]

13. You are given the following incomplete frequency distribution. It is known that the total frequency is 1000 and that the median is 413.11. Estimate by calculation the missing frequencies and find the value of the mode: Value 300-325 325-350 350-375 375-400

14.

Value 400-425 425-450 450-475 475-500

Frequency 5 17 80 ?

Frequency 326 ?

88 9

(a) Find the Median and Mode from the following table: 20-25

Age No. of men

30-35 100

25-30 70

5

35-40 180

40-45 150

45-50 120

50-55 70

55-60 60

[C.U.B.Com. 1991]

(b) Determine the Median and Modal Price of the following distribution: Price of the Commodity (Rs.) Frequency

310319 5

320329 8

330339 12

340349 28

350359 32

360369 9

370379 7

380389 4

Total

105

[C.U. B.Com. 2000]

15. You are given below a certain statistical distribution: Value Less than 100 100-200 200-300 300-400 400 and above Total

Frequency 40 89 148 64 39 380

Calculate the most suitable average giving reasons for your choice. 16. The table below gives the number (F) of candidates obtaining marks x or higher in a certain examination (all marks are given in whole numbers): x F

IO 140

20 133

30 118

40 100

50 75

60 45

70 25

80 9

90 2

100

Calculate the mean and the median marks obtained_by the candidates.

0

312

BUSINESS MATHEMATICS AND STATISTICS

17. The following table shows the marks in Statistics, secured by 60 students:

I Class-interval I Frequency

0-10

10-20

20-30

30--40

40-50

50-60

60-70

5

8

11

15

13

6

2

With the help of a cumulative frequency diagram determine the median mark of the students. [C. U. B.Com. 2002] [Hints: See worked-out Example 24 in Art. 11.8.l.]

18. Form a frequency distribution with 8 classes of equal class-intervals from the following data and work out the mean, median and mode from it: Data: In a workshop employing 30 persons the hourly wages paid are as follows: 2.30,

Rs.

2.30, 4.50,

3.50, 4.50, 2.30,

2.-30,

2.40,

3.20,

5.10, 3.50,

4.50, 3.20,

5.50, 2.40,

4.50,

5.10, 2.40, 5.10,

5.30, 3.20, 2.40,

3.50, 3.40,

2.30,

3.40,

2.30, 5.10,

3.40, 3.40.

[V.U. B.Com. 1997]

ANSWERS A 1.

(a) 41.2;

(b) 69;

(b) 19;

(c) 66.

7.

(c) 21. 2.

(a) G.M.

= 9,

H.M.

= 2*;

(i) 3· ' (ii) 6; (iii) 6' ' (iv) 12; (v) 6. (c) a-E - 4·

(a) A.M. (b) A.M. (c) A.M.

4.

(b)

(e) Mean 9. A.M.

= 20.25, G.M. = 12, H.M. = 7.1; = 23.5, G.M. = 12, H.M. = 7.8; = 87, G.M. = 72, H.M. = 55.74.

= 4.59,

Mode

10. Rs.56.

11.

(a) Rs. 35.41; (b) Rs. 46.84.

12.

(a) 20; (b) 26.

13.

= 162,

(a) 25.71; (b) 167.43 lb.

14.

(a) 51.7; (b) 80% and 20%.

(i) 5; (ii) 3; (iii) 4;

(a) 6.5;

= 3.47,

Median

8. 120 lb;

(i) 6; (ii) 8; (iii) 68;

(c) Mode 6.

= 3,

(c) 3.62, 4;

(a) _2_. n+l' (b) l (a)

Mode

(d) 7 and 7;

n

5.

= 3.33,

(b) 7.4 and 7;

(b)

3.

(a) Mean

15.

(a) 57; (b) 29.

Median

= 455.

16.

(a) 34; (b) 58, 60, 64.

= 6.

= 3,

Mode

= 3.

313

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

17. Median wt. = 47.3 kg, Q1 = 42.4 kg, Q3 = 52.0 kg.

(c) 27.77 years.

22. 30.

18. 149.22 gm.

23. 24.

19.

24.

21. 30.

(a) Rs. 278.19; (b) 445.75 gm.

20.

(a) 25.45. (b) 24.375; 23.91.

(a) 27.14;

25. 37.88; 36.25.

(b) Rs. 290.15;

26. 5 and 45.

B 1.

(a) A.M.

= Rs. 161, G.M. = Rs. 160.30;

(b) A.M.

= Rs. 65.83, G.M. = Rs. 64.21;

10.

(c) 29.

2.

9. Mean = Rs. 1348, Total Profit = Rs. 3,37,000.

(a) 69.08;

(c) 12, 17.

(b) 251.5.

11.

3. 100!6. 4.

(a) x

= 100;

6.

172.85 cm.

Mean

= 339.05, Mode = 342. 73.

(a) 18.83;

(b) 8.

Rs.3.40,

(a) Mean = 46.88, Mode = 50.60;

(b) 7.

12.

(a) Mean =I Rs. 3.48, Median Mode = Rs. 2.30;

(b)

40.8.

(a) Q1 = 37.36, Q2 = 50.3, Q3 Skewness = -0.1027;

(b)

13.56, 17.44, 21.14.

13. 227 and 248, Mode 14.

(a) Median

(b)

= 60.83,

= 413.98.

= 40. 75,

Mode = 38.64;

Rs. 349.32, Rs. 350.98.

15. Median = 241.22.

Rs.1,341.

(a) A.M. = 145.35, Median Mode = 144.06;

(b)

(a) A.M. = 23.18, Median= 23.33, Mode= 24;

(b)

(c) 44.4. 5.

(a) f3 = 42, f4 = 27; (b) 23 and 44;

167.98 cm, 167.56 cm.

= 144.92,

16. Mean = 50.715 or 51.214, Median = 51.166 or 51.67. 17. 34. 18. Rs. 3.58, Rs. 3.40 and 2.30.

EXERCISE ON CHAPTER 11(11) {Miscellaneous Problems) 1.

(a) Find the quartiles of the values: 35, 30, 48, 40, 25, 36, 45. (b) Find the deciles of the following values: 50, 80, 60, 30, 40, 10, 45, 20, 25, 75; 55, 65, 35, 52, 44, 38, 63, 34, 46. How many deciles lie below the first quartile? (c) The numbers 3.2, 5.8, 7.9 and 4.5 have frequencies x, x+2, x-3 and x+6 respectively. If the arithmetic mean is 4876, find the value of x.

314 2.

BUSINESS MATHEMATICS AND STATISTICS

(a) The mean wage of 100 workers working in a factory running two shifts of 60 and 40 workers respectively is Rs. 38. The mean wage of 60 labourers working in the morning shift is Rs. 40. Find the mean wage of labourers working hi' the evening shift. (b) The algebraic sum of the deviations of 25 observations measured from 45 is -55; find the A.M. of the observations. [C.U. B.Com. 1991) 25

[Hints:

L

25

(xi -

45)

= -55 or, L

i=l

25

Xi -

i=l

25

L

45

= -55 or, L

i=l

Xi -

25 x 45

= -55

i=l

25

or,

L

x;

= 1125 -

55

= 1070. :. Mean =

~

=

1

~~0

= 42.8.]

i=l

(c) For some symmetrical distribution, Qi = 24 and Q3 = 42; find Median. [C.U. B.Com. 1998)

3. Find the missing frequency from the following frequency distribution if Mean is 38: Marks No. of students 4.

10 8

20 11

30 20

40 25

50 ?

60 10

70 '3

(a) The mean of 20 observations is ·16.5. If by mistake one observation was copied 12 instead of 21, find the correct value of mean. (b) The mean marks of 100 students was found to be 40. Later on it was discovered that a mark 53 was misread as 83. Find the correct mean mark. [C.U. B.Com. 1990) [Hints: Mean =

1;,"'

or, 40 = fa~ or, Ex= 4000. Correct value of Ex= 4000 - 83 + 53

:. correct Mean= Cor~~~tEx

= 3970.

= 319cii = 39.7.)

5. The mean weekly salary paid to 77 employees in a company was Rs. 78. The mean salary of 32 of them was Rs. 75 and of other 25 was Rs. 82. What was the mean salary of the remaining? 6.

(a) The mean annual salary paid to all employees of a company was Rs. 5000. The mean annual salaries paid to male and female employees were Rs. 5200 and Rs. 4200 respectively. Determine the percentage of males and females employed by the company. (b) The mean monthly salary paid to all employees in a certain company was Rs. 600. The mean monthly salaries paid to male and female employees were Rs. 620 and Rs. 520 respectively. Obtain the percentage of male and female employees in the company.

7. An aeroplane flies along the four sides of a square at speeds of 150, 300, 450 and 600 km per hour respectively. Using an appropriate mean, find the average speed of the plane around the square. 8. Draw an ogive for the following distribution. Locate the median from the graph and verify the result by direct application of the formula. How many workers earned monthly wages between Rs. 60 and Rs. 72?

315

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

Monthly Wages (Rs.) No. of workers 9.

50-55 6

55-60 10

60-65 22

65-70 30

70-75 16

75-80 12

80-100 15

(a) Find the median and mode of the following frequency distribution: j No. of days absent (Jess than)

I No. of students

I5 I 29

10 1 15 224 I 465

20 582

I 25 I 634

30 35 644 650

40 653

45 655

(b) The frequency distributions of marks obtained by 1000 students in an examination are given below. Calculate the Arithmetic Mean: Marks obtained

No. of students

Marks obtained

No. of students

0-9 10-19 20-29 30-39 40-49

25 37 81 290 253

50-59 60-69 70-79 80-89 90-99

225 46 22 17 4 [C.U. B.Com. 1987]

10. The following table gives the distribution of 100 families according to expenditure. If Mode of the distribution is 24, find the missing frequencies x and y : Expenditure No. of families

0-10 14

10-20 x

20-30 27

30-40 y

40-50 15

11. Following is the distribution of marks in Law obtained by 50 students: Marks more than No. of students

0 50

10 46

20 40

30 20

40 10

50 3

Calculate the median marks. If 603 of the students pass this test, find the minimum marks obtained by a pass candidate. 12. The A.M., the mode and the median of a group of 75 observations were calculated to be 27, 34 and 29 respectively. It was later discovered that one observation was wrongly read as 43 instead of 53. Examine to what extent the calculated values of the three averages will be affected by the discovery of the error. 13. The median and the mode of the following daily wage distribution of 230 workers are known to be Rs. 33.5 and Rs. 34 respectively. Three frequency values from the table are, however, missing. Find these missing values?

Wages in Rs. No. of workers 14.

0-10 4

10-20 16

20-30 ?

30-40 ?

40-50 ?

50-60 6

60-70 4

(a) In a distribution, the difference of two quartiles is 2.03; their sum is 72.67 and median 36.18, find the coefficient of skewness.

316

BUSINESS MATHEMATICS AND STATISTICS

(b) For an income distribution of a group of men, 20% of men have income below Rs. 35, 35% below Rs. 75, 60% below Rs. 175 and 80% below Rs. 250. The first and third quartiles are Rs. 55 and Rs. 200. Put the above information in a cumulative frequency distribution and find the median. [Hints: Let the total number of men in the group be 100. Then Income (Rs.) (less than)

35

55 (=Qi)

75

175

200 (= Q3)

250

250 and above

Percentage of men (c. f.)

20

25

35

60

75

80

100

Use simple interpolation to find median.]

15. Show that the weighted arithmetic mean of first n natural numbers whose weights are equal to the corresponding numbers is equal to 2 n} 1 . 16. The following table gives the weekly wages in rupees in a certain commercial organization: Weekly Wages

30-

32-

34-

36-

38-

40-

42-

44-

46-

48-50

3

8

24

31

50

61

38

21

12

2

Frequency

Find (a) the median and the first quartile; (b) the number of wage-earners receiving between Rs. 37 and Rs. 47 per week. 17. Using suitable formulae calculate the mean and the median from the following data:

18.

Mid-value

115

125

135

145

155

165

175

185

195

Total

Frequency

6

25

48

72

116

60

38

22

3

390

(a) Draw an ogive to illustrate the following data and from it determine: (a) median and quartile wages, and (b) the eighth decile wages: Weekly Wages (in Rs.) x 30- 32- 34- 36- 38-- 40- 42- 44-

46- 48-50 Total

NL:rnber of Wage-earners

11

2

9

25

30

49

62

39

20

3

250

(b) The table below gives the marks obtained by 590 \;tudents in a certain examination. Find the number of students obtained less than 50 marks in that examination: Marks No. of students

Less than 20

20-40

40-60

60-80

80-100

250

120

100

70

50 [C.U. B.Com. 2004]

19. Calculate the median and mode of the following: Annual Sales (Rs. '000)

Frequency

Annual Sales (Rs. '000)

Frequency

Less than 10

4

Less than 40

Less than 20

20

Less than 50

55 62

Less than 30

35

Less than 60

67

Is it possible to calculate the arithmetic mean? If possible, calculate it.

317

CHAP. 11: MEASURES OF CENTRAL TENDENCY: MEAN, MEDIAN AND MODE

20. By using the quartiles find a measure of skewness for the following distribution: Annual Sales (Rs. '000)

Number of Firms

Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80 Less than 90 Less than 100

30 225 465 580 634 644 650 665 680

21. Construct the grouped frequency distribution from the following data and hence find its 1 Arithmetic Mean: Marks No. of students

Less than 10

Less than 20

Less than 30

Less than 40

Less than 50

Less than 60

175

360

680

790

900

1000

[C.U. B.Com. 1995]

ANSWERS 1.

(a) 30, 36 and 45; (b) 20, 30, 35, 40, 45, 50, 55, 63, 75; two deciles 20 and 30 lie below the first quartile; (c) 5.

2.

(a) Rs.35; (b) Rs.42.8;

(b) 42.58. 10. 23 and 21. 11. 27.5; 25. 12. Correct A.M. = 27.13; Median and Mode will remain unchanged. 13. 60, 100 and 40. 14.

(c) Rs.33. 3. 18. 4.

5. Rs. 77.80. 6.

(a) 803 and 20%; (b) 80% and 203.

7. 288 km per hour. 8. 67.9, 58. 9.

16.

(a) 16.95; (b) 39.7.

(a) 12.14 days, 11.35 days;

(a) 0.15; (b) Rs.135. (a) Rs. 40.30, Rs. 37. 77; (b) 192.

17. 153.64; 153.8. 18.

(a) Rs. 40.30, Rs. 37.80, Rs. 42.50, Rs. 43.20; (b) 420.

19. Rs. 29,000, Rs. 32, 778. 20. 0.09. 21. 25.45.

Chapter 12

Measures of Dispersion

12.1

Introduction

We have seen how statistical data in the form of a frequency distribution can be represented by a single typical value, known as statistical average (or simply average). An average may give a good idea of the type of data, but it alone cannot reveal all the characteristics of_the data. It cannot tell us in what manner all the values of the variable are scattered (or dispersed) about the average. Two series having the same number of values may have the same mean, but still the values in one may be widely dispersed and the values in the other may be close to one another. Thus it is necessary to know the scatter or dispersion of the values from their mean. In this chapter we shall study the different measures of dispersion of the values of a variable.

12.2

Dispersion

The variation or scattering or deviation of the different values of a variable from their aver11.ge is known as dispersion. Dispersion indicates the extent to which the values vary among themselves.

Measures of Dispersion There are seven measures of dispersion of which four are absolute measures and three are relative measures .

.Absolute Measures 1. Range, 2. Quartile Deviation, 3. Mean Deviation, 4. Standard Deviation.

318

319

CHAP. 12: MEASURES OF DISPERSION

Relative Measures 1. Coefficient of Variation, 2. Coefficient of Quartile Deviation, 3. /Coefficient of Mean Deviation.

12.3

Range

Definition 1. Range is the szmplest absolute mean of dispersion. It is the difference between the largest and the smallest values of a variable. It is simple to understand and easy to calculate, but it is not generally used for practical purposes due to its dependence entirely on the two extreme values (i.e., the largest and the smallest values). Range of the values (in Rs.) 8, 5, 10, 7, 12, 6 is 12 - 5 = 7 (Rs.), since the largest value= 12 and the smallest va1ue = 5. ! In a grouped frequency distribution, the range is measured by the difference between the Jpper boundary of the highest class and the lower boundary of the lowest class. It is also measured by the difference between the mid-values of the highest and the lowest classes. . Range Coefficient of Range= - - - - - - .- - - - - - - - - Sum of the highest and the lowest values

Advantages and Disadvantages Range is simple to understand and easy to calculate. But it has many disadvantages. It is very much affected by the presence of an extremely high or low value. It is not based on all the values of the variable. It cannot be calculated from grouped frequency distribution with open-end classes.

12.4

Quartile Deviation or Semi-interquartile Range

Definition 1. Quartile Deviation (Q) is an absolute measure of dispersion and is defined by the form~la Q = (Qs;Qi), where Qi and Q3 are the first {or lower) and the third (or upper) quartiles respectively. As it is based only on Qi and Q3, it does not take into account the variability of all the values and hence it is not very much used for practical purposes.

Example 1. Find the Quartile Deviation of the daily wages {in Rs.) of 7 persons given: 12, 7, 15, 10, 19, 17, 25. Solution. Arranging the given values in ascending order of magnitudes, we get 7, 10, 12, 15, 17, 19, 25. Here n = 7,

n +1 7+1 3(n + 1) = 3(7 + 1) = . 6 -4- = -4- = 2' 4 4

:. Qi = ntith value= 2nd value= 10 and Q3 =

3

(n4+1)th value= 6th value= 19.

