Business Analytics in Context : An Introduction to Methodologies 9781911635154, 1911635158

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Business Analytics in Context An Introduction to Methodologies

To Lesley and Beatrice, without whom this book would have been completed two years earlier.

Business Analytics in Context An Introduction to Methodologies

Gareth Woods

(G)

Goodfellow Publishers Ltd

(G)

Published by Goodfellow Publishers Limited, Woodeaton, Oxford, OX3 9TJ http://www.goodfellowpublishers.com

British Library Cataloguing in Publication Data: a catalogue record for this title is available from the British Library. Library of Congress Catalog Card Number: on file. ISBN: 978-1-911635-15-4 Copyright © Gareth Woods, 2020 All rights reserved. The text of this publication, or any part thereof, may not be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, storage in an information retrieval system, or otherwise, without prior permission of the publisher or under licence from the Copyright Licensing Agency Limited. Further details of such licences (for reprographic reproduction) may be obtained from the Copyright Licensing Agency Limited, of Saffron House, 6–10 Kirby Street, London EC1N 8TS. All trademarks used herein are the property of their repective owners, The use of trademarks or brand names in this text does not imply any affiliation with or endorsement of this book by such owners. Design and typesetting by P.K. McBride, www.macbride.org.uk Cover design by Cylinder

Contents

About the author vii Acknowledgements vii How to use the book and to study successfully viii

1

Numeracy 1 1.1 1.3 1.4 1.5 1.6 1.7 1.8

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Basic Operations and Order of Operations 1 Factors and Multiples 9 Fractions 12 Powers and Indices 16 Standard Form 20 Ratios 23 Percentages 26 Applying your Knowledge 40 Solutions 45

Algebra 49 2.1 Substitution 49 2.2 Algebraic Equations 51 2.3 Inequalities 55 2.4 Expanding Brackets 57 2.5 Factorisation 60 2.6 Coordinate Geometry 63 2.7 Simultaneous Equations 68 2.8 Optimisation using Linear Programming 71 2.9 Quadratic Equations 76 2.10 Pythagoras’ Theorem 83 2.11 Trigonometry 85 2.12 Logarithms 88 2.13 Sequences and Series 91 Applying your Knowledge 99 Solutions 105

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Probability 111 3.1 3.2 3.3 3.4

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111 117 120 131 141 147

Statistics 151 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11

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Introduction to Probability Conditional Probability and Bayes Theorem Probability Distributions Decision Analysis Applying your Knowledge Solutions

Representing Data 151 Sampling Methods 160 Measures of Location 162 Measures of Spread 171 Correlation 179 Linear Regression 183 Multiple Regression 188 Forecasting 193 Confidence Intervals 198 Hypothesis Testing 203 Type I and II Errors 209 Applying your Knowledge 215 Solutions 221

Calculus 227 5.1 5.2 5.3 5.4

Introduction to Differentiation 227 Differentiating Various Functions 228 Stationary Points 230 Optimisation Using Calculus 236 Applying your Knowledge 244 Solutions 249

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About the author Since completing my PhD in Theoretical Physics, I have been working for the past 11 years teaching students mathematics and statistics. Over this time, I developed a keen interest in education and educational research which led me completing a Masters in Education, where I studied the support seeking behaviours of mature students at university. Currently I am Mathematics Teaching Fellow at Aston University supporting students from different disciplines with understanding the mathematics they need to succeed in their chosen fields. Working one-to-one with students and delivering mathematics workshops has given me first-hand experience of how students learn and the typical problems that they encounter. Helping students to overcome their anxiety around mathematics and showing them that mathematics is not as scary as they think is key to my approach. I have found that many students, especially those on a business degree, have not studied mathematics for several years. This purpose of this book is to firstly remind you of the mathematical techniques that you have learnt previously and to provide you with a business context for them. It will then gradually introduce you to some of the main analytical and statistical techniques required for a business degree, and will also help you after your studies, particularly should you decide to run your own business. Gareth Woods

Acknowledgements Firstly, thank you to Mark Styles for inviting me to write this book and for his drive and determination to get it over the finishing line. I would like to thank Harry Flynn for the office banter, cups of milky coffee and for proofreading the book in exchange for a steak dinner (I haven’t forgotten). Finally, the biggest thank you is saved for my wife Lesley Woods, who has not only loved and supported me over the past seven years, but whilst writing this book, has given birth to our little miracle Beatrice Constance; what a journey it has been for us.

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How to use the book and to study successfully The book has been split into five main chapters, with each chapter structured similarly. The first two chapters are based around numeracy and algebra, with their purpose being to get you up to speed with the mathematical techniques you have studied previously. It is vital not to rush through these sections as they provide you with the tools to be successful with the subsequent sections. From my experience of supporting students, it is actually their numeracy and algebra skills that hold them back rather than the business analysis techniques that that you will learn later on in the book. Each section has a short explanation of the idea we are studying with several examples to follow through. Each example has been written so to not skip even the smallest of steps. You will then get the opportunity to practise the idea yourself in the confidence builder sections. The purpose of the confidence builder sections is to allow you to practise the technique you have previously learnt. Practise is so vital in mathematics, and even though you may think you understand an idea, you should attempt as many questions as you can to cement the idea. At the end of each chapter there are two further types of questions, ‘End of Chapter Questions’ and ‘Applying Business Analytics’. The ‘End of Chapter Questions’ test everything you have learnt in that chapter, i.e. you will have to apply multiple techniques to answer the question. The ‘Applying Business Analytics’ questions also require you to apply multiple techniques to answer them, but they are also based on a problem that you may encounter if you are running your own business.

Online resources This book will be fully supported by a range of online resources that will help students to develop their mathematical skills and build their confidence in being able to apply these skills to real world business situations. Careful consideration has been given to how the online resources integrate with the book to ensure a structured approach to navigating through the content, allowing you to build your confidence in applying the mathematical techniques shown. The learning pathway has been developed so that there are no dead-ends if you find yourself struggling to grasp a particular concept or technique. Mathematics can be a daunting subject for many, and so particular attention has been given to directing you back to content that you may need to engage with again and understand in order to progress further. These resources will be created to meet the changing needs of learners based on feedback from users of the book. Use the link to find out what resources are currently available.

https://www.goodfellowpublishers.com/baic

Whether you are a student or a lecturer, if you have a suggestion on what should be included in these resources, please send an email to [email protected]

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Numeracy

In this section we will learn the basic techniques required to be successful at business analytics. The section will include topics such as order of operations, fractions, percentages and ratios.

1.1 Basic Operations and Order of Operations We will begin by investigating the importance of the order in which we do calculations. To perform calculations we use one of the following four operations +, −, × and ÷.

1.1.1 Addition When we are dealing with the sum of two or more numbers, i.e. 2 + 3 = 5 the order in which we write the numbers down is irrelevant. We can see that 2 + 3 = 5 is equivalent to 3 + 2 = 5. As the order in which we add numbers together does not matter we say addition is commutative. We will not use words such as this throughout this text as I want to try and keep the mathematical language to a minimum. I only mention it so that if you see the word used in other texts or in a lecture you know what it means.

1.1.2 Subtraction When we are dealing with subtraction we are finding the difference between two numbers. For example if we write 13 − 8 we are finding the difference between 13 and 8 which is 5. It is important to note in subtraction that the order in which we write the numbers down is vital, i.e. 13 − 8 = 5 is not the same as 8 − 13 = −5. It should also be noted that when we subtract a negative number it is equivalent to adding a positive number, i.e. 8 − (−5) = 8 + 5 = 13.

1.1.3 Multiplication Multiplication is typically denoted with ×, i.e. 9 × 3 = 27. You may also see it written with a ·, i.e. 4 · 3 = 12 or even with no sign and only with brackets, i.e. (5)(8) = 40. Similarly to addition, the order in which we multiply two or more numbers does not matter, i.e. 7 × 8 = 8 × 7 = 56. Care is required when multiplying positive and negative numbers, you should remember the following rules when multiplying numbers:

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positive × positive = positive positive × negative = negative negative × positive = negative negative × negative = positive

Example 1.1.1 Calculate the value of (−4) × 6. In this example we are multiplying a negative with a positive which we know from above will give us a negative answer. We are therefore required to calculate 4 × 6 = 24 and hence we find (−4) × 6 = −24. Calculate the value of (−7)(−6). In this example we are multiplying a negative with a negative which if we follow the rules will give us a positive answer. We are therefore required to calculate 7 × 6 = 42 and hence we find (−7) × (−6) = 42.

1.1.4 Division Similarly to multiplication, there are several different ways of writing a division. For example, we can write 16 ÷ 4 to mean 16 divided by 4. We could also write this division using 16/4 or as a fraction 16 . When we write the division as a fraction, i.e. 16 we call 4 4 the number on the top the numerator and the number on the bottom is the denominator. It should also be noted that the order in which we write the numbers down is important, i.e. 16 ÷ 4 = 4 whereas 4 ÷ 16 = 0.25. Similarly with multiplication it is important that we take care when dividing positive and negative numbers and you should remember the following rules: positive ÷ positive = positive positive ÷ negative = negative negative ÷ positive = negative negative ÷ negative = positive

Example 1.1.2 Calculate the value of −16 ÷ 8 To calculate −16 ÷ 8 we should first note that we are dividing a negative with a positive, which if we follow the rules from above will give us a negative answer. We are therefore required to calculate 16 ÷ 8 = 2 and we find −16 ÷ 8 = −2. Calculate the value of −16 ÷ (−4) To calculate the value of −16 ÷ −4 we should note that we are dividing a negative by another negative and hence we should have a positive answer. Upon calculating 16 ÷ 4 = 4 we can conclude that −16 ÷ (−4) = 4.

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1.1.5 Order of Operations Now that we are familiar with the four main operations we can perform on numbers, we should also be aware that the order in which we do the operations is vitally important. For example, if we wrote down: (4 + 12) ÷ 3 − 2 × 5 we need a set of rules to follow to ensure that we get an answer that we all can agree upon. To help us recall the order in which we perform calculations we use the acronym BIDMAS. You may also know the alternate acronym BODMAS. BIDMAS tells us the order in which we should perform the operations within an expression or equation. B Brackets → Anything within brackets should be calculated first. I Indices → Any indices, e.g. 33 must be calculated next (you may know them as powers). D Division → Perform all divisions as you move from left to right. M Multiplication → Perform all multiplications as you move from left to right. A Addition → Perform all additions as you move from left to right. S Subtraction → Perform all subtractions as you move from left to right.

Example 1.1.3 Calculate the value of 3 + 8 × (5 − 2). To evaluate the expression, following the order described by BIDMAS, we should first calculate the value of the term within the brackets, i.e. (5 − 2) = 3 to find: 3 + 8 × (5 − 2) = 3 + 8 × 3 Notice that there are no powers or divisions to calculate so we should move on to the multiplication, i.e. 8 × 3 = 24 to find: 3 + 8 × 3 = 3 + 24 The final step is to perform the addition to find: 3 + 24 = 27

Example 1.1.4 Calculate the value of 9 − 25 ÷ (10 − 5)2 × 3 + 7. To evaluate the expression we should follow BIDMAS and calculate any values in brackets. The only bracket here is (10 − 5) = 5, hence we find: 9 − 25 ÷ (10 − 5)2 × 3 + 7 = 9 − 25 ÷ 52 × 3 + 7 The next step is to calculate any powers, which in this example is 52 = 5 × 5 = 25. We can therefore simplify the expression to find: 9 − 25 ÷ 52 × 3 + 7 = 9 − 25 ÷ 25 × 3 + 7 The next part of BIDMAS is to perform divisions; in this example there is one, i.e. 25 ÷ 25 = 1 to find: 9 − 25 ÷ 25 × 3 + 7 = 9 − 1 × 3 + 7.