. . = Q 3 - Qi = 19 - 10 = Rs . 4 . 50 • Hence, Q uart1·1e D.ev1at1on 2 2

_,.-

320

BUSINESS MATHEMATICS AND STATISTICS

Advantage~

and Disadvantages

Quartile Deviation is easy to calculate and its calculation depends on the first and the third quartiles. It is not based on all the values of the variable and so it does not take into account the variation of each observation about the average (central value). It can be calculated from a grouped frequency distribution with open-end classes.

12.5

Mean Deviation (or Average Deviation or Mean Absolute Deviation)

Definition 1. MeAn Deviu,tion of a series of values of a variable is the arithmetic mean of all the absolute de'llioiiom (i.e., difference without regard to sign) from any one of its averages (Mean, Median or Mode, but usually Mean or Median). It is an absolute measure of dispersion. Mean Deviation of a set of n values x1, x2, · · · , n

L lxi -

Xn

n

xi

L ldil

i-1 i=l E!di Mean Deviation = ----- = -- = n n n

where x = A.M. and ldil = lxi For a frequency distribution,

about their A.M. is defined by

xi

i.e.,

=absolute deviation of Xi from

. .

Mean Dev1at1on =

. .

Eld!

Mean Dev1at1on = -·n

x.

E/ la: - xi E/ !di = ---, E/ N

where x = value or mid-value according as the data is ungrouped or grouped and Median.

x=

Mean or

Note; The expression !di is read as mod. d and gives only numerical or absolute value of d without regard to sign. Thus I - 31 = 3, I + 41 = 4, I - 0.561 = 0.56. The reason for taking only the absolute and not the algebraic values of the deviations is that the algebraic sum of the deviations of the values from their mean is zero.

Advantages and Disadvantages Mean Deviation is based on all the values of the variable and sometimes gives fairly good result as a measure of dispersion. However, the practice of neglecting signs and taking absolute deviations for the calculation of Mean Deviation seems rather unjustified and this makes algebraic treatment difficult.

Coefficient of Mean Dispersion The coefficient of Mean Dispersion is defined by the formula:

or,

. t f M ·:-r..· • Mean Deviation from Mean C oe ffi c1en o ean u1spers1on = M ean Mean Deviation from Median Median

321

CHAP. 12: MEASURES OF DISPERSION

Example 2. Calculate the Mean Deviation of the following values about the median in respect of the following data: 8, 15, 53, 49, 19, 62, 7, 15, 95, 77. (C.U.B.Com. 2008 Type]

Solution. Arranging the given values in ascending order of magnitude, we get 7, 8, 15, 15, 19, 49, 53, 62, 77, 95. Median

= =

n+l 10+1 - -th value = - -th value = 5.5th value 2 2 19 + 49 68 Mean of the 5th and 6th values = =2 2

= 34 .

•Absolute deviations of the values from the median 34 are respectively 27, 26, 19, 19 15, 15, '19, 28, 43, 61. Mean Deviation about the Median

= E ldl

27 + 26 + 19 + 19 + 15 + 15 + 19 + 28 + 43 + 61 10

--=~~~~~~~~~~~~~~~~~-

n

272 = =27.2. 10 . Example 3. Find the Mean Devi.ation about the Arithmetic Mean of the numbers 31, 35, 29, 63, 55, 72, 37. (C.U. B.Com. 2003)

Solution. A . h t' M (-) 31+35 + 29 + 63 + 55 + 72 + 37 322 nt me ic ean x = = T = 46 . 7

I TABLE 12.1: CALCULATION OF ABSOLUTE DEVIATIONS I Value :z: 31 35 29 63 55 72 37

Deviation from Mean (d = :z: - if = :z: - 46) -15 -11 -17 17 9 26 -9 /v

Absolute Deviation

ldl 15 11 17 17 9 26 9

104 = ~ ldl

-

Total

:. the required Mean Deviation about the Mean= E

ldl

n

=

Example 4. Find the mean deviation of the following series:

Bus. Math. & Stat. [C.U.}--21

x

10

Frequency

3

11 12

12

13

14

Total

18

12

3

48

104 = 14.86. 7

322

BUSINESS MATHEMATICS AND STATISTICS

Solution. TABLE 12.2: CALCULATIONS FOR MEAN DEVIATION

x

f

fx

Ix-xi

fix-xi

10

3 12 18 12 3

30 132 216 156 42

2 1 0 1 2

6 12

11

12 13 14 Total

3JY-= Ef Ix ~xi

= 'Efx

576

48=N

0

12 6

x = 'Ef x = N

576 48

= 12.

xi

. . a b out t he Mean = 'Ef Ix . . Mean D ev1atlon N-

= 36 = 0. 75 · 48

I TABLE 12.3: CALCULATION OF CUMULATIVE FREQUENCY I x

3 15

3

12 18 12

12 13 14

Median

Cumulative Frequency (less than)

f

10 11

33

45

3

48=N

N;

1

=

(

) th value= (

=

2 (24th value +

1

~9 ) th, i.e., 24.5th value

25th value).

From the 3rd column, we see that 16th to 33rd values are each 12. :. 24th and 25th values are each 12. . 12+12 :. Median= = 12. 2 Hence, Mean Deviation about the Median is also O. 75. Example 5. Calculate mean deviation from the median from the following: Class-intervals Frequency

2-4 3

4-6

.&;8

8-10

4-

2

1

323

CHAP. 12: MEASURES OF DISPERSION

Solution.

I TABLE 12.4: CALCULATION OF CUMULATIVE FREQUENCY I Class-boundary

Cumulative Frequency (less than)

2

0 3

4

Median--+

+-

5= N/2

7

6 8 10

9

lO=N

Median =the value corresponding to Cumulative Frequency J¥-, i.e., 5, and 5 lies between 3 and 7. Therefore, Median class is 4-6;

. Median (M) = li

+

(N/2)-F

.

.

5 -3

. x c = 4 + -- x 2 = 4 + 1 =

f

4

m

5.

TABLE 12.5: CALCULATION OF MEAN DEVIATION Class-intervals 2-4

Mid-value :z:

d = :z: --5

ldl

I

'/ldl

3 5

-2

2

7

0 .2

9

4

0 2 4

3 4 2 1

6 0 4 4

4-6 6-8

8-10 Total

Hence, the required Mean Deviation from the median =

12.6

14 = Y.:.f

lO=N

E~dl

=

~~

ldl

= 1.4.

Standard Deviation

Definition 1. It is the most important absolute measure of dispersion. Standard Deviation (S.D.) of a set of values of a variable is defined as the positive square root of the arithmetic mean of the squares of all the deviations of the values from their arithmetic mean. In short, it may be defined as the Square Root of the Mean of the squares of deviations from Mean. S.D. is usually denoted by the Greek small letter a (pronounced Sigma). If xi, x2, · · · , Xn be a series of values of a variable and x their A.M., then S.D. (a) is defined by , where N = Ef. . Ef(x - 2) = 2' .. N

Ef x Ef2 Efx 2Ef - - - = 2 or -·=2+-=2+2·1=4 N N "'N N .

Hence, mean (x) = 'f-1!- = 4. Otherwise. If mi be the first moment about any value A, then mean (x) = A+ mi. Here A = 2 and mi = 2. Hence, mean (x) = 2 + 2 = 4. ") F' b (u irst moment a out zero

Ex 16 = --:;= 1 + 3 +4 5 + 7 = 4 =

Ex 2 Second moment about zero= -

'

n

2

=1

+ 3 2 + 5 2 + 72 4

4·

= 1 + 9 + 25 + 49 = -84 = 21. 4

4

Example 2. Find the first, second and t'iird moments about the origin 4 for the set of numbers

2, 4, 6, 8.

Solution. rth moment about 4 =

E(:i:4

4

t,

r = 1, 2, 3.

I TABLE 13.1: CALCULATIONS FOR FIRST THREE MOMENTS I :z: 2 4 6 8 Total

:z:-4 -2 0 2 4 4 = E(x - 4)

(~ - 4}3

(:z: - 4)2 4 0 4 16 24 = E(x - 4) 2

-8 0 8 64 64 = E(x - 4) 3

E(x - 4) 4 = 4 = 1. 4 E(x - 4) 2 24 2nd moment about 4 = = 4 = 6. 4 E(x - 4) 3 64 3rd moment about 4 = = 4 = 16. 4 1st moment about 4 =

Example 3. Find the first, second, third and fourth central moments for the set of numbers 2,

4, 6, 8.

. Solution. mr

= rth central moment =

E(x-xr n

, r = 1, 2, 3, 4.

I TABLE 13.2: CALCULATIONS FOR THE FIRST FOUR MOMENTS I :z: 2 4 6 8 20=Ex

:z:-:il=:z:-5 -3 -1 1 3 O = E (x - x)

(:z: - x)2 9

1 1 9

20 = E(x - x) 2

(:z: - x)s -27 -1 1 27 O = E(x - x) 3

(:z: - x)4 81 1 1 81 164 = E(x - x) 4

357

CHAP. 13: MOMENTS, MEASURES OF SKEWNESS AND KURTOSIS

_ Ex 20 .. X=-=-=5.

4

n

2

m = E(x - x) = 20 = S; 2

4

4 4

3

m = E(x - x) = Q = O· 3 4 4 '

m = E(x - x) = 164 = 4 1. 4 4 4

Example 4. The first two moments of a distribution about the value 5 of the variable are 2 and 20. Find the mean and the variance.

Solution. We have by definition,

and

..!:._Ef(x - 5) = 2 N

(1)

~Ef(x -

(2)

5)

2

= 20.

From (1),

1 N

-(Ef x - Ef5) = 2,

or,

1 5Ef NEfx - N = 2,

or,

N

Efx

= 2 + 5 = 7 [·: Ef = NJ

Ef x

N

Mean=

=7.

From (2),

~Ef(x -

5)

2

1

= 20, or, NE! (x 2 - lOx + 25) = 20, 1

2

1

E/

or,

NEfx - N · 10 · Efx + 25 · N = 20,

or,

NEfx 2 -10 x 7 +25 x l =20,

or,

1 2 NEfx = 65.

1

2

• Efx - (Efx) Variance = (S.D.) 2 = ~ N

2

= 65 - 49 = 16.

Otherwise: The mean and the variance can also be determined by using the relations x = A +mi and a 2 = m2 = m2 - m~2 . Here A = 5, mi = 2, m2 = 20. Example 5. The first 3 moments of a distribution about the value 7 calculated from a set of 9 observations are 0.2, 19.4 and -41.0. Find the measures of central tendency and dispersion, and also the third moment about origin.

358

BUSINESS MATHEMATICS AND STATISTICS

Solution. By definition, we have

and

E(x - 7) = 0. 2 9

(1)

E(x - 7)2 = 19.4 9

(2)

E(x - 7)3 = -41.0. 9

(3)

From (1), Ex-7 x 9 Ex Ex = 0.2,, or, g - 7 = 0.2,, or, g = 7.2, 9 :. Mean= 7.2, which is a measure of central tendency. From (2), E (x 2 - l4x + 49) Ex 2 - 14Ex + 49 x 9 = 19.4,, or, = 19.4 9 9 Ex 2 14Ex Ex 2 · or, - - - - - + 49 = 19.4,, or, - - -14 x 7.2 + 49 = 19.4 9 9 9 or,

gE~

E~

= 19.4 + 100.8 - 49,, or, - - = 71.2. 9 Ex2 - (Ex)2 - . 171 2 - (7 2)2 9 9 -

v .

.. S.D. =

.

v'71.2 - 51.84 = v'l9.36 = 4.4,

which is a measure of dispersion. From (3), E (x 3 - 3x2 · 7 + 3 · x · 72 - 73) 9 = -41.0 3 2 or, Ex - 21Ex + ~47Ex - 343 x 9 = _ 4 1.0 or, or, or,

Ex3 - 21 x Ex2 + 147 x Ex - 343 x 9 = -41.0 9 9 9 9 3 Ex - - - 21 x 71.2 + 147 x 7.2 - 343 = -41.0 9 3 Ex3 g = -41.0+343-1058.4+1495.2, or, -Ex9- = 738.8.

:. the third moment about the origin= 738.8. Note: The problem can also be solved by applying the relations obtained in Art. 13.2.1.

13.3

Skewness

A frequency distribution is said to be symmetrical when the values of the variable equidistant from their mean have equal frequencies.

359

CHAP. 13: MOMENTS, MEASURES OF SKEWNESS AND KURTOSIS

If a frequency distributi'bn is not symmetrical, it is said to be asymmetrical or skewed. Any deviation from symmetry is called skewness. In the words of Riggleman and Frisbee: "Skewness is the lack of symmetry. When a frequency distribution is plotted on a chart, skewness present in the items tends to be dispersed more on one side of the mean than on the other." We discuss this important property of data in terms of symmetry or lack thereof without employing any particular measure.

M=Me=Mo Skewness =0 Fig 13.1 Symmetrical Distribution.

Mot M

Mt Mo

Me Fig 13.2 Positively Skewed Distribution.

Me Fig 13.3 Negatively Skewed Distribution.

Skewness may be positive or negative. A distribution is said to be positively skewed if the frequency curve has a longer tail towards the higher values of x, i.e., if the frequency curve gradually slopes down towards the high values of x. For a positively skewed distribution, Mean (M) > Median (Me)> Mode (Mo). A distribution is said to be negatively skewed if the frequency curve has a longer tail towards the lower vi:.lues of x. For a negatively skewed distribution, Mean < Median < Mode. For a symmetrical distribution, Mean = Median = Mode.

13.3.1

Measures of Skewness

The degree of skewness is measured by its coefficient. The common measures of skewness are: -

Pearson's first measure: Mean - Mode Skewness = - - - - - - - Standard Deviation

.f

Illustration 1. If Mean

= 25,

Mode

Skewness =

= 22

25 - 22 9

and Standard Deviation

3

= g=

1

3

,

or, 0.33.

= 9,

then

360 -

BUSINESS MATHEMATICS AND STATISTICS

Pearson's second measure: _ 3(Mean - Median) . Skewness Standard Deviation

./

Illustration 2. If Mean

= 18,

Median

Skewness=

-

= 16.5

and Standard Deviation

= 6,

then

45 3 18 16 5 · ) = · = 0.75. ( -

6

6

Bowley's measure:

where Qi, Q2, Qa are the first, second and third quartiles respectively. For a symmetrical distribution, skewness= O; :. Qi+ Qa - 2Q2 = 0, or, Qi+ Q3 = 2Q2. -

Moment-measure:

ma m3 ma Skewness= 3 = --aj2 = ~· a m2 m2ym2

where m2 and ma are the second and the third central moments and a is the S.D.

All the four measures of skewness defined above are independent of the units of measurement. Note: R.A. Fisher defined Skewness b1) as 'Yl

= +V/'1, where f31 = mVm~.

Example 6. (i) If Qi = 26, Q2 = 46 and Qa = 76, find skewness. [C.U. B.Com. 1996) (ii) For some symmetrical di.2003

79

0

2004 2005 Total

80 40

1 2

0 1 4

80 80

290 = EY

0

10 = EX 2

34 = EXY

XY

0

Using (2), a

= EY N

= 34 = 58 an d b ~ EXY EX 2 10 3·4 ·

= 290 = '

5

From (1), the required equation of the best fitted straight line is Y = 58 + 3.4X. Year

x

2001 2002

-2 -1

2003 2004

0 1 2

2005

· Trend Values (Y = 58 + 3.4X) 58 + 3.4 x (-2) = 51.2 58 + 3.4 )( (-1) = 54.6 58 + 3.4 x 0 = 58.0 58 + 3.4 x 1 = 61.4 58 + 3.4 x 2 = 64.8

Note: Unless otherwise specified, we shall assume that the values of Y refer to mid-year values, i.e., as of July 1. Thus, in Example 9 in Art. 18.8, X = 0 corresponds to July 1, 2003, X = -1 to July 1, 2002, X = 1 to July 1, 2004, etc.

Example 10. Fit a suitable straight line to the following data by the method of least squares and estimate the percentage of insured people in 1997: Year

1989

1990

1991

1992

1993

Percentage of Insured People

11.3

13.0

9.7

10.6

10.7 (C.U. B.Com.(H) 2001]

Solution. Let the equation of the straight line to be fitted to the given data by the method of least squares with the origin at the middle year 1991 and unit of X as 1 year be Y=a+bX.

(1)

By the method of least squares, the values of a and b are given by

/ a =

Here N = Number of years = 5.

EY N

EXY and b = EX2 .

(2)

559

CHAP. 18: TIME SERIES ANALYSIS TABLE 18.11: CALCULATIONS FOR FITTING THE LINE (1) TO THE GIVEN DATA Year

Percentage of Insured People (Y)

x

x2

XY

1989 1990 ---+1991 1992 1993 Total

11.3 13.0 9.7 10.6 10.7 55.3 = EY

-2 -1 0 1 2 0

4 1 0 1 4 10 = EX 2

-22.6 -13.0 0 10.6 21.4 -3.6=EXY -··''/

Using (2), we get a

=

EY N

= 55.3 = 11.06 and b = EXY2 = -3.6 = -0.36. 5

EX

10

. ·

Hence, from (1), the required equation of the best fitted straight line to the given data is Y = 11.06 - 0.36X, with origin at 1991 and unit of X = 1 year. In 1997, X = 6 and¥= 11.06-' 0.36 x 6 = 11.06- 2.16 = 8.9. Hence, estimated percentage of people in 1997 = 8.9. Example 11. The following table gives the annual profits (in thousand rupees) in a factory:

Year Profit ('000 Rs.)