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Business Analytics in Context

After division we should perform any multiplications, i.e. 1 × 3 = 3 to simplify the expression to: 4

Introduction to Sustainable Tourism

9−1×3+7=9−3+7

The next step is to perform any additions. We should be careful at this point when dealing with the −3 term. In expressions such as 9 − 3 + 7 the minus sign belongs to the 3 and should therefore be considered when performing the addition, i.e. when we are performing the addition of −3 + 7 we see that it is equal to 4. If you are struggling with addition with negative numbers, consider a number line. 4

Introduction to Sustainable Tourism -4

-3

-2

-1

0

1

2

3

4

5

On the number line above we are starting at −3. When we add a number using the number line we move that many steps to the right (we move to the left when subtracting). Therefore when we add 7 to −3 we are moving 7 places to the right which takes us to 4.

-4

-3

-2

-1

0

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Using the above result, we therefore find: 9−3+7=9+4 which can finally be simplified by performing the final addition 9 + 4 = 13.

Example 1.1.5 Calculate the value of (4 + 2 × 3) × (2 − 4 + 32). To evaluate this expression we should note that we have two brackets to calculate. We should therefore take each bracket in turn, calculate their value by applying BIDMAS to each one and then finally multiply their answers. If we begin with the bracket (4 + 2 × 3) we see that there are no other brackets within the original bracket, there are also no powers or division signs so we should perform the multiplication, i.e. 2 × 3 = 6 to find: 4 + 2 × 3 = 4 + 6. To complete this bracket we should finally perform the addition to find 4 + 6 = 10 and we have found: (4 + 2 × 3) = 10. We should now tackle the second bracket, i.e. (2 − 4 + 32 ), where we begin by noticing that there are no further brackets to calculate but there is a power, i.e. 32 = 3 × 3 = 9 to find: (2 − 4 + 32) = (2 − 4 + 9). As there are no divisions or multiplications to consider we can move to the final steps of performing the addition/subtraction. We should first consider the addition, remembering that the subtraction sign is attached to the 4 to find −4 + 9 = 5: 2−4+9=2+5

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and by performing the final addition we find: 2 + 5 = 7, i.e. (2 − 4 + 32) = 7. Finally to calculate the value of the expression we are required to multiply the values we found in the two brackets (4 + 2 × 3) × (2 − 4 + 32 ) = 10 × 7 = 70.

Example 1.1.6 Calculate the value of the expression: 2 + 3 × 24 2 − (4 ÷ 2)2 To evaluate this expression we are required to first calculate the value of the numerator and denominator separately by applying the rules of BIDMAS. Beginning with the numerator (2 + 3 × 24) we see that there are no other brackets, but there is a power which should be calculated first, i.e. 24 = 2 × 2 × 2 × 2 = 16 and we find: 2 + 3 × 24 = 2 + 3 × 16. The next step is to perform the multiplication, i.e. 3 × 16 = 48 and we find: 2 + 3 × 16 = 2 + 48. We finally perform the addition to find 2 + 48 = 50, i.e. (2 + 3 × 24) = 50. Now that we have the value of the numerator we should calculate the value of the denominator (2 − (4 ÷ 2)2). The first step in evaluating this expression is to calculate the bracket within the denominator, i.e. (4 ÷ 2) = 2 to find: 2 − (4 ÷ 2)2 = 2 − 22. The next step in the process is to calculate the value of the term involving the power, i.e. 22 = 2 × 2 = 4 to find: 2 − 22 = 2 − 4 and the final step to simplify the denominator is to perform the subtraction to find 2−4 = −2, i.e. 2 − (4 ÷ 2)2 = −2. Now that we have the value of the numerator and denominator we can perform the final division to find: 2 + 3 × 24 50 = = −25 2 − (4 ÷ 2)2 – 2 where we have used the result that the division of a positive number by a negative number gives a negative number.

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Business Analytics in Context

Confidence Builder Questions: Set 1.1 1. −5 × −3 + 2

6. 4 ÷ 22 + 8 ÷ 4

2. 2 + 3 × 4 − 3

7. (2 + 3 × 2)(3 + 2 × 3)

3. 8 ÷ (−2 − 2)

8. (2 + 32 ÷ 9 − 2)(1 + 5 × 10) 9. 1 + 4 2×2 3 2 10. 2 + 3(2 +3 4) 2 × 8 − 5(2 − 7)

4. 7 − 4(22 − 6) 5. −2 + 3 × 42

The Confidence Builder Solutions are at the end of the chapter.

1.2 Rounding It is important when we write a number down, that we write it to an approximate degree of accuracy. For example at a recent football game the attendance was 32,856. For most people who are interested in the attendance figures, an answer of 33,000 would be an appropriate degree of accuracy. It is often unnecessary to give the actual value when dealing with numbers – a rounded version is often appropriate. There are two procedures for rounding numbers that we will consider in this text: decimal places and significant figures. When we rounding it is important to remember that when we write the number down to the appropriate degree of accuracy then: If the next digit is 5 or more we round up. If the next digit is less than 5 then it stays the same.

1.2.1 Decimal Places Instead of rounding a number to the nearest whole number, we are sometimes required to write the number more accurately and consider some of the digits after the decimal point. When we do this we are rounding the number to a certain number of decimal places. Decimal places are counted from the decimal point. Consider the following number: 5.361 We see that the digit 3 after the decimal point is in the position of the first decimal place. The 6 is in the second and finally 1 is in the position of the third decimal place.

Example 1.2.1 Write 2.71623534823 to two and three decimal places. When we are rounding using decimal places the same rules apply, i.e. if the value after the degree of accuracy we are rounding to is 5 or greater then we round up and if it less than 5 then it stays the same. If we want to write 2.71623534823 to two decimal places we go to the second decimal place which in this example is 1. We then look at the value that comes after this one, i.e. 6

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and as this value is 5 or greater then we round up. This means that to two decimal places 2.71623534823 = 2.72 as the 1 gets rounded up to 2. If we want to write 2.71623534823 to three decimal places we go to the third decimal place which in this example is 6. We then look at the value that comes after this one, i.e. 2 and as this value is less than 5 the number remains the same. This means that to three decimal place 2.71623534823 = 2.716 as the 6 remains unchanged.

Example 1.2.2 Write 3.1415926553589793 to five decimal places. To round 3.1415926553589793 to five decimal places we are required to look at the fifth decimal place which is given by 9 and then consider the value that comes after it, 2. As the value 2 is less than 5 then the number remains the same. We can therefore write 3.1415926553589793 = 3.14159 to five decimal places.

1.2.2 Significant Figures In addition to rounding to a certain number of decimal places, we can also round values to a certain number of significant figures. The word significant means “to have meaning”. For example, consider the number: 1,204 where the 1 in this case represents how many thousands we have and is the first significant value. The 2, which for this number represents how many hundreds there are, is the second significant value. The 0, which represents how many tens there are is the third significant value. It should be noted that any leading zeros, i.e. zeros that appear before any non-zero values are not classed as being significant, but as the zero here follows on from a non-zero value then it is significant. Finally the 4 which represents how many units there are is the fourth significant value. If we consider the number: 0.00241 then the first significant value is 2. This is because leading zeros are not classed as being significant. The second significant value is given by the 4 and finally the third significant value is 1.

Example 1.2.3 Round 0.00837 to two significant figures. As we have leading zeros, the first significant figure is the digit 8. As we are rounding to two significant figures we need to look at the second significant figure, which is 3 and then look at the value after to see if we need to round up or stay the same. As the number after the second significant figure is 7 then we are required to round up as it has a value greater than or equal to 5. We can therefore write that 0.00837 = 0.0084, to two significant figures.

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Business Analytics in Context

Example 1.2.4 Round 3,137,494 to four significant figures. To round to four significant figures we are required to look at the fourth significant figure, which is 7, and then look at the value after it to see if we round up or stay the same. The next figure along is 4 which requires us to leave the 7 unchanged. We therefore write down the first four significant figures and replace the subsequent ones with zeros such that 3,137,494 = 3,137,000 to four significant figures.

Example 1.2.5 Round 5,683,421 to two significant figures. To round to two significant figures we are required to look at the second significant figure, which is 6, and then look at the value after it to see if we round up or stay the same. The next figure along is 8 which requires us to round up. We therefore round up the 6 to a 7 and the subsequent digits are replaced with zeros to find 5,683,421 = 5,700,000, to two significant figures.

Example 1.2.6 Round 219,540 to three significant figures. As we are rounding to three significant figures, we should first look at the third significant figure, which is 9 and then look at the digit after it to see if we are required to round up or stay the same. The next number along is 5 which requires us to round up. As the third significant number is 9 when we round it up it actually rounds to 10 which means we need to add one to the second significant figure and the third significant figure becomes 0. We therefore find that 219,540 rounded to three significant figures becomes 220,000.

Confidence Builder Questions: Set 1.2 1. Write 24,021 to three significant figures.

6. Write 0.00088313 to six decimal places.

2. Write 2.6581 to two decimal places.

7. Write −12.789118 to three significant figures.

3. Write 32.482 to three significant figures.

8. Write −2.1118 to one decimal places.

4. Write 0.0651 to three decimal places.

9. Write 0.00034859 to three significant figures.

5. Write 78.024021 to five significant figures. 10. Write 3,278.9006 to three decimal places.

Numeracy

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1.3 Factors and Multiples 1

1.3.1 Factors Factors are numbers that we multiply together to get another number.

Example 1.3.1 List all of the factors of 12. To find all of the factors of 12 we are required to list all of the numbers that multiply together to give 12. Doing this we find we can make 12 in the following ways: 1 × 12   2 × 6    3 × 4. Therefore all of the factors of 12 are: 1, 2, 3, 4, 6, 12.

Example 1.3.2 List all of the factors of 36. To find all of the factors we need to list all of the numbers that multiply to give 36. For example: 1 × 36   2 × 18  3 × 12  4 × 9  6 × 6. Therefore all of the factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36.

1.3.2 Highest Common Factor The highest common factor (HCF) of two numbers is the largest factor that two numbers have in common. To calculate the HCF we list all of the factors of the two numbers and the HCF is the largest one they have in common.

Example 1.3.3 Calculate the highest common factor of 12 and 16. To calculate the highest common factor of 12 and 16 we are first required to list all of their factors. The factors of 12 are: 1, 2, 3, 4, 6, 12. The factors of 16 can be found to be: 1, 2, 4, 8, 16. With our two sets of factors, the HCF is the highest one they have in common. 12 − 1, 2, 3, 4, 6, 12 16 − 1, 2, 4, 8, 16 We can see from above the highest factor in common is 4. We can therefore write that the highest common factor of 12 and 16 is 4, i.e. HCF (12, 16) = 4.

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Business Analytics in Context

Exampe 1.3.4 Calculate the highest common factor of 34 and 51. We first calculate all of the factors of 34 and 51. The factors of 34 can be found to be: 1, 2, 17, 34. and the factors of 51 are: 1, 3, 17, 51. By looking at the list of the two factors, the highest one they have in common is 17. We can therefore write that the highest common factor of 34 and 51 is equal to 17, i.e.      HCF (34, 51) = 17.

1.3.3 Multiples A multiple is the result of multiplying a number by an integer (not a fraction).

Example 1.3.5 Find the first 5 multiples of 15. The first five multiples of 15 are: 1 × 15 = 15 2 × 15 = 30 3 × 15 = 45 4 × 15 = 60 5 × 15 = 75. Therefore the first five multiples of 15 are 15, 30, 45, 60, 75.

Example 1.3.6 Find the first 5 multiples of 23. The first five multiples of 23 are: 1 × 23 = 23 2 × 23 = 46 3 × 23 = 69 4 × 23 = 92 5 × 23 = 115. Therefore the first five multiples of 23 are 23, 46, 69, 92, 115.

1.3.4 Lowest Common Multiple The lowest common multiple (LCM) is the smallest multiple that two numbers have in common.

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Example 1.3.7 Calculate the lowest common multiple of 12 and 16. To calculate the lowest common multiple of 12 and 16 we are required to first write down the first few multiples of each number (I will typically write down the first ten multiples). This may not be enough to find the lowest common multiple but it is typically a good starting point. Writing down the first ten multiples of each number we find: 12 − 12, 24, 36, 48, 60, 72, 84, 96, 108, 120. 16 − 16, 32, 48, 64, 80, 96, 112, 128, 144, 160. Looking at the list of the two numbers we see that the lowest multiple they have in common is 48. We can therefore write that the lowest common multiple of 12 and 16 is equal to 48, i.e. LCM (12, 16) = 48.

Example 1.3.8 Calculate the lowest common multiple of 21 and 24. To calculate the lowest common multiple of 21 and 24 we write down the first few multiples of each number – I will once again write down the first 10. 21 − 21, 42, 63, 84, 105, 126, 147, 168, 189, 210. 24 − 24, 48, 72, 96, 120, 144, 168, 192, 216, 240. We see that the lowest multiple they have in common is 168, i.e. LCM (21, 24) = 168.

1.3.5 Prime Factorisation Prime Factorisation is the process of writing a number as the product of its prime numbers. A prime number is a number that is only divisible by itself and one. The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 . . . You are not expected to know all of the prime numbers as there are so many of them (in fact no one knows all of the prime numbers), but it would be useful to know the first few listed here as it will make the following questions much easier.

Example 1.3.9 Write down the prime factorisation of 105. We begin the prime factorisation process by writing down two factors of 105: 105 = 5 × 21. Now 5 is a prime number but 21 isn’t. We must therefore find two factors of 21, i.e. 21 = 3 × 7. Now both of these numbers are both prime, and hence we have found all of the prime factors of 105. We can therefore write: 105 = 3 × 5 × 7.

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Business Analytics in Context

Example 1.3.10 Write down the prime factorisation of 600. To begin the prime factorisation process we should first write down two factors of 600, for example 600 = 2 × 300. As 2 is a prime number, we need to consider two factors of 300 = 2 × 150. Since 2 is a prime number, we need to repeat the process and find two factors of 150 = 2 × 75. Once again we have one factor which is a prime number (2), so we should look for two factors of 75 = 3 × 25. As 3 is a prime number, we are left to find two factors of 25 = 5 × 5. We have now found all of the prime factors of 600 and we can write 600 = 2 × 2 × 2 × 3 × 5 × 5 = 23 × 3 × 52.

Confidence Builder Questions: Set 1.3 1. Calculate the highest common factor and lowest common multiple of the following pairs of numbers (a) 12 and 15. (d) 10 and 22. (b) 15 and 45. (e) 18 and 32. (c) 12 and 18.

(f ) 21 and 28.

2. Calculate the prime factorisation of the following numbers (a) 210.

(d) 700.

(b) 102

(e) 620.

(c) 350.

(f )  1200.

1.4 Fractions A fraction is made up of two parts, the numerator and the denominator. The numerator is the number on the top of the fraction and the denominator is the bottom number. For 2 example, in the fraction 3 the numerator is 2 and the denominator is 3.

1.4.1 Adding and Subtracting Fractions To add or subtract two fractions we should perform the following steps: „„ If they do not share a common denominator, change the fractions so that they do. „„ Add or subtract the numerators. „„ Simplify the new fraction, if possible. There are many ways of calculating the common denominator, but the simplest way is to multiply the denominators together. We can then cross multiply the numerator and denominators. This is probably easier to understand when we look at an example.

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Example 1.4.1 Add the fractions:

1  2 + 5  3

To add the fractions we need them to have the same denominator. To do this we multiply the denominators together to find 5 × 3 = 15. We are then required to cross multiply the numerator and denominators, i.e. 1 × 3 and 2 × 5 to find: 1  2  × 3  2 ×5 +  =  1   + 5 3 15 15

and then we can add the numerators together to find: 1 + = 2 3 + 10 = 13 5 3 15 15 15 we should at this point check to see if the final expression can be simplified any further, which is not possible in the case. We therefore have our final solution.

Example 1.4.2 Add the fractions:

1  4 + 6  9

Once again, to add the fractions we need them to have the same denominator. To do this we multiply the denominators together to find 6 × 9 = 54. We are then required to cross multiply the numerators and denominators to find: 1  4  × 9  4 ×6 +  =  1   + 6  9   54   54 and then we can add the numerators together to find: 1 + 4 = 9 + 24 = 33 6 9 54 54 54 we should at this point check to see if the final expression can be simplified. We can see for this example that 3 goes into both the numerator and denominator to find: 1 + 4 6 9

33 =     3 × 11  11 = 54 = 18 3 × 18  

Example 1.4.3 Calculate the value of

3  1 – 5  4

To subtract the fractions we need them to have the same denominator. To do this we multiply the denominators together to find 5 × 4 = 20. We are then required to cross multiply the numerator and denominators to find: 3  1  × 4  1 ×5 –  =  3   – 5  4   20   20

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and then we can subtract the numerators to find: 3 – 1 = 12   –  5    = 7 5 4 20  20  20 we should at this point check to see if the final expression can be simplified, but as it is already in its simplest form we have our final answer.

Example 1.4.4 Calculate the value of the following: 7   2 – 30  5 Once again, to subtract the two fractions we need them to have the same denominator. To do this we multiply the denominators together to find 30 × 5 = 150. We are then required to cross multiply the numerators and denominators to find: 7 2 7    × 5 –     2 × 30 =  35 – 60 30 –  5 = 150 150 150 150 The next step is to subtract the numerators to find: 7 2 35 60 –25 – 25   –    30 –  5 = 150 150 =    150 = 150 We should at this point check to see if the final expression can be simplified. We can see for this example that 25 goes into both the numerator and denominator to find: 7 2 –25 –25 × 1 = – 1   =     30 –  5 = 150 25 × 6 6

1.4.2 Multiplying Fractions To multiply two or more fractions together we are required to multiply the numerators together and the denominators and then simplify where necessary.

Example 1.4.5 Calculate

2 × 3 5 7

To multiply the two fractions we must first multiply the numerators, 2 × 3 = 6 and then multiply the denominators 5 × 7 = 35 and we find:

2 × 3 = 6 5 7 35

Once we have multiplied out the fractions we should check that it can’t be simplified any further. This expression can’t be simplified any further therefore we have our final solution.

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Example 1.4.6 Calculate

5 × 1 2 6 5

Before we can multiply these fractions we must turn the mixed fraction 1 2 5 into a top 2 heavy fraction. We see that 1 5 is equivalent to:



1 2 = 1 + 2 5 5 and as 1 may be written as 5 5 2 2 5 2 7 1 = + = + = 5 1 5 5 5 5

This is now a top heavy fraction and we can multiply our two fractions by multiplying the top and the bottom of the fractions as we would do normally:

5 × 1 2 = 5 × 7 = 35 6 5 6 5 30

This expression can be simplified by dividing the numerator and denominator by 5 to find:

5 2 35 5×7 7 6 × 1 5 = 30 = 5×6 = 6

1.4.3 Dividing Fractions To divide two fractions, we are required to flip the second fraction upside down, multiply the two fractions and then simplify as required.

Example 1.4.7 Calculate:

1 ÷ 2 4 5

To perform this division we must flip the second fraction and multiply to find:

1 2 1 5 1 × 5 5 4 ÷ 5 = 4 × 2 = 4 × 2 = 8

As this expression can’t be simplified any further this is our final answer.

Example 1.4.8 Calculate

3 ÷ 9 7 14

To perform the division we are again required to flip the second fraction and multiply the two fractions to find:

3 9 3 14 3 × 14 42 7 ÷ 14 = 7 × 9 = 7 × 9 = 63

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The final step is to see if the expression can be simplified further. In this example we can divide both the numerator and denominator by 7 to find:

3 9 42 =     7 × 6   6 7 ÷ 14 = 63 7 × 9  = 9

The expression though is not fully simplified as there is a common factors of 3 in both the numerator and denominator:

3 3 × 2 2 9 = 6 7 ÷ 14 9 = 3 × 3 = 3

Confidence Builder Questions: Set 1.4 Calculate the value of the following:

2 1 1. + 5 7



9.

3 × 14 7 15



2.

1 – 2 3 9



10.

2 ÷ 3 3 7



4 3 3. + 7 10



11.

5 × 16 8 21



4.

7 – 3 5 8



12.

8 ÷ 7 7 6



3 + 2 5. 2 4 7



1 × 7 13. 2 2 9



3 – 11 6. 1 7 8



4 ÷ 17 14. 2 7 8



4 2 7. + 13 20



1 × 3 15. 12 15



8.



7 ÷ 12 16. 6 9

7 – 11 5 8

1.5 Powers and Indices Indices (or powers) provide us with a convenient notation when we need to multiply a number by itself several times. There are several rules we must follow when dealing with indices which will be presented in this section. Consider the expression: 4 × 4 × 4 × 4 × 4, which can be written as 45 and can be read as “4 to the power of 5”. We call the number 4 the base and the 5 is known as the power. Similarly we could write: 7 × 7 × 7 × 7 = 74 a × a × a × a × a × a × a = a7

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There are six rules that we must follow when dealing with indices. Indices will be used throughout this textbook and in all aspects of mathematics. If you take one thing from this section on numeracy, I would suggest that the set of rules of indices is that thing.

1.5.1 Multiplying Expressions When we multiply two expressions together with the same base, we simply add their indices, i.e. am × an = am+n .

Example 1.5.1 Simplify

a7 × a3.

If we were to write the expression out in full we would have found: a × a × a × a × a × a × a × a × a × a = a10

a7

a3

Alternatively we could have just added the powers together to find: a7 × a3 = a7+3 = a10

1.5.2 Dividing Expressions When we divide two expressions with the same base, we simply subtract their indices, i.e. am ÷ an = am−n .

Example 1.5.2 Simplify

a9 a4 .

If we were to write the expression out in full we would find: a×a×a×a×a×a×a×a×a    a × a × a × a and we would notice that we can cancel four of the a’s from the numerator and denominator to find: a×a×a×a×a×a×a×a×a    a × a × a × a = a ×a × a × a × a = a5. Alternatively we could have used the rule to simplify this expression and subtracted the powers, i.e.

a9 a9–4 a5 – = a4 =

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1.5.3 Raising a Power to a Power To raise one power by another power, we multiply the powers together: (am)n = am×n. We can generalise this to multiple terms multiplying each other in the brackets to find: (a × b)n = an × bn Please note that this result only holds when the terms with the brackets are all either being multiplied or divided by each other, i.e. the result is not true if there is an addition or subtraction sign between the terms, (a ± b)n ≠ an ± bn.

Example 1.5.3 Simplify:

(2a4)3

If we wanted to write this fully we would have three sets of brackets as it is being raised to the power of 3, i.e. (2a4)3 = (2a4) × (2a4) × (2a4) = (2 × a × a × a × a) × (2 × a × a × a × a) × (2 × a × a × a × a), = 2 × 2 × 2 × a × a × a × a × a × a × a × a × a × a × a × a = 8a12 . 23 =8

a12

This is a very cumbersome way of simplifying the expression, so instead we can use the rule given above. To simplify this expression we are required to raise each term in the expression to the power of three to find (2a4)3 = 23 × (a4)3 = 8a4×3 = 8a12 .

1.5.4 Power of Zero Anything to the power of zero is always equal to 1, i.e. a0 = 1

1.5.5 Negative Power When we have a negative power it means we have one over our base to the positive power, i.e. −m 1 a = am

Example 1.5.4 Simplify the expression: 3−5 To simplify the expression we should use the fact that a negative power is the equivalent of calculating “one over”, i.e. 1 3−5 =  135 = 243

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1.5.6 Fractional Indices For fractional powers, the denominator of the fraction is the root of the number and the numerator of the fraction is the power to raise the answer to, i.e. m n

a = ( √a) = √am n

m

n

Example 1.5.5 Simplify:

3

16 2

3

To simplify this expression we are first required to write 16 2 = (2√16)3. We calculate the square root of 16, i.e. 2√16 = 4. We then raise the answer to the power, i.e. raise 4 to the power of 3, i.e. 43 = 64. We therefore find: 3

16 2 = 64

1.5.7 Further Examples With these six rules we are now in a position to deal with most problems involving indices. In the next two examples we will look at more challenging problems where we are required to use multiple rules in the same question.