1991 60

1992 72

1993 75

1994 65

1995 80

1996 85

1997 90

(i) Fit a straight line trend by the method of least squares; (ii) Find the gradient of the fitted trend line; (iii) Calculate the projected profit for 1998. [C.U. B.Com.(H) 1999) Solution. Let the equation of the straight line to be fitted to the given data by the method of least squares with origin at the middle year 1994 and unit of x as 1 year be (1)

y=a+bx.

Then the normal equations giving the values of a and b are Ey Exy a= and b = --. n Ex 2

(2)

I TABLE 18.12: CALCULATIONS FOR THE STRAIGHT LINE TREND I Year

1991 1992 1993 ---+1994 1995 1996 1997 Total

Profit ('000 Rs.) y

60 72

75 65 80 85 90 527 = Ey

:J:

:J:2

-3 -2 -1 0 1 2 3 0

9 4 1 0 1 4 9 28 = Ex 2

:i:y

-180 -144 -75 0 80 170 270 520 - 399 = 121

= Exy

560

BUSINESS MATHEMATICS AND STATISTICS

:. from (2), a

=

527

121

T = 75.29 and b = 2B = 4.32.

Hence, from (1), the required fitted straight line trend is y = 75.29 + 4.32x. (ii) The gradient of the fitted trend line = b = 4.32. (iii) In the year 1998, x = 4 and projected profit for 1998 is

y

= 75.29 + 4.32

= Rs. 92.57 thousand

x 4

= Rs. 92,570.

Example 12. Fit a straight line trend equation by the method of least squares and estimate the

trend values: Year Values

1991 80

1992

1993

1994

90

92

83

1995 94

1996 99

1997 92

1998 104

Solution,. Here N = number of years = 8, which is even.

Let the straight line trend equation by the method of least squares with the origin at the mid-point of 1994 and 1995 and unit of X as ~ year be

Y = a+bX.

( 1)

Then a and b are given by EY

a

=N

EXY and b = EX 2 •

(2)

TABLE 18.13: CALCULATIONS FOR FITTING THE STRAIGHT LINE TREND Values

Year

Y

1991 1992 1993 1994

80 90 92 83

1995 1996 1997 1998 Total

94 99 92 104 734 = ~y

x

x2

XY

-7 -5 -3 -1

49 25 9 1

-560 -450 -276 -83

1

1

94 297 460 728 210 = I:XY

--+

3 5 7 0

9 25 49 168 = I:X 2

and b =

-=2

Using (2), a

EY N

=-

734 8

=-

= 91. 75

EXY EX

210 168

= 1.25.

:. from {1), the required equation of the straight line trend is Y = 91.75 + l.25X.

561

CHAP. 18: TIME SERIES ANALYSIS

I TABLE 18.14: CALCULATIONS FOR TREND VALUES I Year

x

1991

-7

1992 1993

-5

1994

-3 -1

1995 1996

3

Trend Values (Y = 91.75 + l.25X) 91.75 + 1.25 x (-7) = 83.0 91.75 + 1.25 x (-'5) = 85.5 91. 75 + 1.25 x (-3) = 88.0 91.75 + 1.25 x (-1) = 90.5 91.75 + 1.25 x 1 = 93.0 91. 75 + 1.25 x 3 = 95.5 91.75 + 1.25 x 5 = 98.0 91. 75 + 1.25 x 7 = 100.5

1

1997

5 7

1998

Note: If the number of years is even, there is no middle year and in this case the mid-point, which is taken as the origin, lies midway between the two middle years. In example 12 in Art. 18.8, the mid-point (i.e., the origin) lies midway between July 1, 1994 and July 1, 1995, which is January 1, 1995 (or December 31, 1994). To avoid fractions, the units of X are taken as ~ year (or 6 months).

Example 13. The weights of a calf taken at weekly intervals are given below: Age in Weeks (x) Weight (y) in kg Unit

1

2

3

4

5

6

7

8

9

10

52.5

58.7

65

70.2

75.4

81.1

87.2

95.5

102.2

108.4

Fit a straight line using the method of least squares and calculate the average rate of growth per week. [C.U.B.Com.(H) 2002} Solution. Let y = a + bx be the straight line to be fitted to the given data by the method of least squares with unit of x as 1 week without shifting the origin. Then the normal equations . giving the values of a and b are Ey = na+bEx (1) and Exy = aEx + bEx2 • {2) TABLE 18.15: CALCULATIONS FOR FITTING A STRAIGHT LINE TO THE GIVEN DATA Age in weeks

Wt. in kg

x

y

1 2 3 4 5 6 7 8 9 10 55 =Ex Bus. Math. & Stat. [C.U.}--36

52.5 58.7 65.0 70.2 75.4 81.l 87.2 95.5 102.2 108.4 796.2 = Ey ·

x2

1 4 9 16\ 25 36 49 64 81 100 385 = Ex 2

xy

52.5 117.4 195.0 280.8 377.0 486.6 610.4 764.0 919.8 1084.0 4887.5 =Exy

562

BUSINESS MATHEMATICS AND STATISTICS

Here n = 10. From (1) and (2), we get lOa + 55b = 796.2

(3)

55a + 385b = 4887.5.

(4)

and Now, (4) x 2 - (3) x ~ 165b = 1016.8 =} b = 6.162 (approx.) Substituting b = 6.162 in (3), we get

or, or,

lOa + 55 x 6.162 = 796.2, lOa = 796.2 - 338.91 = 457.29, a= 45.729 = 45.73 (approx.)

=

The equation of the best fitted straight line is y 45. 73 + 6.162x. Slope of this line is 6.162 which is the average rate of growth per week.

18.8.1

Non-linear Trend: Second Degree or Quadratic Trend

Method of Fitting Parabolic Curves Let (i) Y = a+bX +cX 2 be the equation of the parabola to be fitted to the given time series data (X1, Y1), (X2, Y2), · · · , ( X N, YN). Using Method of Least Squares, the normal equations for determining the constants a, b, care

EY EXY and EX 2Y

=

Na+ bEX + cEX 2, aEX + bEX 2 + cEX 3 aEX 2 + bEX 3 + cEX 4 .

If the mid-point of time is taken as the origin, then EX = 0, EX 3 normal equations reduce to EY EXY and EX 2Y

=

= 0, and the above

Na+cEX 2 , bEX 2 aEX 2 + cEX 4 •

Solving these equations, we get the values of a, band c, and substituting these values in (i), we obtain the trend equation. Example 14. Fit a parabolic curve of second degree to the data given below and estimate the value for 1999 and comment on it: Years Sales (in Rs. '00000)

1993 10

1994 12

1995 13

1996

1997

10

8

563

CHAP. 18: TIME SERIES ANALYSIS

Solution. Let (1) Y =a+bX +cX 2 be the equation of the parabolic curve to be fitted to the given data. By the method of least squares, the normal equations referred to mid-point of time as the origin are EY = Na+cEX 2 , EXY = bEX 2 and EX 2Y = aEX 2 + cEX 4 • Here N = 5. TABLE 18.16: CALCULATIONS FOR FITTING A PARABOLIC CURVE Sales (in Rs. '00000) Y

x

x2

x4

XY

x2y

1993

10

-2

4

16

-20

40

1994

12

-1

1

1

-12

12

13

0 1

0 1

0

0

0

1

10

10

Years

~

1995

/ /

1996

/

10

1997

8

2

4

16

16

32

Total

53= EY

0

10= EX 2

34 = EX 4

-6= EXY

94 = r;x 2 Y

Using the normal equations, we get 53 = -6 = 94

=

5a + c x 10 bx 10

,(3)

a x 10 + c x 34.

(4)

(2)

From (3), b = - 160 = -0.6. Multiplying (2) by 2 and then subtracting from (4), we get -12

= 14c,,

or, c = -

6

1

= -0.857.

From (2),

5a

= 53 -

IOc = 53 - 10 x (-0.857)

= 61.57;

:. a

= 12.314.

Hence, the required equation of the fitted parabolic curve is Y = 12.314 - 0.6X - 0.857 X 2 with origin at the middle year 1995 and unit of X as 1 year. In 1999, X = 4, y

= 12.314 -

0.6 x 4 - 0.857 x 16 = 12.314 - 16.112 = -3.798.

Negative trend value for the year 1999 indicates that the model (i.e., fitting parabolic curve) is not suitable for the given data.

18.8.2

Exponential Trend

The equation of the exponential curve is (1)

where X represents time, Y represents time series data and a, b are constants.

564

BUSINESS MATHEMATICS AND STATISTICS

In the curve (1), the given Y values form a G.P. while the corresponding X values form an A.P. Data from the fields of biology, banking and economics often exhibit exponential trend. (The growth of bacteria is exponential). In business, sales or earnings may grow exponentially over a short period. Taking logarithms of both sides of (1), we get

= log a + X log b.

log Y

When plotted on a semi-logarithmic graph, the curve will be a straight line. Taking y = log Y, A = log a and B = log b, the above equation reduces to y = A+BX. Using the method of least squares, the normal equations are Ey and EXy

NA+BEX AEX + BEX 2 •

= =

When deviations are taken from the middle year, these equations become Ey =NA and EXy = BEX 2 • A= Ey/N and B = EXy/EX 2 . Example 15. The sales of a company in thousand of rupees for the year 1995 through 2001 are given below: Year Sales

1995 32

1996 47

1997 65

1998 92

1999

2000

2001

132

190

275

Estimate sales figures for the year 2002 using an equation of the form Y = abx, where X = years and Y = sales. Solution. Taking origin at the middle year 1998, the normal equations for the exponential trend Y = abx are log a= Ey/N and logb = EXy/EX 2 ; where y = logY.

I TABLE 18.17: Year 1995 1996 1997 1998 1999 2000 2001 Total

CALCULATIONS FOR FITTING EXPONENTIAL TREND

Sales (Y) (Rs. '000)

x

y = logY

32 47 65 92 132 190 275

-3 -2 -1 0 1 2 3

1.5051 1.6721 1.8129 1.9638 2.1206 2.2788 2.4393

...

. ..

13.7926 '

= Ey

x2 9 4 1 0 1 4 9 28 = EX 2

Xy

-4.5153 -3.3442 -1.8129 0 2.1206 4.5576 7.3179 4.3237 = EXy

I

565

CHAP. 18: TIME SERIES ANALYSIS

Substituting the values of Ey, EX 2 , EX y in the normal equations, we get 13. 7926 4.3237 = 1.9704 and logb = - - = 0.1544. 7 28 [a = antilog (1.9704) = 93.42 and b = antilog (0.1544) From the given equation, we have loga

=

log Y =

For the year 2002, X

log a + X log b 1.9704 + 0.1544X.

= 1.4271

(1)

= 4; therefore, from (1), we get logY = 1.9704 + 0.1544 x 4 = 2.5880. :. Y

= antilog (2.5880) = 387.3.

Hence, the required sale figures for the year 2002 is Rs. 3,87,300.

18.8.3

Computation of Monthly Trend from Annual Tuend

Computation of Monthly Trend depend on the nature of the given annual data.

Case I. Let annual data for odd number of years be given. If Y = a + bX be the equation of the straight line trend fitted to the annual data with the origin at the middle year (on 30th June or 1st July) and unit of X = 1 year, then the equation of the straight line trend fitted to monthly averages is

y =

_!!:._

12

.!!__ x

+ 12

,

where a and b are divided by 12. Now, slope 2 of this trend line is the rate of change in monthly average with respect to 'x' for 12 months. So, monthly change will be 1~ • 1b2 , i.e., 1 ~ 4 and hence the monthly trend equation will be

t

y =

_!!:._

__!!_ x

12 + 144 , with origin at 30th June or 1st July and unit of X = 1 month.

Case II. Let annual data for even number of years be given. If Y = a+ bX be the equation of the straight line trend fitted to the annual qC.R. for Q2) + 100, Q4x C.R. for Q3) + 100, Q1 x C.R. for Q4) + 100.

(1) (2) (3) (4)

573

CHAP. 18: TIME SERIES ANALYSIS

Usually, the 2nd C.R. for Qi wi11 be either higher or lower than the 1st C.R. 100 for Qi dep~riling on the presence of an increase or decrease in trend. We can now adjust the chain

relatives for this trend. If d =2nd C.R. for Qi - 100, i.e., the difference between the 1st and 2nd C.R. for Qi, then we subtract !d, ~d, id, ~d from the chain relatives for Q2, Q3, Q4 and the 2nd C.R. for Qi respectively to obtain the adjusted chain relatives. These adjusted chain''relatives expressed as percentages of their A.M. give the required seasonal indices (See Example 4).

Example 20. Calculate Seasonal Indices by the method of link Example 18. (See data given for the four years 2002-05.}

relativ~s

from the data given in .

Solution.

I TABLE 18.24: CALCULATIONS OF AVERAGE LINK RELATIVES I

~ r

Qi

2002

...

2003

111.48

Link Relatives Q3 89.23 96.55

q:J

,92.65

100.00

Q4 108.93 106.35

2004

104.48

84.29

94.92

92.86

2005

115.38

91.67

92.73

113.73

Total

331.34

357.84

384.20

421.87

Average (A.M.)

110.45

89.46

96.05

105.47

The link relative (L.R.) for the 1st quarter Qi of the 1st year 2002 cannot be determined. For the other three quarters of 2002,

L.R. for Q2 = (58 + 65) x 100, L.R. for Q3 = (56 + 58) x 100, L.R. for Q4 = (61+56) x 100. Similarly, we determine the other link relatives. From the average link relatives obtained in the last row of the above Table 18.24, we now find chain relatives, taking 100 as the chain relative (C.R.) for Qi.

C.R. for Qi

=

100,

C.R. for Q2

=

(89.46 x 100) + 100 = 89.46,

[using definition (1)]

C.R. for Q3

=

(96.05 x 89.46) + 100 = 85.93,

[using definition (2)]

C.R. for Q4

=

(105.47 x 85.93) + 100 = 90.63,

[using definition (3)]

~r.id

=

(110.45 x 90.63) + 100 = 100.10.

[using definition (4)]

C.R. for Qi

:. d = 100.10 - 100 = 0.10; :. id= 0.025, ~d = 0.050, id= 0.075, ~d = 0.10. The adjusted chain relatives are respectively 100, 89.46 - 0.025, 85.93 - 0.05, 90.63 - 0.075, i.e., 100, 89.435, 85.88, 90.555, or, 100, 89.44, 85.88, 90.56.

574

BUSINESS MATHEMATICS AND STATISTICS

A.M. of the adjusted chain relatives The required seasonal indices are

= (100 + 89.44 + 85.88 + 90.56) + 4 = 91.47.

110.407 x 100, 89.44 x 100 85.88 x 100 90.56 x 100 91.47 , 91.47 ' 91.47 ' 9 i.e.,

109.33, 97. 78, 93.89, 99.00.

Note: The total of the seasonal indices is 400.

18.10

Deseasonalization of Data

Deseasonalization of data means elimination of seasonal variation from the original data. If the additive model is used, we first find trend (T) by the moving average method and it is removed from the original data leaving deviations from trend

Y-T=S+C+l. These deviations are then averaged to eliminate the cyclical fluctuation (C) and irregular movement (I). Thus, we obtain seasonal variations (S) which are suitably adjusted so that the total of the seasonal variations is zero. The adjusted seasonal variations are then subtracted from the corresponding data to get deseasonalized data. If the multiplicative model is used, seasonal indices are first determined by the ratio to moving average method and then seasonal effects (= seasonal index + 100) are found. The original data are now divided by the corresponding seasonal effects to give the deseasonalized data. Such data may include trend, cyclical and irregular movements.

In short, original data expressed as percentages of seasonal indices give the deseasonalized Jata. Example 21. A large company estimates its average monthly sales in a particular year to be Rs. 2,00,000. The seasonal indices (S.I.) of the sales data are as follows: Month

S.I.

Jan.

Feb.

Mar.

Apr.

May

June

July

Aug:

Sept.

Oct.

Nov.

76

77

98

128

137

122

101

104

100

102

82

Dec. 73

Using this information, draw up a monthly sales budget for the company (Assume that there is no trend). (Osmanla U. 1997 Type]

Solution. If Rs. x be the average monthly sales in a year and I, the seasonal index of a month, then estimated sales for the month = !% of x = Rs. x x 1{m.

575

CHAP. 18: TIME SERIES ANALYSIS

TABLE 18.25: CALCULATIONS OF MONTHLY SALES BUDGET Month

Seasonal Index

Estimated Sales (Rs.)

January

76

2, 00, 000 x l~~

= 1, 52, 000

February

77

2, 00, 000 x i7rio

= 1, 54, 000

March

98

2,00,000 x :~o

= 1,96,000

April

128

2,00,000 x !~~

= 2,56,000

May

137

2,00,000 x !~~

= 2,74,000

June

122

2,00,000 x !~~

= 2,44,000

July

101

2,00,000 x !g~

= 2,02,000

August

104

2,00,000 x !g~

= 2,08,000

September

100

2, 00, 000 x

October

102

2, oq, ooo x !g~

= 2, o4, ooo = 1,64,000

!gg = 2, 00, 000

"

November

82

2,00,000 x

8 1; 0

December

73

2, 00, 000 x

7 1 0

Totbl

1200

g = 1, 46, 000 24,00,000

Example 22. Deseasonalize the data given in Example 18 by the method of moving average using the additive model. Solution. Seasonal variations are first determined as shown in Example 18 by the method of moving average. Seasonals are 2.83, -0.50, -2.17, -0.17 for the 4 quarters, I, II, III, IV respectively. To deseasonalize the given data, the seasonals are subtracted from the corresponding quarterly data for each year.