Example 1.5.6 Simplify the expression a7 × a4 a9 For questions involving fractions we are required to first simplify the numerator and denominator and then perform the division. As the denominator is already in its simplest form we should look to simplify the numerator. If we take the numerator, i.e. a7 × a4 we see that to simplify it we use the multiplication law, i.e. am × an = am+n, to find: a7 × a4 = a11 We can now re-write the fraction as a7 × a4 a11 = a9    a9 Now by using the division law of indices, i.e am ÷ an = am−n we find a7 × a4 a11 = = a11-9 = a2 a9    a9

Example 1.5.7 Simplify:

1

√m × m4 × n 3 m × n2 The first step is to simplify the numerator and denominator and then divide the two expressions. If we begin with the numerator:

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√m × m4 × n 3 1 The first step is to re-write √m as m 2 to find: 1

1

1

√m × m4 × n 3 = m 2 × m4 × n 3 We then need to use the multiplication law to combine all of the like terms together to find: 1

1+4

1

1

1

9

m 2 × m4 × n 3 = m2 × n 3 = m2 × n 3 1

1

8

9

where we have performed the addition 2 + 4 = 2 + 2 = 2 . We cannot simplify the denominator any further as we cannot combine two numbers with a different base. We therefore find the fraction becomes: 1

9

1

√m × m4 × n 3 = m2 × n 3 m × n2 m × n2 Now to complete the simplification we should divide the terms involving m and n separately to find: 9

1

9

1

1 5 m2 × n 3 m2 n 3 9 7 = × 2 = m2 -1 n 3 -2 = m2 n- 3 2     m×n m n

where we have used the division law of indices.

Confidence Builder Questions: Set 1.5 Simplify the following expressions m7 1. m2

(m2 )3 6. (m2 )3

m6n8 2. m2n3

7.

m2 × m4 3.    m3

2m2 ×(n3)4 ×(m3)−2 8.   3m4 × (n2)3

m7 × m−2 × m3 4.   m3 × m2

(m2 × n3)3 9. (m × n2)2

2√m × n2 × 3√n 5.   4m2 × n3

(2m2 × n4)2 10. (3m2 × n3)4

√m3 m−3

1.6 Standard Form In standard form the number must be written in two parts: a × 10b „„ Where a is a single digit taking a value between 1 (inclusive) and 10 (not inclusive). „„ Followed by a power of 10 that places the decimal point where it should be – the power tells you how many places to move the decimal point.

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1.6.1 Powers of 10 To help us write numbers in standard form, we should be aware of the various powers of 10. I have listed the following powers of 10 which may help you with the following section. 1 10−6 = 1000000 = 0.000001 1 10−5 = 100000 = 0.00001 1 10−4 = 10000 = 0.0001 1 10−3 = 1000 = 0.001 1 10−2 = 100 = 0.01 1 10−1 = 10 = 0.1 100 = 1 101 = 10 102 = 10 × 10 = 100, 103 = 10 × 10 × 10 = 1000, 104 = 10 × 10 × 10 × 10 = 10000, 105 = 10 × 10 × 10 × 10 × 10 = 100000, 106 = 10 × 10 × 10 × 10 × 10 × 10 = 1000000.

Example 1.6.1 Write 7000 in standard form. We can write 7000 as 7 × 1000. If we look at the list of powers of 10 we can see that 1000 can be written as 103 therefore 7000 = 7 × 103.

Example 1.6.2 Write 8,981,000 in standard form. We can write 8,981,000 as 8.981 × 1,000,000 where 1,000,000 can be re-written as 106 so to find 8,981,000 = 8.981 × 106.

Example 1.6.3 Write 0.0004321 in standard form. We can write 0.0004321 as 4.321 × 0.0001. We can see from the powers of 10 above that 0.0001 = 10−4 and we therefore find 0.0004321 = 4.321 × 10−4.

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1.6.2 Calculating the Power of 10 To calculate the correct power of 10 you need to ask the question “how many places do I need to move the decimal point?” When a number is 10 or greater, the decimal place needs to move to the left and the power of 10 is positive.

Example 1.6.4 Write 642 in standard form. To write 642 in standard form we should first write in the decimal point to find 642.0. Then we are required to move the decimal point two places to the left to find: 642 = 6.42 × 102

Example 1.6.5 Write 689000 in standard form. To write 689000 in standard form, we should first introduce the decimal point. This will be placed at the end of the number to find 689000.0. To now write this in standard form, we move the decimal point five places to the left so that we have a number between 1 and 10, and we can therefore write: 689000 = 6.89 × 105 When a number is smaller than 1, the decimal place needs to move to the right and the power of 10 is negative.

Example 1.6.6 Write 0.00542 in standard form To write 0.00542 in standard form we move the decimal point three places to the right to find: 0.00542 = 5.42 × 10−3

Example 1.6.7 Write 0.0000031 in standard form. To write 0.0000031 in standard form we move the decimal point six places to the right to find 0.0000031 = 3.1 × 10−6

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Confidence Builder Questions: Set 1.6 Write the following in standard form 1. 12,000 2. 0.00981 3. 1,414,000 4. 0.0000008721 5. 120.431 6. 0.00187 7. 893.7810

1 8. 76,650,000 9. 0.00911 10. 34,000,000 11. 98.2891 12. 0.0000000223 13. 156,000,000,000 14. 13.981

1.7 Ratios Ratios are used to compare amounts of two or more quantities. For example, if we wanted to make a pastry dough, we would mix two parts of flour to one part fat. We would therefore write the ratio of flour to fat as 2 : 1. We say that as the pastry is made up of two parts flour and one part fat that there are three parts in total. We can also say that two thirds of the pastry is flour and one third of it is fat. In a similar way to fractions, ratios can also be simplified by finding common factors. You should try to divide by the highest common factor first when writing down ratios, but you can divide the ratio multiple times until you get to its simplest form.

Example 1.7.1 There are 21 female students and 15 male students in a lecture. What is the ratio of female and male students in the lecture (simplify your answer as far as possible). We can write down the ratio of female students to male students initially as 21 : 15. We can simplify this by dividing through by the highest common factor of these two values. The highest common factor of 21 and 15 is 3. We therefore divide the ratio through by 3 to find the ratio of female students to male students is 7 : 5, i.e. for every 7 female students there are 5 male students.

Example 1.7.2 If Leanne has 80p and Lesley has £1.20, what is the ratio of Leanne’s money to Lesley’s money in it simplest form? First, we can only compare two quantities when they are written in the same units. We therefore need to convert either the pounds to pence or pence into pounds. For this example, we will convert £1.20 into pence by multiplying by 100 to find £1.20 = 120p. We can therefore write down the ratio of Leanne’s money to Lesley’s money as 80 : 120. This can be simplified by dividing through by the common factor of 40 to find the final ratio 2 : 3. Another way of saying this is that for every 2p Leanne has, Lesley has 3p.

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1.7.1 Using Ratios Ratios are used to solve many real life problems, e.g. in recipes, scale drawings or map work.

Example 1.7.3 Heather is planning a new kitchen. She draws a scale drawing of her kitchen using a scale of 1 : 100. She measures the length of the kitchen as 7.2m. How long would the kitchen appear on the scale drawing? (Give your answer in mm). To answer this, we are required to convert 7.2m into mm. To covert m into mm we multiply by 1000. Thus we find 7.2m is equivalent to 7200mm. The ratio of 1 : 100 tells us that the kitchen is 100 times bigger than the scale drawing. As we know the actual size of the kitchen we are required to divide by 100 to find the size of the scale drawing. We therefore find that the length in the scale drawing should be 7200 ÷ 100 = 72mm.

Example 1.7.4 A recipe to make lasagne for 8 people uses 400 grams of minced beef. How much minced beef would be needed to serve 10 people. The most effective way of tackling this problem is to calculate how much mince is required for one person and then scale it up for ten people. We can see from the recipe that for 8 people, 400 grams of mince are required. We can find that for one person we would need 400 ÷ 8 = 50 grams of minced beef. Therefore, for 10 people we would need 10 × 50 = 500 grams of minced beef.

Example 1.7.5 A recipe for flapjack required 320 grams of oats. This is enough to make 20 flapjacks. How much oats would be required to make 50 flapjacks. Similarly to Example 1.7.4, the best way to proceed is to calculate how many oats are first required to make one flapjack. We know 20 flapjacks requires 320 grams of oats. We can therefore calculate that for one flapjack we require 320 ÷ 20 = 16 grams of oats. If we are required to make 50 flapjacks we would need 50 × 16 = 800 grams of oats.

1.7.2 Dividing in a Ratio We can use ratios to divide up amounts.

Example 1.7.6 Share £2000 in the ratio of 3 : 7. The ratio 3 : 7 tells us that we need to split the money into 3 + 7 = 10 equal parts and to give one person 3 parts and the other person 7 parts. We are therefore required to find how much one part is worth by dividing the amount into 10 equal parts, with each part worth 2000 ÷ 10 = 200. As each part is worth £200 we see that one person gets 3 × 200 = £600 and the other person gets 7 × 200 = £1400.

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Example 1.7.7 Three investors get a return of £12,000 on their investment. They invested the money in a ratio of 2 : 3 : 7. How much does each investor get back? The ratio of 2 : 3 : 7 tells us that we need to split in return into 2 + 3 + 7 = 12 equal parts and to give one investor two parts, the other three parts and the final investor gets seven parts. We are required to find out how much each part is worth by dividing the return into 12 parts, to find each part is worth 12,000 ÷ 12 = 1000. We therefore see that the first investor gets 2 × 1000 = £2000, the second investor receives 3 × 1000 = £3000 and finally the third investor receives 7 × 1000 = £7000.

1.7.3 Exchange Rates Ratios are used extensively when dealing with exchange rates. When we see an exchange rate it is actually a ratio.

Example 1.7.8 The exchange rate from a local travel agent’s office is shown below: GBP (£) 1 = US dollar

1.282

Euro

1.129

1. A business woman is travelling to New York. She changes £600 to US dollars. How many dollars does she get? 2. A German company wants to buy some goods in the UK. They change 2000 euros into GB pounds. How much will they get? (to the nearest pound). Solution 1. The exchange rate tells us that one pound is equivalent to 1.282 US dollars which can be written as the ratio 1 : 1.282.

To convert out £600 into US dollars we multiply by 1.282 as the ratio tells us that for each pound the woman gets 1.282 US dollars. Therefore the woman will receive 600 × 1.282 = 769.2 US dollars. To the nearest dollar this can be writen as $769.

2. From the table the exchange rate is 1 : 1.129, and this means for each pound we have, we get 1.129 euros. As we are converting from euros to pounds we need to reverse the ratio and see how many pounds we get for each euro. To do this we are required to write the ratio in the form n : 1 instead of 1 : n. To re-write the ratio we divide both sides by 1.129 to find   1 : 1 => 0.886 : 1 1.129 This tells us that one euro is equivalent to £0.886, and to convert 2000 euros to pounds we simply multiply by 0.866 to find: 2000 × 0.886 = 1772. and we see that 2000 euros is equivalent to £1772.

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Confidence Builder Questions: Set 1.7 1. Simplify the following ratios (a) 12 : 18

(b) 120 : 40

(c) 13 : 52

(d) 6 : 21

(e) 27 : 15

(f ) 10 : 200

(g) 15 : 18

(h) 24 : 64

(i) 18 : 63

(j) 5 : 15

2. Share the following amounts in the given ratios. (a) Share £1000 in the ratio 1 : 4.

(b) Share £3500 in the ratio 2 : 3.

(c) Share £400 in the ratio 1 : 3.

(d) Share £500 in the ratio 1 : 4 : 5.

(e) Share £875 in the ratio 1 : 1 : 3.

(f ) Share £580 in the ratio 1 : 3 : 6.

3. Calculate the following currency conversion questions using the given exchange rates: GBP (£) 1 = US dollar

1.311

Euro

1.254

(a) Edward buys a watch in the US for 1200 US dollars. How much would the watch cost in GB pounds? (b) Graham wants to exchange £2000 into US dollars. How many dollars will Graham receive? (c) Marie buys a watch in Germany for 290 euros. How much would the watch cost in GB pounds. (d) Colin has 250 US dollars and wants to convert them in pounds. How many pounds should he receive? (e) Christine is travelling from Florida to Austria. She wants to convert 3000 US dollars to euros. How many euros should Christine receive. (f ) Phoebe is travelling from France to Miami. She has 5000 euros and wants to convert them to US dollars. How many dollars should she receive?

1.8 Percentages The word percent means “out of 100”, with percentages written using the symbol %. For example if 75% of the population own a mobile phone, this is equivalent to saying that 75 out of every 100 people own a mobile phone.

1.8.1 Finding Percentages A percentage is defined as a fraction of 100. For example 35% means (35 in every 100) or 35 as a fraction, 100 . 35% can also be written as a decimal by dividing 35 by 100 to find 0.35. To find the percentage of a quantity we first write the percentage as a fraction or as a decimal, then multiply by the quantity.

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Example 1.8.1 Calculate 40% of 250. We first write 40% as a decimal by diving 40 by 100 to find 40 ÷ 100 = 0.4. Then to find 40% of 250 we are required to multiply 250 by 0.4 to find: 0.4 × 250 = 100

Example 1.8.2 Audrey is buying a pair of jeans. The original price was £50, but there is a discount of 15%. How much will the discount be, and what would the new price be? We will begin by writing 15% as a decimal, i.e 15 ÷ 100 = 0.15 and then multiply by the original price to find how much the discount is, i.e. 0.15 × 50 = 7.50. As it is a discount of 15% we are required to subtract £7.50 from the original price to find the final cost of the jeans: 50 − 7.50 = £42.50 Therefore Audrey pays £42.50 for her jeans.

Example 1.8.3 A car costs £12,000 before VAT (charged at 20%). How much will the car cost with VAT? We begin by calculating how much VAT needs to be added. The amount is given by: 20 100 × 12,000 = 2,400 As VAT is added to the price of the car, the final cost of the car will be 12,000 + 2,400 = £14,400

Example 1.8.4 A car costs £15,000 after VAT has been added (at 20%). How much was the car before VAT was added? We must be careful when tackling problems like this. We define the original price of the car (before VAT) as 100%. Therefore with VAT added to the original price, £15,000 represents 120%. We can write this as a ratio now, i.e. 120% : 15,000 We can now find what 1% is equivalent to by dividing both sides of the ratio by 120 to find 15,000 1% : = 125  120 We can now multiply both sides of the ratio by 100 to find out the original price of the car before VAT was added 100% : 125 × 100 = 12,500. We can therefore say that the car cost £12,500 before VAT was added.

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1.8.2 Percentage Increase and Decrease In many cases we are required to calculate the percentage increase/decrease in an item. To calculate percentage change we use: change in amount percentage change = × 100     original amount

Example 1.8.5 Edward is dealing in electrical goods. He buys a TV for £450 and sells it for £500. What was his percentage profit? To work out the percentage profit, we are first required to calculate the change in the amount. For this example, the change in the amount is 500 − 450 = 50. We can then calculate the percentage increase. increase in amount 50 percentage increase = × 100 = × 100 = 11.11%   original amount   450 We therefore see that Edward makes a 11.11% profit.

Example 1.8.6 Jayne buys a new car for £19,500. Three years later the car is worth £12,500. What is the percentage decrease in the value of the car? To calculate the percentage decrease we are required to calculate the decrease in the value of the car, i.e. 19,500 − 12,500 = 7,000 To find the percentage decrease we are required to calculate: decrease in amount 7,000 percentage decrease = × 100 = × 100 = 35.90%  original amount   19,500 We therefore say that Jayne’s car has decreased in value by 35.90%.

1.8.3 Compound Interest and Depreciation Compound Interest Compound interest is the interest calculated on the initial investment in addition to the accumulated interest of previous periods of the investment. It is typically thought of as ‘interest on interest’, which makes the investment grow at a quicker rate than for simple interest, which calculates the interest based only on the initial investment. There are many factors that determines the rate at which compound interest accrues, they are: 1. The frequency of the compounding periods. For example if the interest was compounded once per year you would earn less in interest than if it was compounded each month for a year. 2. The percentage of interest being paid, for example if you earn interest at 10% you will earn more than if the interest was at 5%.

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To calculate compound interest we can utilise the formula: Pn = P0(1 + i)n where P0 is the principal or initial investment, n is the number of compounding periods, i is the nominal annual interest rate in percentage and Pn is the future value of the investment.

Example 1.8.6 Beatrice invested £10,000 into a savings account with an interest rate of 1.5% per annum. After five years how much would the investment be worth? As the interest rate is 1.5% per annum and the money is being invested for five years we can set i = 1.5 ÷ 100 = 0.015 and n = 5. The initial investment P0 = 10,000 and we therefore calculate that after five years the investment will be worth: P5 = 10, 000 × (1 + 0.015)5 = 10,772.84 Therefore Beatrice’s investment will be worth £10,772.84 after five years with 10,772.84 − 10,000 = £772.84 earned in interest.

Example 1.8.7 The bank lends your £15,500 at an interest rate of 3.6% per annum. If interest is compounded monthly, how much will you have to pay back by the end of six years? The principal amount P0 = 15,500. As the interest rate per annum is 3.6%, but we are compounding monthly the nominal interest rate i = 0.036 ÷ 12 = 0.003 (as 12 months are in a year). As we are compounding monthly for 6 years we have n = 12 × 6 = 72 compounding periods. We would therefore calculate the final value of the loan is: P72 = 15, 500(1 + 0.003)72 = 19,230.87

Example 1.8.8 You take out a mortgage of £155,000 over 25 years at an annual interest rate of 2.9%. If the interest is compounded daily how much in total will you be required to pay back? The principal amount for this example is equal to P0 = £155,000. As the interest rate is 2.9% per annum, but compounded daily the nominal interest rate i = 0.029 ÷ 365 = 7.95 × 10−5. As we are compounding daily for 25 years we find the number of compounding periods is given by n = 365 × 25 = 9,125. We therefore calculate the total amount we are required to pay back is given by: P9,125 = 155,000(1 + 7.95 × 10−5 )9,125 = £320,164.13

Compound Depreciation Compound depreciation works in a similar way to compound interest. Compound depreciation is when an item decreases in value by the same percentage over a period of time. For example, you may consider the value of a new car after five years. As this is compound depreciation, the value is decreasing each time. This changes the formula to the form:

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Pn = P0(1 − i)n where P0 is the initial value of the item, i is the nominal interest for which the item is decreasing by per time period and n is the number of time periods.

Example 1.8.9 You purchase a new piece of equipment for your factory at a cost of £56,000. The machine decreases in value by 8% per annum. How much will the machine be worth after five years? The initial cost of the machine P0 = 56,000. The nominal interest i = 8 ÷ 100 = 0.08 and we are considering n = 5 compounding time periods. We can therefore calculate that, after 5 years, the machine will be worth P5 = 56,000(1 − 0.08)5 = 36,908.57.

AER – Annual Equivalent Rate AER is defined as the interest rate which would give you the same money at the end of the period if interest were to be compounded annually. To calculate the AER we use: AER =

[(1 + ni ) – 1] × 100 n

where i is the annual percentage rate and n is the number of compounding periods in a year. It should be noted that if the compounding period for the nominal interest is 1 year then it is equal to the AER.

Example 1.8.10 If you borrow a sum of money with nominal interest rate of 10% which is compounded monthly, what would be the annual equivalent rate. The nominal interest rate i = 10/100 = 0.1. As we are compounding monthly there are 12 compounding periods in a year. We therefore find the AER is given by: 12 AER = 1 + 0.1 – 1 × 100 = 10.5% 12

[(

)

]

1.8.4 Net Present Value (NPV) Net present value (NPV) is defined as the difference between the present value of cash inflows and the present value of cash outflows over some period of time. NPV is predominantly used when we are planning to make an investment, to analyse its potential profitability. To calculate NPV:     n Rt − initial investment NPV = (1 + i)t

Σ

     t=1

where: „„ Rt = net cash inflow during a single time period t. „„ i is equal to the discount rate or return that could be earned in alternative investments „„ n is the number of time periods.

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To fully understand this expression we should familiarise ourselves with summation notation, denoted by Σ.

Summation Notation In business you may be required to add a series of values, for example on a balance sheet. One way of expressing this addition, or summation, would be with what’s called the ‘sigma notation’. It might be easier to explain this by considering a couple of examples: 4

Σn = 1 + 2 + 3 + 4 = 10 n=1 7

Σ(n + 1) = (2 + 1) + (3 + 1) + (4 + 1) + (5 + 1) + (6 + 1) + (7 + 1) n=2

   = 3 + 4 + 5 + 6 + 7 + 8 = 33.

33 below and Introduction to Customer Service written So sigma is just shorthand for ‘add up all the values’ and the numbers above the sigma tell us where to start and finish the addition. This can be summarised in the diagram below:

Sum up to this value

12

Σn

1

What we are summing

n=1

Start value

Let’s try a couple more trickier examples and then you should be ready for the sigmas that you see elsewhere in this book. Consider: 5

Σn

2

n=1

to calculate the value of this sum we should begin with n = 1 and sum up all the values up to n = 5, i.e. n = 1, 2, 3, 4, 5. Upon doing this we find: 5

Σn = 1 + 2 + 3 + 4 + 5 2

2

2

2

2

2

= 1 + 4 + 9 + 16 + 25 = 55

n=1

If we consider a second example: 6

Σ (2n – 1) n=3

We notice that we start the summation from n = 3 this time and finish at n = 6, i.e. n = 3, 4, 5, 6. Upon substituting these values into (2n − 1) and adding them together we find 6

Σ (2n – 1) = (2 × 3 − 1) + (2 × 4 − 1) + (2 × 5 − 1) + (2 × 6 − 1) = 5 + 7 + 9 + 11 = 32 n=3

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We can actually use any letter we like for the summation, for example we could write: 3

Σ i(i + 1) i=1

To calculate this summation we are required to first subsitute i = 1 into i(i + 1) to find 1(1 + 1) and then repeat the process for i = 2 to find 2(2 + 1) and i = 3 to find 3(3 + 1). We should then sum the three values to find: 3

Σ i(i + 1) = 1(1 + 1) + 2(2 + 1) + 3(3 + 1) = 2 + 6 + 12 = 20 i=1

Example 1.8.11 You have two possible investments you can make, the details of the costs and returns are shown in the table below. Initial Investment

Project A

Project B

−10,000

−8,000

Year 1 Return

2,000

1,500

Year 2 Return

2,500

3,500

Year 3 Return

5,000

5,500

Assuming a discount rate of 10% use the formula for NPV to determine the best project to invest in. To determine the best project we must calculate the NPV for each project. We should first note that the project has returns for three years so we can write: Rt

    3

NPV =

Σ (1 + i)

t

− initial investment

     t= 1



=

R1

+

R2

(1 + i) (1 + i)

2

+

R3 (1 + i)3

− initial investment

For project A we find the NPV is given by: 2000 2500 5000 NPV = + + − 10,000 (1 + 0.1) (1 + 0.1)2 (1 + 0.1)3   = −2359.13 indicating that we would expect to make a loss after three years of the investment. For project B we find the NPV is given by: 1500 3500 5300 NPV = + + − 8,000 (1 + 0.1) (1 + 0.1)2 (1 + 0.1)3   = 388.43 therefore you would expect to make a small profit from project B. If you were to choose one of the two projects to consider, using NPV is it clear that project B would be the most suitable.

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Internal Rate of Return – IRR The internal rate of return is defined as the discount rate for a project to have a NPV = 0. Typically, if you are asked to find the IRR, you are being asked to find the discount rate (it was 10% in the last example) that would lead to a zero NPV. IRR represents the intrinsic rate of return that we should expect from an investment considering the amount and timing of the cash flows. If, for example, we have an IRR of 12% it would suggest that the proposed investment will generate an average annual rate of return equal to 12% over the lifetime of the project, taking into account various considerations such as the amount and timing of the expected cash inflows and outflows specific to that particular investment. To approximately calculate the IRR we use the following equation: N P V1 × (R2 − R1) IRR = R1 + N P V1 − N P V2 where R1 is equal to a discount rate which gives a negative NPV which is equal to NPV1 and R2 is the discount rate which gives a postive NPV equal to NPV2.