576

BUSINESS MATHEMATICS AND STATISTICS

TABLE 18.26: CALCULATIONS OF DESEASONALIZED DATA Year/Quarter

Output ('000 tons) Y

I II III IV I II III IV I II III IV I

2002

2003

2004

2005 '

Seasonal Variations S

Deseasonalized Data Y - S

65

2.83

62.17

58

-0.50

58.50

56

-2.17

58.17

61

-0.17

61.17

58

2.83

55.17

63

-0.50

63.50

63

-2.17

65.17

67

-0.17

67.17

70

2.83

67.17

59.

-0.50

59.5

56

-2.17.

58.17

52

-0.17

52.17

60

2.83

57.17

II

55

-0.50

55.50

III

51

-2.17

53.17

IV

58

-0.17

58.17

Example 23. Deseasonalize the following sales data using a multiplicative model:

Quarter

Sales (Rs. '00000)

Seasonal Index

I

23.7

78

II

25.2

124

III

21.4

50

IV

65.4

148

. . Original Data Original Data Solution. Deseasonal1zed Data= S al Ef£ or S al d x 100, eason ect eason 1n ex where seasonal effect= (Seasonal Index)

+ 100.

TABLE 18.27: COMPUTATIONS OF DESEASONALIZED DATA ~

Quarter

-

Sale (Rs. 100900( Seasonal Index

Seasonal Effect

Deseasonalized Data

I

23.7

78

0.78

~~.;~ = 30.38

II

25.2

124

1.24

~~2~ = 20.32

III

21.4

50

0.50

lli 0.50 = 42 •80

IV

65.4

148

1.48

~~4: = 44.19

400

4.00

Total

Note: If the original data were not affected by seasonal variation, then sales in the four q)iarters I, II, III, IV would have been 30.38, 20.32, 42.80, 44.19 lakh rupees respectively.

577

GHAP. 18: TIME SERIES ANALYSIS

18.11

Measurement of Cyclical Fluctuation

To measure cyclical fluctuation, we first find trend (T) and seasonal variation (S) by any suitable method, say, by the method of moving average, and then eliminate them from the original data by using additive or multiplicative model. We are now left with only cyclical fluctuation (C) and irregular movement (/). Irregular movement (I) is then removed by using moving average of appropriate period depending on the average duration of I leaving only the cyclical fluctuation.

18.11.1

Estimation of Irregular Movement

Irregular Movement can be obtained by eliminating trend (T), seasonal variation (S) and cyclical fluctuation (C) from the original data. In practice, irregular movements are found to be of small magnitude.

18.11.2

Business Forecasting

Successful business activity demands a reasonably accurate forecasting of future business conditions upon which decisions regarding production, inventories, price-fixation, etc., depend. To eliminate guesswork, modern statistical methods are employed as a very useful tool of forecasting. The methods essentially consist of a knowledge of the past and present conditions and a proper analysis of time series data, isolating them into the four standard components, viz., Trend, Seasonal, Cyclical and Irregular movements which we have discussed in this chapter. Another useful aid in business-prediction is the construction of Index Numbers (discussed in detail in Chapter 17) associated with various business activities.

18.11.3

Business Barometers

A useful aid in practical f~casting is the construction of Index Numbers associated with different business activities, e.g., Wholesale prices, Consumer prices, Stock prices, Industrial production, etc. These terms are loosely referred to as Barometers in Business Statistics. Sometimes, the term Barometer is used to mean an indicator of the present economic situations or to designate an indicator of future conditions. Index numbers of different business activities may be combined to yield a Composite Business Activity Index. However, the study of general business conditions as revealed by this Index should be supplemented by special studies of individual businesses based on their respective Index Numbers.

EXERCISE ON CHAPTER 18 1 - - - - - - - - Theory 1. What do you mean by time series? Mention the chief components of time series. [C.U. B.Com.(H) 1989, '99, 2003;B.U. B.Com.(H) 2008]

2.

(a) Define trend. Enumerate the methods of determining trend in time series. (b) Explain what is meant by Secular Trend in time series analysis. Briefly mention the important tvpes of forces which influence an economic time series.

Bus. Math. & Stat. [C.U.]-37

578

BUSINESS MATHEMATICS AND STATISTICS

(c) Explain the object and utility of Time Series Analysis. 3.

[C.U. B.Com.(H) 2000]

(a) What is the time series? What is the need to analyze a time series? Enumerate the different methods of finding the secular trend. (b) Explain clearly what is meant by Time Series Analysis. [C.U. B.Com. (H) 1986]

(c) Enumerate the factors that explain the utility of analyzing time series. 4.

(a) What is a 'moving average'? What are its uses in Time Series? [C.U. B.Com. 2003]

(b) What are the merits and limitations of moving average technique for computing the 1 trend values? jC.U. B.Com.(H) 1988, '99, 2001] 5. What are the components of Time Series? With which component of a time series would you mainly associate each of the following: (a) A fire in a factory delaying production by four weeks; (b) An after-Puja sale in a department store; (c) The increased food production due to a constant increase in population; (d) A recession; (e) General increase in the sale of TV sets? [Ans. (a) Irregular Movement; (b) Seasonal Variation; (c) Secular Trend; (d) Cyclical Fluctuation; (e) Secular Trend.]

6.

(a) What do you mean by Seasonal Variation? Explain with a few examples the utility of such a study. [C.U. B.Com. 2008] (b) State the various methods of finding seasonal indices. (c) What is a seasonal index? Briefly explain the various methods of constructing such an index.

7. Give an account of the moving average method of trend determination. How does one choose the length of the moving average? 8. Briefly describe Link Relative Method for finding the seasonal variations. 9.

(a) What are the objectives of analyzing a time series?

[C.U. B.Com.(H) 1986, '87]

(b) Describe briefly the different components of a time series.

[C.U.B.Com. 1997]

10. Define Time Series and describe what is meant by analysis of time series. Briefly state the causes that bring about changes and the utility of analysis of time series. 11. What do you mean by Seasonal Fluctuations in time series? Give an example. What are the major uses of Seasonal indices in time series analysis? [C.U. B.Com.(H) 1990] 12. Explain the meaning of deseasonalizing data. What purpose does it serve? 13. Define a Time Series and state its main components. What are the different methods of determining the trend component of a time series?

579

CHAP. 18: TIME SERIES ANALYSIS

...

14. Distinguish between Seasonal, Cyclical and Random fluctuations. Describe any method of eliminating their influence. 15. (a) Give reasons for selecting a particular model for analyzing time series data. Explain how seasonal indices will be determined for a time series data. (b) Explain briefly, how the seasonal element in a Time Series data is isolated and eliminated. [CA May 1986] 16. Explain the nature of cyclical variations in a time series. How do seasonal variations differ from them? Give an outline of the moving average method of measuring seasonal variations. 17. Discuss some of the adjustments for population changes, calendar variation and price changes, which are necessary to make the time series data homogeneous and comparable. ' 18. Describe the ratio to moving average and the ratio to trend methods of estimating seasonal indices. Compare the two methods. 19. Write short notes on: (a) Moving Average;

[CA May 1987]

(b) '!fend; (c) Seasonal Variations; (d) Seasonal Indices;

[C.U. B.Com. 2004]

(e) Business Forecasting; (f) Business Barometers;

[CA May 1986]

(g) Components of a Time Series; (h) ·Usefulness of Seasonal Index.

[C.U. B.Com.(H) 1991]

Problems A 1. Fit a trend line to the following data by the free-hand method: Year Sales of a Firm (in million Rs.)

2.

1998

1999

2000

2001

2002

2003

2004

2005

(a) Draw a trend line by the semi-average method using the following data: Year Production of Steel (in '000000 tonnes)

2000 350

2001 360

2002 355

2003 365

2004 358

2005 363

(b) Plot the following data on a graph paper and ascertain trend by the method of semi-a~rages:

Year Production (in million to1mes)