Example 1.8.12 Calculate the IRR of project B from example 1.8.11. We first need to find a discount rate that gives us a negative NPV. To do this we simply choose a random value for the discount rate and use trial and error. I will begin by choosing a discount rate of 15% which gives a NPV equal to £-432.81. You should check this for yourself. If this value had not been negative we would have had to choose a different discount rate, but we can now write down R1 = 15% and NPV1 = £-432.81. We know from example 1.8.11 that a discount rate of 10% leads to a positive NPV so we can write R2 = 10% and NPV2 = 388.43 and the IRR can be calculated to be: N P V1 × (R2 − R1) IRR = R1 + N P V1 − N P V2    

−432.81 × (10 − 15) = 15 + = 12.36% −432.81 − 388.43 This is just an approximate value. If we wanted to calculate an exact value we would have to use software such as Excel using the syntax IRR(values). For this example, using Excel, we find the actual value is equal to 12%.

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Confidence Builder Questions: Set 1.8 1. (a) Calculate 25% of 120.

(b) Calculate 30% of 480.



(c) Calculate 74% of 1300.

(d) Calculate 57% of 1450.



(e) Calculate 28% of 6780.

(f ) Calculate 21% of 1340.

2. (a) A pair of jeans cost £80 before being reduced by 20%. What is the new price?

(b) A pair of shoes cost £40 before being reduced by 30%.What is the new price of the shoes?



(c) A car costs £20,000 before VAT is added. What is the new cost of the car?



(d) A car costs £25,500 after VAT is added. What was the cost before VAT was added?



(e) A TV costs £350 after being reduced by 20%. What was the original cost before the sale?



(f ) A car costs £21,000 after VAT is added. What was the cost before VAT was added?



(g) A Blu-Ray player costs £150 after being reduced by 25%. What was the original cost of the Blu-Ray player before the sale?

3. For the following questions give your answers to two decimal places if required.

(a) The cost of a laptop has decreased from £600 to £400. What is the percentage decrease in the cost of the laptop?



(b) The cost of a new phone has increased from £799 to £999. What is the percentage increase in the cost of the phone.



(c) A car is on sale at a price of £12,000. The original price of the car was £14,500. What is the percentage decrease in the cost of the car?



(d) The weight of a chocolate bar has decreased from 80grams to 60grams. What is the percentage change in the weight of the chocolate bar?



(e) The cost of a TV subscription service has increased from £58.99 to £63.99. What is the percentage change in the subscription cost?



(f ) The number of people attending a football match since their team was relegated has decreased from 27,500 to 22,400, What is the percentage decrease in the attendance?

4. For the following questions give your answers to the nearest pound. (a) You invest £10,000 in a savings account with a nominal rate of interest of 5%. Assuming the interest is compounded yearly how much would the investment be worth after 5 years?

(b) You invest £12,500 in a savings account with a nominal rate of interest of 1.5%. Assuming the interest is compounded monthly how much would it be worth after 8 years?



(c) Assuming a nominal interest rate of 0.5% compounded yearly, how much do you need to invest now to have £8,000 in 10 years time?

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(d) A new piece of machinery cost £590,000. It is known that its value will depreciate by 12% per year, how much will the machinery be worth in 10 years time?



(e) A new car costs £23,500. It is known that the car will lose 23% of its value per year. How much will the car be worth in three years time?

5. For the following questions, give your answers to two decimal places if required.

(a) If you borrow a sum of money with nominal interest rate of 15% which is compounded monthly, what would be the annual equivalent rate?



(b) If you borrow a sum of money with nominal interest rate of 22% which is compounded monthly, what would be the annual equivalent rate?



(c) If you borrow a sum of money with nominal interest rate of 18% which is compounded quarterly, what would be the annual equivalent rate?



(d) If you borrow a sum of money with nominal interest rate of 12% which is compounded daily, what would be the annual equivalent rate?

6. A project with a 3 year life at a cost of £28,000 will generate revenue of £8000 in the first year, £12,000 in the second year and £17,000 in the third year. If the discount rate is 3% what is the NPV of the project giving your answer to the nearest pound.Calculate an approximate value for IRR. 7. A project with a 4 year life at a cost of £22,000 will generate revenue of £6700 in the first year, £13,000 in the second year, £12,000 in the third year and £10,000 in the fourth year. If the discount rate is 2%, what is the NPV of the project giving your answer to the nearrest pound. Calculate an approximate value for IRR.

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My notes

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Applying your Knowledge Applying Business Analytics 1. A potential business partner has approached you with a plan to set up a new business.

To set it up will require a capital investment of £120,000. Your partner has £30,000 available to invest and a venture capitalist is willing to provide an additional £55,000.

(a) How much additional capital are you required to invest to set the business up? (b) What percentage of the business would you own based on the initial investment that you provided to start the business?

The venture capitalist is unhappy about ownership being based on the amount of initial investment. They feel that as they are putting down a greater amount of initial capital they should have a greater say in how the business is run. They wish the ownership to follow a ratio of 2 : 3 : 8 with respect to your partner, yourself and the capital invester.

(c) What percentage of the business would you own based on this ratio? (d) After the first year of the business being up and running, you make a profit of £6,000. What is this value as a percentage of your initial investment? (e) Over the next four years the profit the business makes increases by 12.5% each year. How much profit will the business make by the end of the next four years?

You propose an expansion to your business and have developed three possible projects. The follow table outlines the initial investment required for each project and the expected returns for the next three years. Initial Investment Year 1 Profit Year 2 Profit Year 3 Profit

Project A −10, 000 4, 000 5, 500 6, 000

Project B −15, 000 5, 000 4, 500 7, 000

Project C −13, 500 3, 500 4, 500 8, 000

(f ) Assuming a discount rate of 12.5%, use a NPV approach to calculate the most profitable project.

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Solutions 1

Confidence Builders Set 1.1 1. 17 6. 3

2. 11 7. 72

3. −2 8. 51

4. 15 9. 4

5. 46 10. 2

Set 1.2 1. 24, 000

2. 2.66

3. 32.5

4. 0.065

5. 78.024

6. 0.000883

7. −12.8

8. −2.1

9. 0.000349

10. 3, 278.901

Set 1.3 1. (a) HCF = 3 and LCM = 60.

(d) HCF = 2 and LCM = 110.



(b) HCF = 15 and LCM = 45. (e) HCF = 2 and LCM = 288.



(c) HCF = 6 and LCM = 36.

(f ) HCF = 7 andLCM = 84.

2. (a) 210 = 21 × 31 × 51 × 71.

(d) 700 = 22 × 52 × 71 .



(b) 102 = 21 × 31 × 171

(e) 620 = 22 × 51 × 311



(c) 350 = 21 × 52 × 71.

(f ) 1200 = 24 × 31 × 52 .

Set 1.4

1. 19 35

7. 53 130



35 = 1 17 13. 18 18

2. 1 9

8. 11 40



48 = 1 13 14. 35 35

3. 61 70

9. 2 5

15. 1 60



4.

41 = 1 1 40 40



16. 21 22



5.

85 = 3 1 28 28

11. 10 21

6. 17 56

12. 48 49

10. 14 = 1 5 9 9

Set 1.5 1. m5 4. m3 9

7. m2 4 10. 81 × m−4 × n−4

2. m4 × n5 3 2 1 5. 2 ×m−2 × n−3 2 8. 3 × m−8 × n6

3.

m3

6. m0 = 1 9. m4 × n5

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Business Analytics in Context

Set 1.6 1. 1.2 × 104

2. 9.81 × 10−3

3. 1.414 × 106

4. 8.721 × 10−7

5. 1.20431 × 102

6. 1.87 × 10−3

7. 8.937810 × 102

8. 7.665 × 107

9. 9.11 × 10−3

10. 3.4 × 107

11. 9.82891 × 101

12. 2.23 × 10−8

13. 1.56 × 1011

14. 1.3981 × 101

Set 1.7 1. (a) 2 : 3 (f ) 1 : 20

(b) 3 : 1 (g) 5 : 6

2. (a) 200, 800 (d) 50, 200, 250

(b) 1,400, 2,100 (c) 100, 300 (e) 175, 175, 525 (f ) 58, 174, 348

3. (a) £915.33

(b) $2622

(c) e231.26



(e) e2869.57

(f ) $5227.27

(d) £190.69

(c) 1 : 4 (h) 3 : 8

(d) 2 : 7 (i) 2 : 7

(e) 9 : 5 (j) 1 : 3

Set 1.8 1. (a) 30

(b) 144

(c) 962

(d) 826.5

2. (a) £64

(b) £28

(c) £24,000 (d) £21,250 (e) £437.5 (f ) £17,500 (g) £200

3. (a) 33.33% (b) 25.03% (c) 17.24% (d) 25% 4. (a) £12,763 (b) £14,093 (c) £7611

7. NPV = £17610, IRR = 30%

(e) 8.48% (f ) 18.55%

(d) £164,316 (e) £10,729

5. (a) 16.08% (b) 24.36% (c) 19.25% (d) 12.75% 6. NPV = £6636, IRR = 13%

(e) 1898.4 (f ) 281.4

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Applying Business Analytics 1. (a) Total capital required = £120, 000. To calculate the additional capital we are required to calculate the difference between the initial capital requirement and what is invested by your partner and venture capitalist, i.e.

120, 000 − 30, 000 − 55, 000 = £35, 000

(b) Percentage of the business is equal to 35,000 × 100 = 29.17% 120,000 (c) Using the ratio 2 : 3 : 8 we see that you own a total of 3 parts out of a total of 2 + 3 + 8 = 13 parts. Based on this ratio you own approximately: 3 × 100 = 23.08.% 13 (d) Calculating this value as a percentage of your initial investment which is 35,000 we find: 6,000 35,000

= 17.14%.

(e) As the profit increases by 12.5% each year we find that we are looking at compounding the amount over a four year period, i.e.

( )

12.5  Profit = 6000 × 1 + 100

4

= £9, 610.84

(f ) We are required to calcuate the NPV for each project with a discount rate of 12.5%. NPVA = −10,000 +

4,000 5,500 6,000 + + = 2,115.23 1.125 1.1252 1.1253

NPVB = −15,000 +

5,000 4,500 7,000 + + = –2,083.68 1.125 1.1252 1.1253

NPVC = −13,500 +

3,500 4,500 8,000 + + = –1,214.68 1.125 1.1252 1.1253

It is clear from the NPV calculation that the most profitable project would be A.

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Business Analytics in Context

2

Algebra

In this section you will be reminded about the different topics you need to be familiar with to be successful in a module on business analytics. Many, if not all of these topics you will have studied previously, but you may need a little refresher. It is important that you don’t skip this section as the concepts we will be looking at are fundamental and will help you immensely in the later sections.

2.1 Substitution We use formulae all of the time in business analytics. For example we can use a formula to calculate the supply and demand of an item. Consider the following supply and demand equations. Qd= 30 – 3P Qs= –40 + 2P where Qd is the demand of the item and Qs is the supply of the item. You can see that they are both dependent on another variable P which in this example is the price of the item. We can see that if we change the value of P both the values of Qd and Qs will change as a consequence. We therefore say that P is the independent variable and both Qs and Qd are the dependent variables as they change as a consequence of P changing. For example, I may want to know the demand on an item at a certain price. I might ask myself “what is the demand going to be if I price the item at £2.50?” To find the demand I would replace the P in our demand equation with the value 2.50 and the resulting answer would give the demand. The demand would therefore be: Qd = 30 – 3×2.50 = 22.50 Generally when we are performing substitution we replace a letter with a specific value.

Example 2.1.1 You are given the expression y = x2 + 3x – 2 Calculate the value of y for the following values of x. 1. x = 2 2. x = 0 3. x = –3

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Business Analytics in Context

Solution 1. When x = 2 we swap any ‘x’s in our expression with the value 2, i.e. y = x2+3x – 2 = 22+3×2 – 2 = 4+6 – 2 = 8. When x = 2, y = 8. 2. When x = 0 we are again required to swap any ‘x’s in our expression with the value 0. y = x2+3x – 2 = 02+3×0 – 2 = 0+0 – 2 = –2. When x = 0, y = –2. 3. When x = –3 we are still required to swap any ‘x’s in our expression with the value –3, but we should take extra care. When subsituting a negative value into an expression you should write the value inside brackets. Whilst this may not seem to be significant, you will find that if using your calculator and you forget to put your brackets in, your calculator will give you the wrong answer. If we use the brackets carefully we will find y = x2+3x – 2 = (–3)2+3(–3) – 2 = 9 – 9 – 2 = –2. When x = –3, y= –2.