1999 100

2000 120

2001 95

2002 105

2003 108

2004 102

2005 112

580

BUSINESS l\lATllEl\IATICS AND STATISTICS

~~~~~~~~~~~~~~~~~~~~~~~

3. Given the numbers 1, 0, - 1, 0, 1, 0, - 1, 0, 1; find the moving average of order four. 4.

(a) Obtain the 5-year moving averages for the following data: Year Annual Sales (in crore Rs.)

1997

1998

1999

2000

2001

2002

2003

2004

2005

36

43

43

34

44

54

34

24

14

Construct also the 4-year centred moving averages. (b) Construct 5-yearly moving averages of the number of students studying in a college shown below: Year

1991

1992

1993

1994

1995

1996

1997

1998

1999

2000

Number

332

317

357

392

402

405

410

427

405

431 [I.S.C. 2000]

5.

(a) Given the numbers 2, 6, 1, 5, 3, 7, 2. Obtain moving averages of order 3.

(b) Find the trend for the following series using a three-year moving average: Year Values

1

2

3

4

5

6

2

4

5

7

8

10

[B.U. B.Com.(H) 2008

(c) Using 3-year moving average method determine the trend and short-term fluctuations for the following data: Year Values

1991

1992

1993

1994

1995

1996

1997

21

34

45

28

40

57

73 [C.U. B.Com.(H) 1999]

(d) Find the 3-year weighted moving average with weights 1, 4, 1 for the following series: Year Values 6.

1

2

2

6

3

4

5

6

7

5

3

7

2

(a) The table below gives the figures of production of a commodity during the years 2001-05 in the State of Punjab: Year (X) Production (Y) (in '000 tons)

2001

2002

2003

2004

2005

10

12

8

10

14

Use the method of least squares to fit a straight line to the data. From this result find the trend values for different years (Take 2003 as origin of X -series). (b) Fit a least square trend line to the following data: Year Volume of Sale (in suitable units) Estimate the volume of sale for 2000.

1994

1995

1996

1997

1998

1999

12

15

17

22

24

30

[C.U. B.Com.(H) 2000]

581

CHAP. 18: TIME SERIES ANALYSIS

(c) Fit a suitable straight line to the following data by the method of least squares and_ estimate the percentage of insured people in 1997: Year

1989

1990

1991

1992

1993

Percentage of Insured People

11.3

13.0

9.7

10.6

10.7

[C.U. B.Com.(H) 2001] [Hints: See worked-out Example 10. ttere-n = 5, a=~= 5 ~· 3 = 11.06, b = ~~1;'

= -~·g = -0.36.]

(d) Fit a linear trend equation by the method of least squares from the following data and estimate the trend value for the year 2004: Year

1997

1998

1999

2000

2001

2002

Price (Rs.)

250

207

228

240

281

392 [C.U. B.Com.(H) 2004]

7.

(a) Find the values of the trend ordinates by the method of least squares from the data given below: Year

1991

1992

1993

1994

1995

1996

1997

Sales (Rs. '000)

125

128

133

135

140

141

143

[C.U. B.Com.(H) 1999 Type] [Hints: Fit the line Y = a+ bX to the given data, taking origin at the middle year 1994 and unit of X as 1 year.]

(b) Fit a straight line trend by the method of least squares (taking 1995 as the year of origin) to the following data concerning sales of a certain firm: Year

1991

1992

1993

1994

1995

1996

1997

1998

1999

48

55

63

65

72

84

90

87

82

Sales ('000 Rs.)

[Hints: Fit the line Y =a+ bX to the given data, taking origin at the middle year 1995 and unit of X as 1 year.]

(c) Fit a straight line trend equation by the method of least squares from the following data and estimate the profit for the year 2005: Year Profit (Rs. lakh)

1975

1980

1985

1990

1995

2000

10

13

15

20

22

28 [C.U. B.Com. 2005]

8. Calcu1ate the seasonal indices in the cases of the following quarterly data in certain units using-the method of averages: (a) Total Production of Paper ('000 tons) Year

2003 2004 2005

Quarters

I 37 41 35

II 38 34 37

III 37 25 35

IV 40 31 41

BUSINESS MATHEMATICS AND STATISTICS

Year

(b)

2002 2003 2004 2005

Qi

39 45 44 53

Quarters Q2 Qa

21 23 26 23

52 63 69 64

Q4

81 76 75 84

[Hints: S.I. = (average for each quarter x 100)

+ grand average)]

9. Calculate trend values from the following data relating to the production of tea in India by the moving average method, on the assumption of a four-yearly cycle: Year

Production (million lb)

1996 464

1997 515

1998 518

10. Fit a second degree parabola (Y = a + bX Year

1999 467

2000 502

2002 557

2003 571

2004 586

2005 612

+ cX 2 ) to the following data:

2001

2002

1

5

Purchases [in crore (Rs.)]

2001 540

2003 10

2004 22

2005 38

11. Find the seasonal indices by the m2thod of moving averages from the following series of

observations:

Quarter

Sales of Woollen Yarn ('00000 Rs.) 2002 2003 2004 2005

100 93 79 98

97 88 76 94

I II

III IV

106 96 83 103

110 101 88 106

12. Find the trend by the method of moving averages from the following table:

I Total Production of cement (1,00,000 tons) I Year

1990 1991 1992

Quarter

I

34 37 43

II

III

IV

32 34 40

31 33 38

36 41 48 [C.U. B.Com. 1994]

583

CHAP. 18: TIME SERIES A~ALYSIS

B 1. Fit a trend line to the following data by the Least Squares Method:

Year Production (in '000 tonnes)

1995

1997

1999

2001

2003

18

21

23

27

16

Estimate the productions in 2000 and 2005. [Hints: Take 1999 as the origin and unit of X as 2 years.]

2. Fit a straight line trend to the data and estimate the profit for the year 1987:

Year Profits (Rs. lakh)

1980

1981

1982

1983

1984

1985

1986

60

72

75

65

80

85

95 [ICWAI June 1989]

3. Obtain the straight line trend equation and tabulate against each year after estimation

of the trend and short-term fluctuation: Year Value

1997

1998

1999

2000

2001

2002

2003

2004

2005

380

400

650

720

690

620

670

950

1040

4. Fit a straight line trend equation by the method of least squares and estimate the value for 2005.

Year Value

1996

1997

1998

1999

2000

2001

2002

2003

380

400

650

720

690

600

870

930

[C.U.B.Com. 2008 Type]

5. The following series of observations is known to have a business cycle with a period of 4 years. Find the trend values by the moving average method: Year

1995

1996

1997

1998

1999

2000

2001

2002

2003

2004

2005

Production- ('000 tonnes)

506

620

1036

673

588

696

1116

738

663

773

1189

6. Calculate the Seasonal Index from the following data using the average method: Year

1st Quarter

2nd Quarter

3rd Quarter

4th Quarter

2001

72

68

70

2002

76

70

80 82

2003

74

84 84

80

86

2004

76

66 74

2005

78

74

74 78 82

584

BUSINESS MATHEMATICS AND STATISTICS

7. Using 4-quarterly moving average in respect of the following data, find: (a) the trend, {b) short-term fluctuations, (c) seasonal variations: Year

1st Quarter

2nd Quarter

3rd Quarter

4th Quarter

2001 2002 2003 2004 2005

35 38 47 61 72

86 109 158 177 206

67 91 104 134 141

124 176 226 240 307

8. Obtain the seasonal indices by the ratio to moving average method from the following data: Year

I

2002 2003 2004

101 106 110

Quarters II III

IV

93 96 101

98 103 106

79 83 88

9. Construct indices of seasonal variations from the following time series data on consumption of cold drinks which contain only seasonal and irregular variations, using multiplicative model: Year

I

2002 2003 2004 2005

90 75 80 85

Quarters II III IV

75 80 75 82

87 78 75 80

70 75 72 81

(in '000 bottles). 10. Fit a straight line trend by the method of least squares to the following data and find by which year the production will reach 63 million tons: Year

Production (million tons)

1998 50.3

1999 52.7

2000 49.3

2001 57.3

2002 56.8

2003 60.7

11. Obtain seasonal indices for the following data:

~ n

Summer Rain Winter

Output in Thousand Units 2001 2002 2003 2004 2005

31 39 45

42 44 57

49 53 65

47 51 62

51 54 66

2004 62.1

2005 58.6

585

CHAP. 18: TIME SERIES ANALYSIS

12. Bank debits giving a good measure of volume of transactions are given below for 2002-05. Find the seasonal indices by the method of moving averages using additive model: Year

Bank Debits in million Rs. 4th 1st 2nd 3rd

2002

16.00

2003

15.90

2004 2005

16.30 17.10

13.50 12.20

14.70 15.60

17.00 18.10

11.90

16.90

19.20

13.20

15.00

18.70

13. Agricultural outputs, in million tonnes, for 5 years are given below: Years Outputs

1999

2000

2001

2002

2003

80

85

87

93

100

Obtain a least square linear estimate of the output for the year 2005. 14. Fit a linear trend equation to the following data. Hence, estimate the value of sales for the year 2005:

15.

Years

1999

2000

2001

2002

2003

Sales (in lakh of Rs.)

100

120

140

160

180

(a) Fit an equation of the form Y =a+ bX + cX 2 to the data given below:

I; I

1

2

25

f

14~ I

3

28

f

3313:

(b) Fit a quadratic trend to the following series of production data: Years Production (Y)

1999

2000

2001

2002

2003

2004

2005

37

38

37

40

41

45

50

Y-values being the average production in thousand tons. 16.

(a) Fit an exponential trend of the form Y figures below:

= abx to the data giving the population

Census Year

1945

1955

1965

1975

1985

1995

2005

Population (in crore)

25.0

25.1

27.9

31.9

36.1

43.9

54.7

Estimate the population figures for the year 2015 using the fitted exponential trend equation. (b) The following table gives the profits of a concern for 5 years ending 2005: Years Profit (in Rs. thousand)

Fit an equation of the type Y = abx.

2001

2002

2003

2004

2005

1.6

4.5

13.6

40.2

125.0

586

BUSINESS MATHEMATICS AND STATISTICS

c 1. Fit by the method of least squares a linear equation y

= mx + c to the following data:

Years

1998

1999

2000

2001

2002

2003

2004

2005

Production ('00000 tons)

9.7

10.1

10.2

10.7

11.9

12.9

12.4

14.8

Obtain the trend values for the years 2002 and 2003. 2. Construct a 4-year centred moving averages from the following data: Year

1945

1955

1965

1975

1985

1995

2005

Imported Cotton consumption in India (in '000 bales)

129

131

106

91

95

84

93

3. Calculate the trend values by the method of moving averages, assuming a four-yearly cycle, from the following data relating to sugar production in India: Year

1994 1995

1996 1997 1998

1999 2000 2001

2002

2003

2004 2005

Sugar Production (Jakh tonnes)

37.4

38.7

42.6

58.4

38.6

51.4

31.1

39.5

47.9

48.4

64.6

84.4

4. Fit a parabolic curve of second degree y = a + bx + cx 2 to the data given below by the method of least squares:

Year (x) Import (y) (in '000 bales)

2001

2002

2003

2004

2005

10

12

13

10

8

[Take 2003 as origin and unit of x as 1 year. J 5. Deseasonalize the following data with the help of the seasonal data given against: Month

Jan.

Feb.

Mar.

Apr.

May

June

Cash Balance ('00000 Rs.)

360

400

360

360

550

Seasonal Index

120

80

350 110

90

70

100

[Hints: Deseasonalized value= ~ x 100.J

6. Obtain the seasonal indices by the method of moving averages (using additive model) from the following data:

Quarter

Quarterly Output ('00000 tons) 2002 2003 2004 2005

I

31

42

49

47

II

39

44

53

51

III

45

57

65

62

IV

36

45

55

50

\

587

CHAP. 18: TIME SERIES ANALYSIS

7. Using additive model, estimate the seasonal indices by the method of moving averages from the table given below. Deseasonalize the given production figures with the help of the seasonal indices obtained, and explain the significance of the deseasonalized data: Total Production {'00000 tons) Quarters

Year

2003 2004 2005

I

II

III

IV

37 41 35

38 34 37

37 25 35

40 31 41

8. D.eseasonalize the following production data by the method of moving averages:

Quarters

2002

Quarterly Output ('000 tons) 2003 2004 2005

49 50 61 20.

30 49 50 35

I II III

IV

75 79 65 70

35 62 60 25

9. A company estimates its average monthly sales in a particular year to be Rs. 2,00,000. The seasonal indices (S.I.) of the sales data are as follows: Month

S.I.

Feb. 77

Jan.

76

Mar.

April

May

98

128

137

June 122

Assuming that there is no trend, draw up a monthly sales budget for the months January to June. [C.U. B.Com.(H) 1986] 10. A large manufacturing company estimates its average monthly sales in a particular year

to be Rs. 20,00,000. The seasonal indices of the sales data are given below: Month

S.I.

Jan.

78

Feb. 75

Mar.

April

May

June

July

Aug.

Sept.

Oct.

Nov.

Dec.

100

126

138

121

101

104

99

103

80

75

Ignoring the possible existence of a trend, use the above information to draw up a monthly sales budget for the company. 11. The following data gives....the value of sales of a company for the years 1995-2005. Year (X)

1995

1996

1997

1998

1999

2000

2001

2002

2003

2004

2005

Sales (Y) (Rs. '000)

50.0

36.5

43.0

44.5

38.9

38.l

32.6

38.7

41.7

41.l

33.8

Use the method of least squares to fit a straight line trend to the data given above. Compute the trend values for 1998 and 2003 (Take X = 0 for the year 1999 and the unit of X as 1 year).

588

BUSINESS MATHEMATICS AND STATISTICS

12. Calculate the quarterly seasonal indices in respect of the following data by using the simple average method: Year

Quarter I

Quarter II

Quarter III

Quarter IV

2001 2002 2003 2004 2005

71

68 69 66 73 74

79 82 84 84 86

71

76 74 76 78

74 80 78 82

13. Use the method of monthly averages to find the monthly indices for the following data of production of a commodity for the years 2003, 2004, 2005: PRODUCTION IN LAKHS OF TONNES Months January February

2003

2004

2005

12

15 14 13 16 16 15 17 12 13 12 13 14

16 15 14 16 15 17 16 13

11

March

10

April

14 15 15 16 13

May June July August September

11

October

10

November

12 15

December

10 10 11

15

[Hints: First find the total for each of the 12 months and divide each of these totals by 3 (no. of years) to find the monthly averages Xi, X2, · · · , X 12 for 12 months. Then find the grand average G of these 12 monthly averages. The required monthly indices are X1 - G, X2 - G, · · ·, X12 - G.J

14. Calculate the seasonal indices by the method of link relatives from the following data:

Quarter I II III

IV

[Hints: See worked-out Example 20.J

Year 2004

2002

2003

75 60 54 59

86 65 63 80

90 72 66 82

2005

100 78 72 93

589

CHAP. 18: TIME SERIES ANALYSIS

ANSWERS A 3. 4.

5.

o, o,

0, 0, 0.

(d) y = 266.33 + 13.49x; Rs. 387.74.

(a) 40.0, 43.6, 41.8, 38.0, 34.0, 40.0, 42.38, 40.25, 35.25;

1.

(b) 360, 374.6, 393.2, 407.2;•409.8, 415.6 for 1953-58.

(b) Y = 71.778 + 5.0833X; Rs. 97.1943 thousand;

(a) 3, 4, 3, 5, 4;

(c) y = 18 + 1.743x; unit of x = 2.5 years; Rs. 30.20 lakh.

(b) 3.67, 5.33, 6.67, 8.33, 10.33;

6.

(a) Y = 135 + 3.107X, 125.679, 128.786, 131.893, 135, 138.107, 141.214, 144.321,

(c) -, 33.33, 35.67, 37.67, 41.67, 56.67, -; -, 0.67, 9.33, -9.67, -1.67, 0.33, -;

8.

(a) 1.75, 0.42, -3.59, 1.42;

(d) 4.5, 3.5, 4.0, 4.0, 5.5.

9. 495.8, 503.6, 511.6, 529.5, 553.0, 572.5.

(b) 86.4, 44.4, 118.4, 150.8. 10. Y = 10.78+9.lX +2.21X 2 with origin at 1996.

(a) Y = 10.8 + 0.6X; 9.6, 10.2, 10.8, 11.4, 12.0;

11. 10.12, 0.13, -14.08, 3.83 using additive model.

(b) Y = 20 + l.74X; 32.2 units;

12. 33.625, 34.25, 34.75, 35.625, 37, 38.5, 39.875, 41.375 in lakh tons.

(c) y = 11.06 - 0.4x; 8.18;

B 1. Y = 21 + O.lX; 21,100 tons and 21,600 tons.

30.50, -31.50, 86.37, -84. 75, 25. 75, -20.38, 80.62, -91.88, 32.87;

2. y = 76 + 4.857 x; Rs. 95.428 lakh.

(c) Seasonal indices: -74.625, 25.345, -19.155, 68.435.

3. Y = 680 + 70.5X, -18.0, -68.5, 111.0, 110.5, 10.0, -130.5, -151.0, 58.5, 78.0.

8. 110.9, 99.9. 84.9, 104.3.

4. Yi = 655 + 35.83X referred to mid-point of 1983-84 as origin and unit of X as 6 months; 1049.13.

9. 104.8, 99.0, 101.6, 94.6. 10. Y = 55.975 + 0.825X; Year 1994.

5. 719.00, 738.75, 758.25, 776.38, 793.88, 812.88, 831.63. The trend values for the 1st two and the last two years cannot be determined.

12. 0.77, -3.31, 0.11, 2.43 in million Rs.

6. 96.4, 92.1, 106.9, 100.5.

13. 108.2.

1.

14. Y = 140 + 20X; 220 lakh rupees.

11. -4.49, -2.27, 6.76, in '00 units.

(a) Trend: 78.38, 81.63, 87.50, 97.00, 104.63, 111.88, 119.33, 127.50, 135.50, 139.63, 145. 75, 151.25, 154.38, 159.38, 163.88, 173.13;

15.

(a) Y = 22. 78 + 1.46X + 0.64X 2 ; (b) Y = 39.181+2.036X + 0.488X 2 •

16.

(b) Short-term fluctuations: -11.38, 42.37, -49.50, 12.00, -13.63, 64.12, -72.63,

(a) Y = 33.56(1.142)x; 157.15 crore. (b) Y = 13.79(2.982)x.

c 1. y = 0.336x + 11.588 referred to mid-point of 1995-96 as origin and unit of x as 6 months; 11.92, 12.60 (in '00000 tons). 2. 110.000, 99.875, 92.375 (in '00000 bales). 3. 151.95, 162.95, 173.55, 190.95, 208. 75, 212, 211.50, 222.90.

4. y = 12.314 -0.6X -0.857X 2 with origin at 1975 and unit of x as 1 year. 5. 300, 500, 500, 400, 500, 550 (in '00000 Rs.) (using Multiplicative Model). 6. -3.655, -1.615, 8.385, -3.115 (in '00000 tons).

590 7. Seasonal indices ('00000 tons): 3.12, 0.69, -4.25, 0.44. Deseasonalized data: 33.88, 37.31, 41.25, 39.56, 37.88; 33.31, 29.25, 30.56, 31.88, 36.31, 39.25, 40.56. Significance: If the climatic conditions due to change of season or the festivals am:! customs in a year had not affected the production, the expected production would be the above deseasonalized data. 8. Deseasonalized data ('00000 tons): 28.28, 38.70, 39.28, 57.75, 47.28, 39.70, 50.28, 42.74, 33.28, 51.70, 49.28, 47.74, 73.28, 68.70, 54.28, 92.74. 9. Rs. 1,52,000, 1,54,000; 1,96,000; 2,56,000,

BUSINESS MATHEMATICS AND STATISTICS

2,74,000, 2,44,000; 10. Estimated monthly sales (Rs. lakh): 15.6, 15.0, 20.0, 25.2, 27.6, 24.2, 20.2, 20.8, 19.8, 20.6, 16.0, 15.0. 11. Y = 39.9-0.767x, where x = X -1973; 41.434, 37.599 (in Rs. '000).

12. -1.25, -6.25, 6. 75,

o. 75.

13. 104.9, 97.5, 90.3, 112.3, 112.3, 114.6, 119.5, 92.5, 82.9, 78.0, 87.8, 107.3. 14. 124.2, 93.5, 82.5, 99.8.

Chapter 19

Measures of Association of Attributes

19.1

Introduction: Attributes

We know that statistics deals with quantitative phenomenon only and quantitative character may arise in any of the two ways given below. (i) We may measure magnitude or size of some phenomenon, e.g., we may measure the height of students of a first year class, their weight, etc. Similarly, we may study the wage structure of the .workers of a factory, the amount of rainfall in a month. We can measure these data quantitatively. These types of data are known as statistics of variables. The various statistical techniques like measures of central tendency, dispersion, correlation deal with such variables. (ii) Certain phenomena like blindness, deafness, etc. are not capable of direct quantitative measurement. In s_uch cases, we use quantitative character only indirectly in the process of counting. For example, we can find out of 2000 persons, how many are blind and how many are not blind, but we cannot measure blindness (or deafness). In such a case, we can study only the presence or absence of a particular characteristic called statistics of attributes. If we deal with statistics of attributes, we have to classify the data. This classification is done on the basis of presence or absence of particular .attributes or characteristics. When we study only one attribute, two class are formed-one possessing that attribute and another not possessing it. For example, if we study the attribute employment, two classes shall be formed: those who are employed and those who are not employed. If two attributes are studied, four classes shall be formed. If, besides employment, we study the sex-wise distribution, four classes shall be formed: number of males employed, number of females employed, number of males unemployed and number of females unemployed.

19.2

Notation and Class Frequencies of Attributes

We use certain symbols to represent different classes and their frequencies. Capital letters A and B are used to represent the presence of the attributes and the Greek letters 'o:' (alpha) and 591

59~

BUSINESS MATHEMATICS AND STATISTICS

'/3' (beta) respectively to represent absence of the attributes. Thus, 'o:' =not A and '/3' =not B. For example, if A represents males, then 'o:' will represent females. Similarly, .if B represents literates, then f3 will denote illiterates. The different attributes are denoted by AB, A/3, o:B and o:/3. In this example, (AB) will mean number of literate males and (o:/3) the number of illiterate females. The number of observations in each class is called 'class frequency'. Thus, if the number of literate males is 50, the frequency of class (AB) is 50. Class frequencies of AB, A/3, o:B and o:/3 are denoted respectively by (AB), (A/3), (o:B) and (o:/3) in which attributes are written in first brackets. Here,

(A) denotes number of individuals possessing attribute A.

(AB) denotes the number of individuals possessing attributes A and B. (a/3) denotes number of individuals possessing attributes o: and /3, i.e., not possessing attributes A and B. (A/3) denotes the number of individuals possessing attribute A but not B, etc. The number of attributes in a class gives the order of the class. A class having one attribute is of first order, two attributes is of second order and so on. We shall denote total numbet1w observations by N which is of zero order. Thus, (A), (B), (o:), (/3) are frequencies of first order; (AB), (o:B), (A/3), (o:/3) are frequencies of second order. We may say (A), (B) are frequencies of positive class or attributes and (o:), (/3) are that of negative class or attributes.

Number of Class Frequencies In a study of n attributes, the total number of class frequencies is 3n. For one attribute, the number of frequencies is 3 1 = 3, for two attributes, the number frequencies is 32 = 9 which are in the order 1+4 + 4 = 9. We can express any class frequency in terms of class frequencies of higher order, since N =total number of observations= (A)+ (o:) or (B) + (/3). Similarly, (A) = (AB)+ (A/3) and (o:) = (o:B)

+ (o:/3)

which are ultimate class frequencies with two attributes A and B. The classes of the highest order are the ultimate class-frequencies. A given data can be completely determined if only the ultimate class frequencies are known. The total number of classes of ultimate order of n attributes is 2n. For two attributes, the number of classes of ultimate order is 22 = 4, for three attributes, it is 23 = 8. The frequencies of the positive, negative and ultimate classes can be found with the help of a 2 x 2 contingency table or nine square table shown below. [It is called nine square table, because .nine squares are formed in it.] A

Total

B

(AB)

(o:B)

(B)

/3

(A/3)

(o:/3)

(/3)

Total

(A)

(o:)

N

CHAP. 19:

MEASUR~F

593

ASSOCIATION OF ATTRIBUTES

7 From this table, we obtain certain relationships

Vertical total

Horizontal total

+ (A/3) = (A) (o:B) + (o:/3) = (a) (B) + (/3) = N

(AB)+ (o:B) = (B)

(AB)

(A/3)

+ (o:/3) =

(/3)

(A)+ (a) = N.

From the above results, we can write

(AB) = (A) - (A/3), (A/3) = (A) - (AB) N - (/3) = B, (o:B) = (B) - (AB) N - (a)

N

= =

= (AB)+ (A/3) + (o:B) + (a/3) (A) +(a), or, N = (B) + (/3). (A), N

To find out whether the given data are consistent or not, we have to find whether any one or more of the ultimate class frequencies is negative or not. If all the ultimate class frequencies are positive, then the given data are consistent. But if any of the ultimate class frequencies is found to be negative, then the given data are inconsistent.

19.3

Illustrative Examples

Example 1. From the following data find out the missing frequencies:

(AB)= 300, (A)= 500, N

= 1400,

(B)

= 800.

Solution. Putting these values in the nine-square table, we get

A

0:

Total

(A1'1)

(o:B)

(B)

300

500

800

(A/3)

(o:/3)

(/3)

200

400

600

(A)

(a)

N

500

900

1400

B

/3

Total Calculations for finding missing frequencies are shown below:

(A/3) = (A) - (AB) = 500 - 300 = 200 (o:B) = (B) - (AB) = 800 - 300 = 500

(o:/3) = (/3) - (A/3) = 600 - 200 = 400 (o:) = N - (A) = 1400 - 500 = 900 I (/3) = N - (B) = 1400 - 800 = 600. !us. Math~ & Stat. [C.U.)-38

594

BUSINESS MATHEMATICS AND STATISTICS

Example 2. Prom the following two cases, find out whether the data ar.e consistent or not. .

= 100, = 100,

Case I. (A) Case II. (A)

{B)

(B)

= 150, = 150,

(AB) (AB)

= 60, N = 500 = 140, N = 500. [B.Com. Madras, 1993)

Solution. Case I. Given, (A) = 100, (B) = 150, (AB) in the nine-square table, we get

= 60,

A

N

= 500.

Substituting the values

Total

B

(AB) 60

(aB) 90

(B) 150

/3

(A/3) 40

(a/3) 310

(/3) 350

Total

(A) 100

(a) 400

N 500

From the above table, we get

=

(A/3)

(A) - (AB)

= 100 -

60 = 40

(aB) = (B) - (AB) = 150 - 60 = 90 (a/3) = (a) - (aB) = 400 - 90 = 310 (/3) = (A/3)

+ (a/3) =

(a) = (aB)

+ (a/3)

40 + 310 = 350

= 90 + 310 = 400.

Since all the ultimate class frequencies are positive, the given data are consistent. Case II. Given, (A) = 100, (B) = 150, (AB) = 140, N = 500. Putting these values in the nine-square table we can find out the missing frequencies as shown below:

A

Total

B

(AB) 140

(aB) 10

(B) 150

/3

(A/3) -40

(a/3) 390

(/3) 350

Total

(A) 100

(a) 400

N 500

From the table, we get the missing frequencies as shown below.

(A/3) (aB)

= (A) =

(AB) = 100 - 140 = -40 (B) - (AB)= 150 -140 = 10

(a/3) = (a) - (aB) = 400 - 10 (/3)

= (A/3) + (a/3)

(a) = (aB)

= 390

= -40 + 390 =

+ (a/3) =

350

10 + 390 = 400.

From the above values, we see that one of the ultimate class frequencies, i.e., (A,8) is negative and hence the given data are inconsistent. 1

595

CHAP. 19: r.{EASURES OF ASSOCIATION OF ATTRIBUTES

19.4

Association and Disassociation

Usually, we say that two attributes A and B are associated if they appear together in a number of cases. In statistics, A and B are said to be associated only if they are independent. Otherwise, they are disassociated. In case of second order frequencies, two attributes A and B are

> (A,B)(aB); if (AB)(a,8) < (A,B)(aB).

(i) associated if (AB)(a,8) and (ii) disassociated

Types of Association Two attributes .A. and B will have (i) Positive association if (AB)

> (A~B).

• t•ion 1"f (AB) (1·1·) N egat•1ve assoc1a

A and B will

b~


and (iii) negatively associated if (a{3)

(a~);

< (a~).

Expectation Expectation is the product of the probability and the number of observations. Probability of (A) =

(~).

Similarly for B:

Probability of (A) and (B) combined=

(~)

x

(!).

Expectation of (A) and (B) combined = N x (A) x (B) = (A)

N

N

X

N

(B)

Example 3. Find out whether attributes (i) AB, (ii) A{3, (iii) aB and (iv) a{3 are independent,

associated or disassociated from the following data: N = 1000, (A) = 400, (B)

= 800 and (AB) = 300.

596

BUSINESS MATHEMATICS AND STATISTICS

Solution. Let us construct the nine-square table by using the given .information: A B

f3 Total

Total

(AB)

(aB) 500

(B)

300

(Af3) 100

(af3) 100

((3) 200

(A)

(a) 600

N 1000

400

800

From the above table, we get

(Af3) = 100, (aB) = 500, (af3) = 100, (a) = 600, ((3) = 200. (i) For A and B, we have (AB)(af3) = 300 x 100 = 30, 000. 4 (A)(B) = oo x 800 = 320 and (AfJ)(aB) = 100 x 500 = 50, 000. N 1000 We see that (AB)(a,8)

< (A,8)(aB), since 30,000 < 50,000.

Hence, A and J3 are disassociated or negatively associated. Now (AB)

= 300 and

(A)(B)

N

= 320 => (AB)

-:j:. (A)(B).

N

Hence, A and B are not independent.

(ii) For A and (3, we have (Af3) = 100 and

We see that {A,13)

>

(A)(,13) N

(A~f3) =

40