Example 2.1.2 You are given the equation: s = (12 – 4.9t)t Calculate the value of s for the following values of t. 1. t = 4 2. t = –2 3. t = –6 Solution 1. When t = 4 we swap any ‘t’s in our expression with the value 4. s = (12 – 4.9t)t = (12 – 4.9×4)×4 = –30.4 When t = 4, s = –30.4. 2. When t = –2 we are again required to swap any ‘t’s in our expression with the value –2 and recall that when we substitute in a negative value, we should write it within brackets. s = (12 – 4.9t)t = (12 – 4.9×(–2))×(–2)= –43.6 When t = –2, s = –43.6. 3. When t = –6 we swap any ‘t’s in our expression with the value –6 to find s = (12 – 4.9t)t =(12 – 4.9×(–6))×(–6) = –248.4.

When t = –6, s = –248.4.

2.1.1 Expressions Sometimes we may be required to derive our own expression and then subsitutute a value into it. Consider the following example.

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Example 2.1.3 A taxi firm charges £0.80 per mile in addition to a fixed charge of £4.20. Write down a formula for the cost C of hiring a taxi to travel m miles and calculate how much it would cost for a 12 mile journey. If we travel just one mile in the taxi it would cost us C = 4.20 + 0.80 = £5. If we were to travel two miles in the taxi it would cost us C = 4.20 + 2 × 0.80 = £5.80. If we were to travel three miles it would cost a total of C = 4.20 + 3 × 0.80 = £6.60. The rule to find the cost of the taxi ride is £4.20 + 0.8 multiplied by the number of miles we have travelled, leading to the following for the cost of the taxi:

C = 4.20 + m × 0.80

To answer the question with regards to the cost of the taxi ride for a 12 mile journey we subsitute m = 12 into the expression to find:

C = 4.20 + 12 × 0.80 = 4.20 + 9.60 = £13.80.

Confidence Builder Questions: Set 2.1 1. Given that a = 3 and b = 2, calculate the value of c = a2 – 2b. 2. Given that x = –3 and y = 2 calculate the value of z = x2 + y2. 3. Given that f = –5 and g = –2 calculate the value of h = –2f – 3g. 4. Given that x = 4 and y = –3 calculate the value of z = (x2 – y2)/(x – y) 5. The demand function of a certain product us given by Qd = 40 – 2P. What would the demand be equal to if the price P = 2.50? 6. You are charged 50p per unit of electricity used in your factory with a fixed charge of £22. Write down a formula for the cost C of using n units of electricty and calculate how much it would cost to use 1200 units of electricity. 7. Work out the value of p(q – 3)/4 when p = 2 and q = –7. 8. Given that A = h(x+10)/2, with A = 27 and h = 4, calculate the value of x. 9. Given that h = 5t2 + 2, calculate the value of h when t = –2. 10. Given that h = 5t2+2, calculate the value of t when h = 47.

2.2 Algebraic Equations An algebraic equation is an equation involving powers of an unknown variable, typically represented by x but any other letter can be used with t, p and q commonly used in business analytics. For example, the following are classed as algebraic expressions:

x5 + 4x2 – 3x + 2 = 0, q = p2 – p + 2, s = t3 – 2t + 4 y + 3 = 12

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You may be wondering where the power is in the equation y + 3 = 12. When there is no power written it means it is a power of one. We could write y1 + 3 = 12, but this is not standard practice. There are many different types of equations, we will begin by considering linear equations.

2.2.1 Linear Equations A linear equation is one of the form:

ax + b = 0 a ≠ 0

where a and b are known constants and x represents the unknown quantity that we are trying to calculate. In this expression a is known as the coefficient of x and b is the coefficient of x0, i.e. the constant term. The equation is classed as linear as the highest power of x is 1. The following are examples of linear equations: 2x + 3 = 0, –2 + 3x = 0, 1 + 3 x = 12 2 4 We can also have linear equations that appear in non–standard form, for example : 3x = 2x – 4,

4g = 5g + 2, z – 5 = 3z

In each of the equations above, they can be rearranged into the form ax + b = 0.

2.2.2 Quadratic Equations Another type of equation you will meet throughout this text is a quadratic equation. Quadratic equations are of the form:

ax2 + bx + c = 0, a ≠ 0

where a is known as the coefficient of x2, b is the coefficient of x and c is the constant term or the coefficient of x0. When solving a quadratic equation it is vital to write it in the form above. We won’t be solving quadratic equations in this section, but we will be looking at this topic later.

2.2.3 Solving a Linear Equation To solve a linear equation we try to make the unknown quantity the subject of the equation. To do this we may use any/all of the following: 1. Add the same quantity to both sides of the equation. 2. Subtract the same quantity from both sides of the equation. 3. When multiplying, we need to multiply both sides of the equation by the same quantity. 4. When dividing, we must divide both sides of the equation by the same value. 5. Take functions of both sides; for example when taking a square root we are required to square root both sides, or cube root both sides. To summarise these rules we can say : Whatever we do to one side of the equation, we must also do to the other side.

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Example 2.2.1 Solve the equation:

x + 13 = 17.

To rearrange for x we remove the 13 from the left-hand side of the equation. To do this we must subtract 13 from both sides: x + 13 = 17 x + 13 –13 = 17 –13 x=4 It is always a good idea to check your answers if possible. In this example we have found x = 4, so if we substitute this back into the original equation we should find that both sides are equal. Upon doing this we find the left side of the equation is 4 + 13 = 17 which is equal to the right hand side.We can, therefore be confident that we have the correct solution.

Example 2.2.2 Solve the equation:

5x –13 = 2

To solve this equation we first wish to get the term 5x on its own and to do this we add 13 to both sides to find: 5x –13 = 2 5x –13 + 13 = 2 + 13 5x = 15 We now have an expression for 5x, but we want to know the value of x, i.e. 1x.To do this we divide both sides of the equation by 5 to find: 5x = 15 5 5 x=3 so the solution to our equation is x = 3. Now that we have our answer, we should check that it is correct. By substituting x = 3 back into the original equation we find the left hand side becomes 5 × 3 – 13 = 2 which is equal to the right hand side of the equation indicating that we have the correct answer.

Example 2.2.3 Solve the equation:

1 x –3 = 2 5 First, aim to get the term 1 x on its own. To do this we add 3 to both sides to find: 5 1 x –3 = 2 5 1 x –3 + 3 = 2 + 3 5 1x=5 5

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Business Analytics in Context

We now have an expression for 1 x, but we want to know the value of x. So we multiply 5 both sides of the equation by 5 to find: 1x×5=5×5 5 x = 25 As in previous examples we should check our answer is correct by substituting it back into the original equation. Upon substituting x = 25 into the left hand side of the equation we find 1 × 25 – 3 = 5 – 3 = 2 which is equal to the right hand side. We can therefore be 5 confident that we have found the correct answer.

Example 2.2.4 Solve the equation: 3x + 4 = 7x – 8 This example is a little bit tricker that the other examples considered so far as we have two terms involving x. Our first task when solving a problem in this form is to get all of the terms involving x together. In this example the left hand side has a 3x and the right hand side has a 7x. Whilst it is not a strict rule, it is typically best to move all the x’s to the side that has the most on to begin with, i.e. the right hand side in this example. To do this we subtract 3x from both sides to find 3x + 4 = 7x – 8 3x + 4 – 3x = 7x – 8 – 3x 4 = 4x – 8 Now that we have collected all of the terms involving x together we need to rearrange to find 4x. To do this we need to add 8 to both sides of the equation to find 4 + 8 = 4x – 8 + 8 12 = 4x The final step is to divide through by 4 to find the value of x, 12 = 4x 4 4 3=x We therefore find x = 3. As usual we should check our answer by subtituting x = 3 back into the original equation. Upon doing this we find the left hand side of the equation becomes 3 × 3 + 4 = 9 + 4 = 13. The right hand side becomes 7 × 3 – 8 = 21 – 8 = 13. As the two sides are equal we can be confident that the answer x = 3 is correct.

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Confidence Builder Questions: Set 2.2 Solve the following equations for the unknown letter 1. 2x + 4 = 10 5. 7x + 3 = –2x + 12 2. 4x + 2 = 14 6. 1 x + 1 = –5 4 3. 7x + 12 = 2x –3 7. 2 x + 2 = 5 3 4. –2x + 3 = x + 9 8. 1 x + 2 = 1 x + 3 3 4

2.3 Inequalities In mathematics, we use inequalities to compare the size of two or more quantities.

2.3.1 Symbols used in Inequalities In an inequality, two expressions get joined together using a relation symbol. There are four symbols that you need to be aware of in inequalities. Symbol ≥ > ≤ <

Meaning is greater than or equal to is strictly greater than is less than or equal to is strictly less than

2.3.2 Solving Single Inequalities An example of an inequality is 2x – 5 ≤ 9, which means that when we take the value of x, multiply it by 2 and subtract 5 we should get an answer less than or equal to 9. To solve the inequality we treat it in a similar way to the equation 2x – 5 = 9, i.e. what we do to one side, we must do to the other side. The difference though between solving an equation and an inequality is that for an inequality our answer is a range of possible values of x rather than a specific value when solving an equation. One note of warning when dealing with inequalities. If you multiply or divide by a negative number, you must flip the inequality sign.

Example 2.3.1 Solve the inequality: 2x – 5 ≤ 9 To solve this inequality we first rearrange the inequality to get 2x on its own. To do this we are required to add 5 to both sides to find 2x –5 + 5 ≤ 9 + 5, 2x ≤ 14

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We now have the values that 2x satisfies, but we want to know the values x itself satisfies. To do this we must divide both sides by 2 to find: x≤7

Example 2.3.2 Solve the inequality: 3x + 4 ≤ 4x – 5 To solve this inequality we are required to get all of the unknowns on one side. It doesn’t matter which side we move the unknowns to, but depending on which way you move them you may need to divide by a negative number. To illustrate this point I will subtract 4x from both sides to find: 3x + 4 ≤ 4x –5 3x + 4 – 4x ≤ 4x –5 –4x –x + 4 ≤ –5 At this point, subtract 4 from both sides: –x ≤ –9 We now have an expression for –x, so we need to divide through by –1 to find an expression for x. Recall that by doing this we are required to flip the inequality sign to find: –9 x ≥ –1 = 9 and so we find x ≥ 9.

2.3.3 Solving Double Inequalities Some examples require us to solve a double inequality such as: 4 < 2x – 4 < 12 For examples like this we are required to solve the inequality in exactly the same way as we have done previously.

Example 2.3.3 Solve the inequality:

4 < 2x – 4 < 12

As with solving just a single inequality, we are required to get the term involving x on its own. To do this we are required to add 4 to each part of the inequality to find: 4 + 4 < 2x – 4 + 4 < 12 + 4 which becomes:

8 < 2x < 16

upon simplification. We now have the values for 2x and to find the values for x we divide both sides by 2, to find: 4 0 the curve has a minimum point which is the lowest point on the curve. When a < 0 the curve has a maximum point which is the greatest value the function takes. We call this point the vertex of the curve. The position of the vertex is given by: 4ac − b2 b x = − 2a y= 4a These steps are probably best understood with a couple of examples.

Example 2.9.7 Sketch the function y = x2 + 5x + 4. We begin by writing down the values of a = 1, b = 5 and c = 4. Following the steps described above: 1. The value of a is positive therefore we have a ‘happy face’ quadratic equation. 2. The next step is to calculate where the curve crosses the axis. The curve crosses the y-axis when x = 0, i.e. when y = 02 + 5(0) + 4 = 4, and it crosses the x-axis when y = 0, i.e. when: x2 + 5x + 4 = 0 which can be solved via the method of factorisation to find (x + 4)(x + 1) = 0 and therefore x = −4 and x = −1.