~~~00

= 80.

.

, smce 100 > 80.

Hence, attributes A and ,13 are positively associated. (iii) For a and B, we have

(aB) = 500 600 x 800 (a)(B) . = 480 => (aB) > N , smce 500 1000 Hence, the attributes a and B are positively associated. (a)(B)

and ~ =

. (a)(f3) 600 x 200 (1v) For a and (3, (af3) = 100 and ~ = = 120. 1000 (a)(,13) . :. (a,8) < N , smce 100 > 120. Hence, a and ,13 are negatively associated or disassociated.

> 480.

597

CHAP. 19: MEASURES OF ASSOCIATION OF ATFRIBUTES

19·:5

Yule's Coefficient of Association

Yule's Coefficient of Association is the most popular tool in studying association between two attributes. This formula (or measure) is used to determine the nature of association, i.e., whether the attributes are eositively or negatively associated or independent. It also indicates the degree or extent to which the two attributes are associated. The Yule's Coefficient of Association is denoted by the symbol Q and is given by the formula: (AB){o:/3) - (A/3)(o:B) Q = (AB)(o:/3) + (A/3)(o:B). If the value of Q lies between -1 and. +1. If Q = 1, then there is perfect positive association between the attributes. If Q = -1, there is perfect negative association (or perfect disassociation) between the two attributes. If Q = 0, then the two attributes are

independent.

Illustrative Examples Example 4. Find Yule's Coefficient of Association by using the following information and interpret the result:

(AB)

= 40,

(A,B)

= 3,

(aB)

Solution. Yule's Coefficient of Association (Q) = (AB)(a,B) - (A,B)(aB) (AB)(a,B) + (A,B)(aB) 40 x 12 - 3 x 33 480 - 99 = 40 x 12 + 3 x 33 = 480 + 99

= 33,

(a,B)

= 12.

-'----'-~--'---'-~'--'-__.;..

381

= 579 = +o. 66

(approx.)

This result indicates that there is a positive association between the two attributes A

andB. Example 5. You are given for two attributes A and B the following data:

(AB)

= 8,

(A)

= 18,

(a,B)

=5

and N

= 35.

Calculate the Coefficient of Association between A and B and interpret your result. Solution. We first construct the nine-square table using the given information.

A B

Total

Total

8 (AB)

12 (aB)

20 (B)

10 (A,B)

5 (a,B)

15

(,B)

18 (A)

17 (a)

35 N

598

BUSINESS MATHEMATICS AND STATISTICS

(A,6) = (A) - (AB) = 18 - 8 = 10

(,6) = (A,6) + (a,6) = 10 + 5 = 15

(B) = N - (,6) = 35 - 15 = 20 (aB) = (B) - (AB) = 20 - 8 = 12

(a)= (aB)+(a,6)=12+5=17 Yule's Coefficient of Association is Q = (AB)(a,6) - (A,6)(aB) = 8 x 5 - 10 x 12 = 40 - 120 = -80 = -O.S. (AB)(a,6) + (AB)(aB) 8 x 5 + 10 x 12 40 + 120 160 This result shows that there is negative association between A and B. Example 6. Given N = 2500, (A) = 420, (AB) = 85, (B) = 670. Prepare 2

x 2 contingency

table and compute Yule's Coefficient of Association and interpret the result.

Solution. Let us construct the 2 x 2 contingency table by using the given data. (A)

(a)

Total

(B)

85 (AB)

585 (aB)

670 (B)

(,6)

335 (A,6)

1495 (a,6)

1830 (,6)

Total

420 (A)

2080 (a)

2500 N

Q = (AB)(a,6) - (A,6)(aB) = (85 x 1495) - (335 x 585) (AB)(a,6) + (A,6)(aB) (85 x 1495) + (335 x 585) 127075 - 195975 -68900 = 127075 + 195975 = - 323050 = -0. 2 1. The negative value of Q indicates that there is a negative association between the attributes A and B. Example 7. Eighty-eight residents of an Indian city, who were interviewed during a sample survey, are classified below according to their smoking and tea drinking habits. Calculate Yule's Coefficient of Association and comment on its value.

Smokers

Non-Smokers

40

33 12

Tea Drinkers Non-tea Drinkers

3

Solution. Let A denotes habit of smoking. Then a will denote habit of non-smoking. Let B denotes habit of tea drinking. Then

/3 will denote habit of not drinking tea.

599

CHAP. 19: MEASURES OF ASSOCIATION OF ATTRIBUTES

The given data in terms of these symbols are:

(AB) = number of smokers and tea drinkers= 40 (Aj'.3) = number of smokers and non-tea drinkers= 3

(aB) = number of tea drinkers and non-smokers= 33

(aj'.3) = number of non-smokers and non-tea drinkers= 12. Yule's Coefficient of Association is

Q = (AB)(aj'.3) - (Aj3)(aB) (AB)(aj'.3) + (Aj3)(aB). Substituting the values of (AB), (Aj'.3), (aB) and (aj'.3) in this formula, we get (40 x 12) - (3 x 33) 480 - 99 381 0 658 Q = (40 x 12) + (3 x 33) = 480 + 99 = 579 = · · This shows that the attributes tea-drinking and smoking are positively associated, i.e., most of the smokers are tea drinkers. Example 8. The following data relate to literacy and unemployment in a group of 500 persons. You are required to calculate Yule's Coefficient of Association between literacy and unemployment. Interpret it.

Literate unemployed

220

Literate employed

20

Illiterate employed

180

Solution. Let A denote literacy, B denote unemployment. Then a denotes illiteracy and j3 denotes employment. Let us construct 2 x 2 contingency A Total table. Here, (AB) = 220, (Aj'.3) = 20 and (aj'.3) = 180. 220 80 300 Also N = 500. Now B (AB) (aB) (B) (/3) = (Aj'.3) + (aj'.3) = 20 + 180 = 200 20 180 200 (B) = N - (/3) = 500 - 200 = 300 j3 (Aj'.3) (aj'.3) (/3) (aB) = (B) - (AB) = 300 - 220 = 80 240 260 500 (A) = (AB)+ (Aj'.3) = 220 + 20 = 240 Total (A) (a) N and (a) = (aB) + (aj'.3) = 80 + 180 = 260. Yule's Coefficient of Association, Q = (AB)(aj'.3) - (At1)(aB) (AB)(aj'.3) + (Aj3)(aj3) = (220 x 180) - (20 x 80) = 39600 - 1600 = 38000 = 0.92. (220 x 180) + (20 x 80) 39600 + 1600 41200 Yule's Qoefficient of Association = 0.92, which indicates that there is a positive associa-

tion between the attributes A and B.

600

BUSINESS MATHEMATICS AND STATISTICS

--------1IEXERCISES ON CHAPTER 191-I- - - - - (Measures of Association of two Attributes) 1. Find the missing frequencies from the following data:

(a) (AB) = 100, (A) {b) (AB) = 480, (A)

= 400, N = 1600 and (B) = 900 ~

710, (a(3)

= 640 and

N = 1870.

2. From the following information find out whether the data are consistent or not: (a) (A) = 250, (B) = 300, (AB) = 90 and N = 800 (b) (A)= 200, (B) = 300, (AB)= 250 and N = 700. 3. From the following data determine whether two attributes A and B are independent, positively associated or negatively associated: (a) (A) = 470, (B) = 620, (AB) = 320 and N = 1000 {b) (AB)= 256, (a(3) = 768, (A(3) = 48 and (a(3) = 144. 4. Gifn the following information, (A) = 490, (AB) = 294, (a) = 570, (aB) = 380. Show whether attributes A and B are positively associated, negatively associated, or independent. 5. Show whether A and B are independent, positively associated or negatively associated. (AB) = 256, (aB) = 768, (A(3) = 48 and (a(3) = 144.

6. Find the coefficient of association by using the following data: (a) (AB) = 200, (A(3) = 140, (aB) = 100 and (a(3) = 130 (b) (A(3) = 50, (AB)= 80, (aB) = 20 and (a(3) = 60 (c) (A(3) = 90, (AB) = 120, (aB) = 140 and (a(3) = 150 (d) (AB)= 120, (A(3)

= 90, (a(3) = 80 and (aB) = 150.

7. Find Yule's coefficient of association when (AB) = 8, (A) = 18, (a(3) = 5 and N = 35. Interpret the result. 8. In an experiment on immunization of cattle from tuberculosis, the following results were obtained: Inoculated Not inoculated

Affected

Unaffected

12

26

16

6

By calculating Yule's Coefficient of Association, examine the effect of vaccine in controlling the disease. [Hints. Let A denote affected and B denote inoculated. Then a denote unaffected and

f3 denote not

inoculated.]

9. Calculate the coefficient of association between the intelligence of fathers and sons from the following data:

"This page is Intentionally Left Blank"

APPENDIX FOUR-FIGURE LOGARITHMIC TABLES LOGARITHMS 0 1 2 3 4 10 0000 0043 0086 0128 0170

5

6

7

8

9

0212 0253 0294 0334 0374 11

0414 0453 0492 0531 0569 06o7 0645 0682 0719 0755

12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 13· 1139 1173 1206 1239 1271 ·14 1461 1492 1523 1553 1584

1303 1335 1367 1399 1430 1614 1644 1673 1703 1732

15 1761 1790 1818 1847 1875

-

1903 1931 1959 1987 2014 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 17 2304 2330 2355 2380 2405 18 2553 2577 2601 2625 2648

2430.. 2455 2480 2504 2529 2672 2695 2718 2742 2765

19 2788 2810 2833 2856 2878 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

3010 3222 3424 3617 3802 3979 4150 4314 4472 4624 4771 4914 5051 5185 5315 5441 5563 5682 5798 5911 6021 6128 6232 6335 6435 6532 6628 6721 6812 6902

3032 3243 3444 3636 3820 3997 4165 4330 4487 4639 4786 4928 5065 5198 5328 5453 5575 5694 5809 5922 6031 6138 6243 6345 6444 6542 6637 6730 6821 6911

3054 3263 3464 3655 3838 4014 4183 4346 4502 4654 4800 4942 5079 5211 5340 5465 5587 5705 5821 5933 6042 6149 6253 6355 6454 6551 6646 6739 6830 6920

3075 3284 3483 3674 3856 4031 4200 4362 4518 4669 4814 4955 5092 5224 5353 5478 5599 5717 5832 ·5944 6053 6160 6263 6365 6464 6561 6656 6749 6839 6928

3096 3304 3502 3692 3874 4·048 4216 4378 4533 4683 4829 4969 5105 5237 5366 5490 5611 5729 5843 5955 6064 6170 6274 6375 6474 6571 6665 6758 6848 6937

2900 2923 3118 3139 3324 3345 3522 3541 3711 3729 3892 3909 4065 4082 4232 4249 4393 4409 4548 4564 4698 4713 4843 4857 4983 4997 5119 5132 5250 5263 5378 5391 5502 5514 5623 5635 5740 5752 5855 5866 5966 5977 6075 6085 618016191 6284 6294 6385 6395 6484 6493 6580 6590 6675 6684 6767 6776 6857 6866 6946 6955

2945 2967 3160 3181 3365 3385 3560 3579 3747 3766 39 27 3945 4099 4116 4265 4281 4425 4440 4579 I 4594 4728 4742 4871 4886 5011 5024 5145 5159 5276 5289 5403 5416 5527 5539 5647 5658 5763 5775 5877 5888 5988' 5999 6096 6107 6201 6212 6304 6314 6405 6415 6503 6513 6599 6609 6693 6702 6785 6794 6875 6884 6964 6972

2989 3201 3404 3598 3784 3962 4133 4298 4456 4609 4757 4900 5038 5172 5302 5428 5551 5670 5786 5899 6010 6117 6222 6325 6425 6522 6618 6712 6803 6893 6981

1 5 4 4 4 3 3 3 3 3 3 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 I 2 l

2 9 8 8 7 7 7 6 6 6 6 6 6 6 5 5 5 5 4 4 4 4 4 4 4 4 3 3 3 3 3

l

3

l

l

3 3 3 3 2

1

2

1 l

l

1

2

1

2 2 2 2 2 2 2 2 2 2 2 2

l

·1 l 1

1 1 l

1 1 l

1

3 13 12 12 11 11

10 10 10 9 9 9 8 8 8 8 8 7 7 7 6 6 6 6 6 5 5 5 5 5 4 4 4 4 4

4

4 4 3 3 3 3 3 3 3 3 3 3 3 3 3

4 17 16 16 15 14 14 13 13 12 12 11 11 11 10 10 10 9 9 9 8 8 8 8 7

7 7 7 6 6 6 6 6 5 5 5 5 5 5 5 4 4 4 4 4 4 4 4 4 4 4

5 21 20 20 18 18 17 16 16 15" 14

14 14 14 13 13 12 12 11

11 11

11 10 10 9 9 9 8 8 8 7 7 7 7 6 6 6 6 6 6 5 5 5 5 5 5 5 5 5 4 4

6

26 24 23 22 21 20 19 19 19 17 17 17 16 16 15 15 14 14 13 13 13 12 12 11 11 10 10 9 9 9 9 8 8 8 8 7 7 7 7 7 6

6 6 6 6 6 6 5 5 5

7 30 28 27 26 25 24 23 22 22 20 20 19 19 18 18 17 17 16 16 15 15 14 14 13 12 12

8 34 32 31 29 28 27 26 25 25 23 23 22 22 21 20 20 19 18 18 17 17 16 15 15 14

11

13 13 12 12 11 11 11 10 10 10 10 9 9 9 9 8 8 8 8 8 7 7

11 11

10 10 10 9 9

9 9 8 8 8 8 8 7 7 7 7 7 7 6 6 6

14

9 38 36 35 33 32 31 29 29 28 26 26 25 24 23 23 22 21 21 20 19 19 18 17 17 16 15 15 14 14 13 13 12 12 12 11 11 11 10 10 10 10 9

9 9 9 9

7

8 8 8

7

b

LOGARITHMS 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71

72 73 74 ,:75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

0

1

2

3

4

5

6

7

8

9

1

2

3

4

5

6

7

8

9

6990 7076 7160 7243 7324 7404 7482 7559 7634 7709 7782 7853 7924 7993 8062 8129 8195 8261 8325 8388 8451 8513 8573 8633 8692 8751 8808 8865 8921 8976 9031 9085 9138 9191 9243 9294 9345 9395 9445 9494 9542 9590 9638 9685 9731 9777 9823 9868 9912 9956

6998 7084 7168 7251 7332 7412 7490 7566 7642 7716 7789 7860 7931 8000 8069 8136 8202 8267 8331 8395 8457 8519 8579 8639 8698 8756 8814 8871 8927 8982 9036 9090 9143 9196 9248 9299 9350 9400 9450 9499 9547 9595 9643 9689 9736 9782 g827 9872 9917 9961

7007 ,7093 7177 7259 7340 7419 7497 7574 7649 7733 7796 7868 7938 8007 8075 8142 8209 8274 8338 8401 8463 8525 8585 8645 8704 8762 8820 8876 8932 8987 9042 9096 9149 9201 9253 9304 9355 9405 9455 9504 9552 9600 9647 9694 9741 9786 9832 9877 9921 9965

7016 7101 7185 7267 7348 7427 7505 7582 7657 7731 7803 7875 7945 8014 8082 8149 8215 8280 8344 8407 8470 8531 8591 8651 8710 8768 8825 8882 8938 8993 9047 9101 9154 9206 9258 9309 9360 9410 9460 9509 9557 9605 9652 9699 9745 9791 9836 9881 9926 9969

7024 7110 7193 7275 7356 7435 7513 7589 7664 7738 7810 7882 7952 8021 8089 8156 8222 8287 8351 8414 8476 8537 8597 8657 8716 8774 8831 8887 8943 8998 9053 9106 9159 9212 9263 9315 9365 9415 9465 9513 9562 9609 9657 9703 9750 9795 9841 9886 . 9930 9974

7033 7118 7202 7284 7364 7443 7520 7597 7672 7745 7818 7889 7959 8028 8096 8162 8228 8293 8357 8420 8482 8543 8603 8663 8722 8779 8837 8893 8949 9004 9058 9112 9165 9217 9269 9320 9370 9420 9469 9518 9566 9614 9661 9708 9754 9800 9845 9890 9934 9978

7042 7126 7210 7292 7372 7451 7528 7604 7679 7752 7825 7896 7966 8035 8102 8169 8235 8299 8363 8426 8488 8549 8609 8669 8727 8785 8842 8899

7050 7135 7218 7300 7380 7459 7536 7612 7686 7760 7832 7903 7973 8041 8109 8176 824:1. 8306 8370 8432 8494 8555 8615 8675 8733 8791 8848 8904 8960 9015 9069 9122 9175 9227 9279 9330 9380 9430 9479 9528 9576 9624 9671 9717 9763 9809 9854 9899 9943 9987

7059 7143 7226 7308 7388 7466 7543 76i9 7694 7767 7839 7910 7980 8048 8116 8182 8248 8312 8376 8439 8500 8561 8621 8681 8739 8797 8854 8910 8965 9020 9074 9128 9180 9232 9284 9335 9385 9435 9484 9533 9581 9628 9675 9722 9768 9814 9859 9903 9948 9991

7067 7152 7235 7316 7396 7474 7551 7627 7701 7774 7846 7917 7987 8055 8122 8189 8254 8319 8382 8445 8506 8567 8627 8686 8745 8802 8859 8915 8971 9025 9079 9133 9196 9238 9289 9340 9390 9440 9489 9538 9586 9633 9680 9727 9773 9818 9863 9908 9952 9996

1

l

5 5 5 5 5 5 5 5 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

7 7 7

l

3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 ·3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

6 6

l

2 2 2 2 2 2 2 2

8 8 7 7 7 7 7 7 7 7 6 6 6 6 6 6 6 6 6 6 6 5 5 5 5 5 .5

~954

9009 9063 9117 9170 9222 9274 9325 9375 9425 9474 9523 9571 9619 9666 9713 9759 9805 9850 9894 9939 9983

ii

l

l

1

1

3 3 2 2 2 2 2 2 2 2 2 2 2

l

l

2

1 1 1 1 1 1 1 1 1

1 1

l

1 l l l l

l

l

l

1

l l

1 l l l l l l l

1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0

2 2 l 2 1 2 2 l 1 2 1 2 1. 2 2 l 1 2 2 l 1 2 1 2 l 2 l 2 l 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 l 1 1 1 l l 1 1 l l

l

1 1 l 1 1. 1 1 l 1 l l

1

6 6 6 6 6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 5 5 5 5 5 5 5 5 5 5 4 5 4 5 4 5 4 5 4 5 4 5 4 5 4 5 4 5 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 •'4 3 3

5

5

P. 5 5 5 5 5 5 5 4 4 4 4 4 4 4 4 4 4 4 4 4

FOUR-FIGURE ANTILOGARITHMIC TABLES ANTILOGARITHMS .00 .01 .02 .03 .04 .05. .06 .07 .08 .09 .10 .11

.12 .13 .14 .15 .16 .17 .18 .i9 .20 .21 .22 .23 .24 .25 .26 .27 .28 .29 .30 .31 .32 .33 .34 .35 .36 .37 .'38 .39 .40 .41 ,42 .43 .44 .45 .46

.47 .48 .49

0 1000 1023 1047 1072 1096 1122 1148 1175 1202 1230 1259 1288 1318 1349 i380 1413 1445 1479 1514 1549 1585 1622 1660 1698 1738 1778 1820 1862 1905 1950 1995 2042 2089 2138 2188 2239 2291 2344 2399 2455 2512 2570 2630 2692 2754 2818 2884 2951 3020 3090

1 1002 1026 1050 1074 1099 1125 1151 1178 1205 1233 1262 1291 1321 1352 i384 1416 1449' 1483 1517 1552 1589 1626 1663 1702 1742 1782 1824 1866 1910 1954 2000 2046 2094 '2143 2193 2244 2296 2350 2404 2460 2518 2576 2636 2698 2761 2825 2891 2958 3027 3097

2 1005 1028 1052 1076 .1102 1127 1153 1180 1208 1236 1265 1294 1324 1355 1387 1419 1452 1486 1521 1556 1592 1629 1667 1706 1746 1786 1828 1871 1914 1959 2004 2051 2099 2148 2198 2249 2301 2355 2410 2466 2523 2582 2642 2704 2767 2831 2897 2965 3034 3105

3 1007 1030 1054 1079 1104 1130 1156 1183 1211 1239 1268 1297 1327 1358 1390 1422 1455 1489 1524 1560 1596 1633 1671 1710 1750 1791 1832 1875 1919 1963 2009 2056 2104 2153 2203 2254 2307 2360 2415 2472 2529 2588 2649 2710 2773 2838 2904 2972 3041 3112

4

1009 1033 1057 1081 1107 1132 1159 1186 1213 1242 1271 1300 1330 1361 1393 1426 1459 1493 1528 1563 1600 1637 1675 1714 1754 1795 1837 1879 1923 1968 2014 2061 2109 2158 2208 2259 2312 2366 2421 2477 2535 2594 2655 2716 2780 2844 291r 2979 3048 3119

5 1012 1035 1059 1084 uo9 u35 1i61 n89 1216 1245 1274 ~303

1334 1365 1396 1429 1462 1496 1531 1567 1603 1641 1679 1718 1758 1799 1841 1884 1928 1972 2018 2065 2113 2163 2213 2265 2317 2371 2427 2483 2541 2600 2661 2723 2786 2851 2917 2985 3055 3126

6

1014 1038 1062 1086 1112 1138 1164 ll91 1219 1247 1276 1306 1337 1368 1400 1432 1466 1500 1535 1570 1607 1644 1683 lf.22 1762 1803 1845 1888 1932 1977 2023 2070 2118 2168 2218 2270 2323 2377 2432 2489 2547 2606 2667 2729 2793 2858 2924 2992 3062 3133

7 1016 1040 1064 1089 1114 1140 1167 ll94 1222 1250 1279 1309 1340 13TL

1403 1435 i469 1503 1538 1574 1611 1648 1687 1726 _1766 1807 1849 1892 1936 1982 2021;\ 2075 2123 2173 '2223 2275 2328 2382 2438 2495 2553 2612 2673 2735 2799 2864 293.1 2999 3069 3141 iii

8 1019 1042 1067 1091 1117 1143 1169 ll97 1225 1253 1282 1312 11343 1374 1406 1439 1472 1507 1542 1578 1614 1652 1690 • l,. 1730 1770 1811 1854 1897 1941 1986 2032 2080 2128 2178 2228 2280 2333 2388 2443 2500 7559 2618 _2679 2742 2805 2871 2938 3006 3076 3148

9 1021 1045 1069 1094 1119 1146 1172 1199 1227 1256 1285 1315 1346 1377 1409 1442 i476 1510 1545 1581 1618 1656 1694 1734 1774 1816 1858 1901 1945 1991 2037 2084 2133 2183 2234 2286 2339 2393 2449 2506 2564 2624 2685 2748 2812 2877 2944 3013 3083 3155

1

.

3 1

0 0 0 0 0 0 0 0 0 0 0 0

2 0 0 0 0 1 1 1 1 1 1 1 1 1

I

J..

2

Q

1

:l

1

2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 l 1 1 l 1 1 1 1 1 1 1 1 1 2 l 2 1 2 l 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 2 3 2 3 2 3 2 3 2 3 2 3

0

1

4

5

6

l

l

l

l

1

l

1

l

1

l

1 1 1 1 2 2 2 2 2

-1 1 1 l 1 _1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 l 2 1 I 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4

2

2 2 2 2 2 2 2 2 2 2 2 2 2 2 ·2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4

7 2 2 2 2 2 2 2 2 2 2 2

8 2 2 2 2 2 2 2 2 2 2 2 2

2 2 2 2 3 2 3 2 3 2 3 2 3 2 3 3 3 3 3 3 3 3 3 .3 3 3 3 3 3 3 3 3 3 3 4 3 4 3 4 3 4 3 4 3 4 4 4 4 4 4 4 4 4 4 4 4 5 4 5 4 5 4 5 4 5 4 5 5 5 5 5 5 5 5 6 5 6

9 2 2 2 2 2 2 2 2 3 3 3 3 3 3

3 3 3 3 3 3 3, 3 3 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6

ANTILOGARITHMS .50 .51 .52 .53 .54 .55 .56 .57• .58 .59 .60 .61 .62 .63 .64 .65 .66 .67 .68 .69 .70 .71 .72

.73 .74 .75 .76 .77 .78• .79 .80 .. 81 .82 .83 .84 .85 .86 .87 .88 .89 .90 .91 .92 .93 .94 .95 .96 .97 .98 .99

0 3162 3236 3311 3388 3467 3548 3631 3715 3802 3890 3981 4074 4169 4266 4365 4467 4571 4677 4786 4898 5012 5129 5248 5370 5495 5623 5754 5888 6026 6166 6310 6457 66o7 6761 6918 7079 7244 7413 7586 7762 7943 8128 8318 8511 8710 8913 9120 9333 9550 9772

1 3170 3243 3319 3396 3475 3556 3639 3724 3811 3899 3990 4083 4178 4276 4375 4477 4581 4688 4797 4909 5023 5140 5260 5383 5508 5636 5768 5902 6039 6180 6324 6471 6622 6776 6934 7096 7261 7430 7603 7780 7962 8147 8337 8531 8730 8933 9141 9354 9572 9795

2 3177 3251 3327 3404 3483 3565 3648 3733 3819 3908 3999 4093 4188 4285 4385 4487 4592 4699 48.08 4920 5035 5152 5272 5395 5521 5649 5781 5916 6053 6194 6339 6486 6637 6792 6950 7112 7278 7447 7621 7798 7980 8166 8356 8551 8750 8954 9162 9376 9594 9817

3 3184 3258 3334 3412 3491 3573 3656 3741 3828 3917 4009 4102 4198 4295 4395 4498 4603 4710 4819 4932 5047 5164 5284 5408 5534 5662 5794 5929 6067 6209 6353 6501 6653 6808 6966 7129 7295 7464 7638 7816 ·7998 8185 8375 8570 8770 8974 9183 9397 9616 9840

4 3192 3266 3342 3420 3499 .3581 3664 3750 3837 3926 4018 4111 4207 4305 4406 4508 4613 4721 4831 4943 5058 5176 5297 5420 5546 5675 5808 5943 6081 6223 6368 6516 6668 6823 6982 7145 7311 7482 7656 7834 8017 8204 8395 8590 8790 8995 9204 9419 9638 9863

. ;; 3199 3273 3350 3428 3508 3589 3673 3758 3846 3936 4027 4121 4217 4315 4416 45 19 4624 4732 4842 4955 5070 5188 5309 5433 5559 5689 5821 5957 6095 6237 6383 6531 6683 6839 6998 7161 7328' 7499 7674 7852 8035 8222 8414 8610 8810 9016 9226 9441 9661 9886

6 3206 3281 3357 3436 3516 3597 3681 3767 3855 3945 4036 4130 4227 .4325 4426 4529 4634 4742 4853 4966 5082 5200 5321 5445 5572 5702 5834 5970. 6109 6252 6397 6546 6699 6855 7015 7178 7345 7516 7691 7870 8054 8241 8433 8630. 8831 9036 9247 9462 9683 9908

7 3214 3289 3365 3443 3524 3606 3690 3776 3864 3954 4046 4140 4236 4335 4436 4539 4645 4753 4864 4977 5093 5212 5333 5458 5585 5715 5&18 5984 6124 6266 6412 6561 6714 6871 7031 7194 7362 7534 7709 7889 8072 8260 8453 8650 8851 9057 9268 9484 9705 9931 iv

8 3221 3296 3373 3451 3532 3614 3698 3784 3873 3963 4055 4150 4246 4345 4446 4550 4656 4764 4875 4989 5105 5224 5346 5470 5598 5728 5861 5998 6138 6281 6427 6577 6730 6887 7047 7211 7379 755 1 7727 7907 8091 8279 8472 8670 8872 9078 9290 ·9506 9727 9954

9

3228 3304 3381 3459 3540 3622 3707 3793 3882 3972 4064 4159 4256 4355 4457 4560 4667 4775 4887 5000 5117 5236 5358 5483 5610 5741 5875 6012 6152 6295 6442 6592 6745 6902 7063 7228 7396 7568 7745 7925 8110 8299 8492 8690 8892 9099 9311 9528 9750 9977

1 2 1 1 1 2 1 2 1 2 1 2 1 .2 1 2 1 2 1 2 1 2 1 2 1 2 l 2 l 2. l 2 l 2 l 2 1 2 l 2 1 2 l 2 1 2 l 2 1 3 1 3 1 3 l 3 1 3 1 3 1 3 1 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 5

3 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 5 5

5 5 5 5 5 5 5 6 6 6 6 6 6 6 7 7 7

4 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 7 7 7 7 7 7 8 8 8 8 8 8 9 9 9

5 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 7 7 7 7 7 7 8 8 8 8 8 8 9 9 9 9 9 10 10 10 10 11

11 11 11

6 4 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 7 7 7 7 7 7 8 8 8 8 8 8 9 9 9 9 9 10 10 10 10 11 11 11 11

12 12 12 12 13 13 13 14

7 5 5 5 6 6 6 6 6 6 6 6 7 7 7

7 7 7 8 8 8 8 8 8 9 9 9 9 10 10 10 10 11 11 u

11 12 12 12 12 13 13 13 14 14 14 15 15 15 16 16

8 6 6 6 6 6 7 7 7 7 7 7 8 8 8 8 8 9 9 9 9 9 10 10 10 IO 10 11

11 11

11 12 12 12 13 13 13 13 14 14 14 15 15 15 16 16 17 17 17 18 i8

9

7 7 7 7 7 7 8 8 8 8 8 9 9

9 9 9 10 10 10 10 11 11 11 11

12 12 12 12 12 13 13 14 14 14 15 15 15 16 16 16 17 17 17 18 18 19 19 20 20 20

FIVE-FIGURE LOGARITHMIC-TABLES LOGARITHMS 0

1

2

3

4

5

6

7

8

9

Mean Differences

1 2 3 4 10 00000 00432 00860 01284 01703 42 85 127 170 02H9 02531 02938 03342 03743 40 81 121 162 11 04139 04532 04922 05308 05690 37 77 u6 154 06070 06446 06819 07188 07555 .37 74 111 148 12 07g18 0827g 08636 08gg1 09342 36 71 106 142 og6g1 10037 10380 10721 1105g 34 68 102 136 13 11394 11727 12057 12385 12710 33 66 98 131 13033 13354 13672 i3988 14301 32 63 95 126 14 14613 14922 15229 15;-,34 15836 30 61 g1 122 16137 16435 16732 17026 17319 29 59 88 118 15 17609 178g8 18184 18469 18752 28 57 85 114 19033 19312 19590 19866 20140 28 55 83 110 16 20412 20683 30951 21219 21484 27 53 Bo 107 21748 22011 2227z 22531 22789 26 52 78 104 17 23045 23300 23553 23805 24055 26 50 76 101 24304 24551 24797 25042 25285 25 49 73 98 18 25527 25768 26007 26245 26482 24 48 71 95 26717 26951 27184 27416 27646 23 46 69 93 19 27875 28103 28330 28556 28780 23 45 68 go 29003 29226 29447 29667 29885 22 44 66 88 20 30103 30320 30535 30750 30963 31175 31387 31597 31806 3201) 21 43 64 85 21 32222 32428 32634 32838 33041 33244 33445 33646 33846 34044 20 41 61 81 22 34242 34439 34635 34830 35025 35218 35411 35603 35793 35984 20 39 58 77 23 36173 36361 36549 36736 36922 37107 37291 37475 37658 37840 19 37 56 74 24 38021 38202 38382 38561 38739 38917 39094 39270 39445 39620 18 35 53 71 25 39794 3g967 40~40 40312 40483 40654 40824 40993 41162 41330 17 34 51 68 26 41497 41664 41830 41996 42160 42325 42488 42651 42813 42975 16 33 49 66 27: 43136 43297 43457 43616 43775 43933 44091 44248 44404 44560 16 32 47 63 28 44716 44871 45025 45179 45332 45484 45637 45788 45939 46090 15 30 46 61 29 46240 46389 46538 46687 46835 46982 47129 47276 47422 47567 15 29 44 59 30 47712 47857 48001 48144 48287 48430 48572 48714 48855 48996 14 29 431 57 31 4g136 4g276 49415 4g554 496g3 49831 4996g 50106 50243 50379 14 28 41 55 32 50515 50650 50786 5og20 51054 51188 51322 51455 51587 51720 13 27 40 54 33 51851 51g83 52114 52244 52375 52504 52634 52763 528g2 53020 13 26 3g 52 34 53148 53275 53403 5352g 53656 53782 53go8 54033 54158 54283 13 25 38 50 35 54407 54531 54654 54777 54900 55023 55145 55267 55388 55509 12 24 37 4g 36 55630 55751 55871 55991 56uo 5622g 56348 56467 56585 56703 12 24 36 48 37 56820 56g37 57054 57171 57287 57403 57519 57634 5774g 57864 12 23 35 46 38 57978 58og2 58206 58320 58433 58546 5865g 58771 58883 58g95 11 23 34 45 39 5g106 5g218 5g329 59439 59550 59660 59770 59879 59988 60097 11 c22 33 44 40 60206 60314 60423 60531 60638 60746 60853 60959 61066 61172 11 21 32 43 41 61278 61384 614go 615g5 61700 61805 61go9 62014 62118 62221 10 21 31 42 42 62325 62428 62531 62634 62737 6283g 62g41 63043 63144 63246 10 20 31 41 43 63347 63448 63548 6364g 63749 63849 63g4g 64048 64147 64246 10 20 30 40 44 64345 64444 64542 64640 64738 64836 64933 65031 65128 65225 10 20 29 39 45 65321 65418 65514 65610 65706 65801 65896 65992 66087 66181 10 1g 29 38 46 66276 66370 66464 66558 66652 66745 6683g 66932 67025 67117 9 lg 28 37 47 67210 67302 673g4 67486 67578 67669 67761 67852 67943 68034 9 18 27 36 48 68124 98215 68305 68395 68485 68574 68664 68753 68842 68931 9 18 27 36 49 69020 69108 69197 6g285 6g373 6g461 6954~. 69636 Gg723 69810 9 18 26 35 [

v

J

5

212 202 193 185 177 170 164 158 152 147 142 138 134 130 126 122 11g 116 113 110 106 101 97 93 89 85 82 79 76 74 72 69 67 65 63 61 60 58 57 55 54 53 51 50 49 40 47 46 45 44

6 7 254 297 242 283 232 270 222 259 213 248 204 238 197 229 190 221 183 213 in.206 171 lgg 165 193 160 187 156 182 151 176 147 171 143 167 139 162 135 158 132 154 127 148 121 141 116 135 111 130 106 124 102 119 98 115 95 111 91 107 88 103 86 100 83 g1 80 g4 78 g1 76 88 73 85 71 83 70 81 68 79 66 77 64 75 63 74 61 71 60 70 59 68 57 67 56 65 5"5 64 53 63 53 62

9 381 364 348 333 319 307 295 284 274 265 256 248 240 233 227 220 214 208 203 198 190 182 174 167 159 153 148 142 137 132 129 124 121 117 113 110 107 104 102 99 97 g5 g2 go 88 86 84 82 72 81 70 79

8 339 323 309 296 284 272 262 253 244 236 228 221 214 208 201 196 190 185 180 176 170 162 154 148 142 136 131 126 122 118 114 110 107 104 101 g8 95 g3 go 88 86 84 82 80 78 76 74 73

LOGARITHMS 0

1

2

3

69897 69984 70070 70157 70757 70842 70927 71012 52 71600 71684 71767 71850 53 72428 72509 72591 72673 54 73239 73320 73400 73480 55 74036 74115 74194 74273 56 74819 74896 74974 75051 57 75587 75664 75740 75815 58 76343 76418 76492 76567 59 77085 77159 77232 77305 60 77815 77887 77960 78032 61 78533 786o4 78675 78746 62 79239 79309 79379 79449 63 79934 80003 80072 8o140 64 80618 80686 80754 80821 65 81291 81358 81425 81491 66 81954 82020 82086 82151 67 82607 82672 82737 828o2 68 83251 83315 83378 83442 69 83885 83948 84011 84073 70 84510 84572 84634 84696 'Tl 85126 85187 85248 85309 72 85733 85794 85854 85914 73 86332 86392 86451 86510 74 86923 86982 87040 87099 75 87506 87564 87622 87679 76 88081 88138 88195 88252 T7 88649 88705 88762 88818 18 89209 89265 89321 89376 79 89763 89818 89873 89927 .80 90309 90363 90417 90472 81 go848 90902 90956 91009_ 82 91381 91434 91487 91540 83 91908 91gtlo 92012 92064 84 g2428 92480 92531 92583 85 92942 92993 93044 93095 86 93450 93500 93551193601 87 93952 94002 94052 94101 88 94448 94498 94547\94596 89 94939 94988 95036 95085 90 g5424 95472195521 g556991 95904 95952 95999 96047 92 96379 96426 96473 96520 93 96848 96895 96942 96988 94 m13 ~7.3,?9 97405 974.51 95 97772 9·1818 97864 97909 96 98227 98272 98318 98363 97 98677 98722 98767 98811 98 99123 99167 99211 99255 99 g9564 99607 99651 99695 50 51

4

5

7

6

70243 70329 70415 71096 71181 71265 71933 72016 r2099 72754 72835 72916 73560 73640 73719 74351 74429 74507 75128 75205 75282 75891 75967 76o42 76641 76716 76790 77379 77452 77525 78104 78176 78247 78817 78888 78958 79518 79588 79657 80209 80277 80346 80889 80956 81023 81558 81624 81690 82217 82282 82347 82866 82930 82995 83506 83569 83632 84136 84198 84261 84757 84819 84880 85370 85431 85491 85974 86034 86094 86570 86629 86688 87157 87216 87274 87737 87795 87852 88309 88366 88423 88874 88930 88986 89432 89487 89542 89982 90037 90091 90526 go580 90634 91062 91116 91169 9'1593 g1645 91698 92117 92169 92221 92634 92686 92737 g3146 I93197 93247 93651,93702 93752 94151 94201 94250 94645 94694 94743 95134 91}182 95231 95617 1 9,5665 95713 96095 96142 96190 96567 96614,96661 97035 97081 97128 97491 97543 97589 97955 g8ooo 980.16 98408 98453 98498 98856198900,98945 99300,99344,99388 99739 99782 99826

8

9

--1

2

70501 70.')85 70672 9 71349 71433 71517 8 77281 72263 72346 8 72997 73078 73159 8 73799 73878 73957 8 74586 74663 74741 8 75358 75435 75511 8 76118 76193 76268 8 76864 76938 77012 7 77597 77670 77743 7 78319 78390 78462 7 79029 79099 79169 7 79727 79796 79865 7 80414 80482 80550 7 81ogo 81158 81224 7 81757 81823 81889 7 82413 82478 82543 7 83059 83123 83187 6 836g6 83759 83822 6 84323 84386 84448 6 84942185003 85065 6 85552 85612 85673 6 86153 86213 86273 6 86747 86806 86864 6 87332 87390 87448 6 87910 87967 88024 6 88480 88536 88593 6 89042 89098 89154 6 89597 8965,3 89708 6 90146 90200 90255 6 go687 90741 90795 5 91222 91 2 75 913 28' 5 91751·91803 91855 5 92273192324 92376 5 g2788 92840 92891 5 93298 1 93349 93399 5 93802 93852 93902 5 94300 94349 94399 5 94792 94841 948go 5 95279 95328 95376 5 95761 95809 95856 5 96237 96284 96332.5 96708 96755 968o2 5 97174,97220 97267 5 97635 g7681 97727 5 9809il98137198182 5 98543 98588 9863215 98989199034 99078 4 99432 99476 99520 4 99870i99913199957 4

17 17 17 16 16 16 15 15 15 15

vi

14

14 14

14 13 13 13 13 13 12 12 12 12 12 12 12 11 11 11 11 11 11 11

10 10 10

10 10 10 10 10 9 9 9 9 9 9 9 9 9

Mean Differences 3

4

5

26 34 43 25 34 42 .,. -;:i 33 42 24 32 41 24 32 40 23 31 39 23 31 39 23 30 38 22 30 37 22 29 37 22 29 36 21 28 36 21 28 35 20 27 34 20 27 34 20 26 33 20 26 33 19 26 32 19 25 32 19 25 31 19 2,5 31 18 24 31 18 24 30 18 24 30 17 23 29 17 23 29 17 23 29 17 22 28 1/ 22 28 17 22 28 16 22 27 16 21 27 16 21 27 i6 21 26 15 20 26 15 20 26 15 20 25 15 20 25 15 20 25 15 19 24 14 19 24 14 19 24 14 lg 24 14 18 23 14 18 23 14 18 23 14 18 23 13 18 22 13 18 22 13 17 22

7

8

9

52 60 50 59 50 58 49 57 48 56 47 55 46 54 45 53 44 52 44 51 43 50 43 50 41 48 41 48 40 47 40 46 39 46 38 45 38 44 37 43 37143 37 13 36142 35 41 35J41 35141 34140 34 39 33139 33 39 32j38 32137 32 t.37 31136 31136 31136 30135 30 35 29 34 29 _34 29 34 28 33 28 33 28 32 28 32 27 32 27 32 27 31 26 31 26 31

69 67 66 65 64 63 62 60 59 58 58 57 55 54 54 53 52 51 50 50 50 49 48 47 46 46 46 45 44 44 43 42 42 42 41 41 40 40 39 39 38 38 38 38 37 36 36 36 35 35

77 76 75 73 72 70 69 68 67 66 65 64 62 61 60 59 59 58 57 56 56 55 54 53 52 52 51 50 50 50 49 48 48 47 46 46 45 45 44 44 .43 42 42 42 42 41 41 40 40 39

6

1

FIVE-FIGURE ANTILOGARITH1\1IC TABLES ANTILOGARITHMS

.. 00 .01 .02 .03 .04 .05 .06 .07 .08 .09 .10 .11

.12 .13 .14 .15 .16 .17 .18 .19 .20 .21 .22 .23 .24 .25 .26 .27 .28 .29 .30 .31 .32 .33 .34 .35 .36 .37 .38 .39 .40 .41 .42 .43 .44 .45 .46 .47 .48 .49

0

1

10000 10233 10471 10715 10965 11220 11482 11749 12023 12303 12589 12882 13183 13490 13804 14125 14454 14791 15136 15488 i5849 16218 16596 16982 17378 17783 18197 18621 19055 19498 19953 20417 20893 21380 21878 22387 22999 23442 23988 24547 25119 25704 26303 26915 27542 28184 28840 29512 30200 30903

1002·3 10257 10495 10740 10990 11246 11508 11776 12050 12331 12618 12912 13213 13521 13836 14158 14488 14825 15171 15524 i5885 16255 16634 17022 17418 17824 18239 18664 19099 19543 19999 20464 20941 21429 21928 22439 2296! 23496 24044 24604 25177 25763 26363 26977 27606 28249 28907 29580 30269 30974

2

3

10046 10069 10280 10304 10520 10544 10765 10789 11015 11041 11272 11298 ll535 11561 11803 11830 12078 12106 12 359 12388 12647 12677 12942 12972 13243 13274 13552 13583 13868 13900 14191 14223 14521 14555 14859 14894 15205 15241 15560 15596 15922 15959 16293 16331 16672 16711 17061 17100 17458 i7498 17865 17906 18281 18323 18707 18750 19143 19187 19588 19634 20045 20091 20512 20559 20989 21038 21478 21528 21979 22029 22491 22542 23014 23067· 23550 23605 24099· 24155 24660 24717 25236 25293 25823 25882 26424 26485 27040 27102 27669 27733 28314 28379 28973 29040 29648 29717 30339 30409 31046 31117 . 'I

J

Bus. Math. & Stat. ·rc.U.}-39

4

5

6

7

10093 10116 10139 10162 10328 10351 10375 10399 10568 10593 10617 10641 10814 10839 10864 10889 11066 11092 11117 11143 11324 11359 11376 11402 n588 11614 11641 11668 11858 11885 11912 11940 12134 12162 12190 12218 12417 12445 12474 12503 _12706 12735 12764 12794 13002 13032 13062 13092 13305 13335 13366 13397 13614 13646 13677 13709 13932 13964 13996 14028 14256 14289 14322 14355 14588 14622 14655 14689 14928 14962 14997 15031 i5276 15311 15346 15382 15631 15668 15704 15740 15996 16032 16o6g_ ~106 16368 16406 16444 16482 16749116788 16827 16866 17140 17179 17219 17258 17539 17579 i7620 17660 17947 17989 18030 18072 18365 18408 18450 18493 18793 18836 18880 18923 19231 19275 19320 19364 19679 19724 19770 19815 20137 20184 20230 20277 20606 20654 207r 20749 21086 21135 211 4 21232 21577 21627 21677 21727 22080 22131 22182 22233 22594 22646 22699 22751 23121 23174 23227 23281 23659 23714 23768 23823 24210 24266 24322 24378 24774 24831 24889 24946 25351 25410 25468 25527 25942 26002 26062 26122 26546126607 26669 26730 27164 27227 27290 27353 27797 27861 27925 27990 28445 28510 28576 28642 29107 29174 29242 29309 29785 29854 29923 29992 30479 30549 30620 30690 31189 31261 31333 31405 vii

8 10186 10423 10666 10914 11169 11429 11695 11967 12246 12531 12823 13122 13428 13740 14060 14388 14723 15066 15417 i5776 16144 16520 16904 17298 17701 18113 18535 18967 19409 19861 20324 20797 21281 21777 22284 22803 23336 23878 24434 25003 25586 26182 26792 27416 28054 28708 29376 30061 30761 31477

Mean Differences 9

1 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5

10209 10447 10691 10940 11194 11455 11722 11995 12274 12560 12853 13152 13459 13772 14093 14421 14757 15101 15453 15812 16181 16558 16943 17338 17742 18155 18578 19011 19454 19907 20370 20845 21330 21827 22336 22856 23388 23933 24491 fQ. 25061 6 25645 6 26242 6 26853 6 27479 6 28119 6 28774 7 29444 7 30130 7 30832 7 31550 7

2 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 7 7 7 7 7 7 8 8 8 8 8 8 9 9 9 9 10 10 10 10 10 11

11 11 11 12 12 12 13 13 13 13 14 14 14

3 7 7 7 8 8 8 8 8 8 9 9 9 9 9 10 10 10 10 11 1.1 11 11 12 12 12 12 13 13 13 14 14 14 15 15 15 16 16 16 17 17 18 18 18 19 19 20 20 21 21 22

4 9 10 10 10 10 11 11 11

11 11 12 12 12 13 13 13 13 14 14 14 15 15 15 16 16 17 17 17 18 18 19 19 19 20 20 21 21 22· 22 23 23 24 24 25 26 26 27 28 28 29

5 12 12 12 13 13 13 13 14 14 14 15 15 15 16 16 16 17 17 18 18 i8 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 27 27 28 29 29 30 31 31 32 33 34 34 35 36

6 14 14 15 15 15 16 16 16 17 17 i8 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 28 29 29 30 31 31 32 33 34 34 35 36 37 38 39 39 40 41 42 43

7 16 17 17 18 18 18 19 19 20 20 21 21 21 22 22 23 24 24 25 25 26 26 27 28 28 29 30 30 31 32 32 33 34 35 36 37 37 38 39 40 41 42 43 44 45 46 47 48 49 50

8 19 19 20 20 20 21 21 22 22 23 24 24 25 25 26 26 27 28 28 29 30 30 31 32 32 33 34 35 35 36 "37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 54 55 56 58

9

21 21 22 23 23 24 24 25 25 26 26 27 28 28 29 30 30 31 32 32 33 34 35 36 36 37 38 39 40 41 42 43 44

45 46 47 48 49 50 51 53 54 55 56 58 59 6o 62 63 65

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