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3. As the value of a is greater than 0 the function has a minimum point, the coordinates of which are 5 5 b x = − 2a = − 2 × 1 = − 2 4 × 1 × 4 − 52 9 4ac − b2 y= = =– 4 4a 4×1

therefore the vertex has the coordinate (−

5 2

, − 94 ).

Bringing all this information together the sketch of the function is given by:

Example 2.9.8 Sketch the function f (x) = −x2 + 2x − 4. As before we begin by writing down the values of a = −1, b = 2 and c = −4. Following the steps as described previously we can sketch the function: 1. As the value of a is negative we have a ‘sad face’ quadratic. 2. The curve crosses the y-axis when x = 0, i.e. when y = −4. To find where the curve crosses the x-axis we are required to solve −x2 + 2x − 4 = 0 using the quadratic equation:



x = −2 ± √22 – 4(–1) (– 4) = – 2 ± √–12 2 × –1 –2 as we can’t square root a negative number then we can see there is no solution. This means that the curve doesn’t cross the x−axis.

3. As the value of a is negative then the curve has a maximum point which is given by (1, −3) Bringing all of this information together we can sketch the function:

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Confidence Builder Questions: Set 2.9 1. Solve the following equations using the method of factorisation. (a) x2 + 4x = –3 (b) x2 + x = 2x + 2 (c) 2x2 – 3x = 2

2

2. Solve the following equations using the quadratic formula. (a) 2x2 – 3x –12 = 0 (b) –2x2 –5x + 2 = 0 (c) 2x2 –5 = 0 3. Solve the following equations using the method of completing the square. (a) x2 + 4x – 3 = 0 (b) –x2 + 2x + 4 = 0 (c) 3x2 + 9x – 2 = 0

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2.10 Pythagoras’ Theorem Pythagoras’ Theorem states that for all right angled triangles: The square of the hypotenuse is equal to the sum of the squares of the other two sides The hypotenuse is the longest side of the triangle, opposite the right angle.

c a

b In the triangle above we can write: a2 + b2 = c2

Example 2.10.1 Find the length of the hypotenuse, correct to 1 decimal place.

4cm

7cm

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Using Pythagoras’ Theorem we can set a = 7 and b = 4 and we therefore find c2 = 72 + 42 = 49 + 16 = 65.

Introduction to Customer Service By taking the square root of both sides we can find the value of the hypotenuse,

67

c = √65 ≈ 8.1cm

Example 2.10.2 Find the length of the missing side correct to 1 decimal place.

12cm

9cm Using Pythagoras’ Theorem we can see that the hypotenuse (the length opposite the right angle) is 12cm. Therefore c = 12. The other length we can denote with either a or b, but we will set a = 9. Using Pythagoras’ Theorem we find 122 = 92 + b2 2 and re-arranging for b we find b2 = 122 – 92 = 144 – 81 = 63 By taking the square root of both sides we can find the value of the missing length, a = √63 ≈ 7.9cm You should remember that when we square root a number we actually get a positive and a negative number. We take the positive number here as we can’t have a negative length.

Confidence Builder Questions: Set 2.10

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Calculate the missing lengths, correct to 1 decimal place.

1 12cm

1)

15cm

14cm

12cm

7cm

2)

9cm

3)

13cm 11cm

4)

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2.11 Trigonometry 68 Business Analytics in Context As we have seen in the previous section on Pythagoras’ Theorem, for a right angled triangle, the side opposite the right angle is called the hypotenuse. If we define one of the angles as θ then we can label the remaining sides as shown in the diagram below:

2 hypotenuse

opposite

θ adjacent

The length next to the angle is called the adjacent and the length opposite the angle is called the opposite. The three lenghts are related by three trigonometric ratios that you need to be familiar with; sine, cosine and tangent. We typically shorten these to sin, cos and tan. The ratios are defined in the following way: opposite sin θ =      hypotenuse adjacent cos θ =     hypotenuse 68

opposite Business Analytics tan in Context θ=   

adjacent

Example 2.11.1 Write down the three trigonometric ratios for the following triangle:

68

5

3 Business Analytics in Context

4

θ

We must first label the sides opposite (opp), adjacent (adj) and hypotenuse (hyp) to find:

opp = 3

hyp = 5

adj = 4

θ

We can now apply the definitions of sine, cosine and tangent to find:

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opposite    3 sin θ = =      hypotenuse  5 adjacent    4 cos θ = =        hypotenuse  5 tan θ =   

opposite    3 =   adjacent   4   

2.11.1 Calculating the Size of an Angle We can use the following trigonometric ratios to calculate the size of an angle in a right angled triangle. To do this we use the inverse trig button on the calculator. The inverse trig button68 is usually found by pressing Shift and either sin, cos or tan to get sin–1, cos–1 Business Analytics in Context or tan–1.

Example 2.11.2 Calculate the angle denoted by θ for the following triangle.

hyp = 7

opp = 3

θ

We must first label the sides of the triangle. The opposite is equal to 3 and the hypotenuse is equal to 7. As we have the opposite and hypotenuse the ratio we should use is sin θ: opposite    3 sin θ = =      hypotenuse  7

3 To calculate weAnalytics are required to calculate the inverse sin, sin–1 of 7 to find 68 the angle Business in Context 3 θ = sin–1 7 = 25.4°

( )

Example 2.11.3 Calculate the angle denoted by θ for the following triangle.

opp = 5 θ adj = 9

We must first label the sides of the triangle and see that the opposite is equal to 5 and the adjacent is equal to 9. As we have the opposite and adjacent the ratio we should use is tanθ: tan θ =     

opposite    5 = adjacent  9

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5 To calculate the angle we are required to calculate the inverse tan, tan–1 of 9 to find 5 θ = tan–1 9 = 29.1°

( )

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In addition to calculating the angle of a triangle we can also use the trigonometric identities to calculate the length of one of the missing sides of a right angled triangle.

2

Example 2.11.4 Calculate the unknown side y in the following triangle. y

35 adj = 12

As with finding the unknown angle, we should correctly label the triangle. For this triangle we can see that the hypotenuse is equal to y and the adjacent is equal to 12. As we have the adjacent and hypotenuse we must use the trigonometric ratio. adjacent   12 cos(35) = =      hypotenuse  y re-arranging to find the value of y: y=

 

12   = 14.6 cos(35)

i.e. y = 14.6.

Confidence Builder Questions: Set 2.11

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Calculate the value of x in each of these triangles, giving your answer to two decimal places. 1)

3)

5) x

x

6

5

2

2)

4) x

70

4

6)

30 8

x

7 x

40 2

1

7)

8) x

12

4

8

7 25 x

x

2

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2.12 Logarithms Logarithms are another way of dealing with powers/indices. For example, if we ask the question “what is 3 raised to the power of 4?”, we could write the answer as 34 = 81 We could have asked this question as: “3 raised to what power is equal to 81?” The answer would be 4. When dealing with this type of question it might be helpful to use logarithms. We would express the question as log3 (81) = 4, which is to say “log base 3 of 81 is equal to 4”. The two questions are equivalent but asked in a slightly different way, i.e. 34 = 81

⇔ log3(81) = 4.

Example 2.12.1 Fill in the missing gaps Logarithmic Form log2 (16) = 4 log5 (625) = 5

Equivalent To ⇔ ⇔ ⇔

Index Form 103 = 1000

By using the equivalance of powers/indices and logarithms we can fill in the gaps, Logarithmic Form log2 (16) = 4 log10 (1000) = 3 log5 (625) = 5

Equivalent To ⇔ ⇔ ⇔

Index Form 24 = 16 103 = 1000 55 = 625

2.12.1 Common Bases There are typically two main bases used in business analytics, base 10 and base e. When we see a logarithm such as log(200) it usually means that it is base 10 and it is equivalent to log10 (200). On your typical calculator if you want to calculate log(200) = log10 (200) you use the ‘log’ button. You should ensure that you know where this button is on your calculator so you should find that log(200) = 2.301. Another base that is commonly used is e, which is Euler’s number, which has the approximate value of 2.71828. When we have a logarithm to the base e we call it the natural logarithm and denote it as loge (20), often written as ln(20) ≈ 2.996.

2.12.2 Rules of Logarithms There are several laws of logarithms which we can use to combine different terms.

Addition Rule When we have two or more logarithms being added together, e.g. log(2) + log(3), we can combine them to a single logarithm by multiplying their arguments, i.e. log(2) + log(3) = log(2×3) = log(6)

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We can generalise this result to find: loga (m) + loga (n) = loga (m×n) It should be noted that we can only combine logarithms that have the same base and there should be no other numbers in front of the logarithm, e.g. you wouldn’t be able to combine the expression: log2(3) + 3log3(4) as they have different bases and there is a 3 in front of the second logarithm.

Subtraction Law When we are calculating the difference of two logarithms, e.g. log(12) – log(3) we can combine them to a single logarithm by dividing their arguments, i.e. log(12) – log(3) = log( 12 ) = log(4) 3 Once again, as when adding logarithms together they must have the same base and there should be no other numbers in front of the logarithm.

Power Law When we are taking the logarithm of a number to some power, we can move the power of the number to the front of the logarithm, i.e. log(an) = nlog(a) We can use these rules to simplify expressions involving logarithms and to solve equations.

Example 2.12.2 Simplify the expression: log(x) + 2log(x2) – log(x3) and write it as a single logarithm. To combine these logarithms we first want to remove the 2 in front of the term log(x2). To do that we shall use the power law to find log(x) + 2log(x2) – log(x3) = log(x) + log((x2)2) – log(x3) = log(x) + log(x4) – log(x3) We can now use the addition and subtraction laws to simplify the expression. Using the addition law we find: log(x) + log(x4) – log(x3) = log(x×x4) – log(x3) = log(x5) – log(x3) We can now use the subtraction law to find:  x 5 log(x5) – log(x3) = log( 3 ) = log(x2) x

Example 2.12.3 You deposit £10,000 into a bank account and after n years you have £15,346.87. Given that the interest rate was 5.5%, calculate how many years you invested your money.

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As we are dealing with a compound interest problem we can employ the equation: p n final value = initial value 1 + 100 where p is the interest rate and n is the number of years. We can substitute the given values into this equation to find: 5.5 n 15,346.87 = 10,000 1 + 100 = 10,000 × 1.055n To solve this equation we are required to first divide through by 10,000 to find:

(

(

)

)

1.534687 = 1.055n Now to solve this equation requires the use of logarithms as the unknown is within the power. If we take logs of both sides we find: log(1.534687) = log(1.055n) Using the power law of logarithms we can bring the power to the front of the expression to find

log(1.534687) = nlog(1.055) and rearrange to find n n=

log(1.534687) log(1.055)

=8

The money was invested for eight years.

Example 2.12.4 Solve the equation

2x = 3x+2

We begin by taking logs of both sides of the equation to find log(2x) = log(3x+2) and by using the powers law of logs we find: x log(2) = (x + 2) log(3) = x log(3) + 2 log(3) and by collecting up the terms involving x we find x log(2) − x log(3) = x(log(2) − log(3)) = 2 log(3) To get the x term on its own we need to divide through by (log(2) − log(3)) to find: 2 log(3) x= ≈ −5.42 log(2) − log(3)

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Confidence Builder Question: Set 2.12 1. Simplfy the following expressions down to a single logarithm: (a) log(2x) + 2log(3x)

(d) –2log(y) + 3log(2x – 3)

(b) log(3x + 1) – 3log(2x – 1)

(e) 5log(z) – 2log(z + 1) + 3log(z + 4)

(c) 3log(a) + 4log(a + 1) – 2log(a + 4)

(f) 2log(y + 2) – 4log(2y –3)

2. Solve the following equations using logarithms, giving your answers to 2 decimal places. (a) 2x = 5

(d) 6x = 32x + 3

(b) 32x = 8

(e) 22x = 5x + 1

(c) 7–3x = 18

(f) 2x = 3x + 1

2.13 Sequences and Series There are many different types of sequences and series. A sequence is defined as an ordered list of numbers that follow a particular rule. The numbers in this ordered list are known as terms of the sequence. We denote the nth term of a sequence with un. For example, consider the following sequence, 1,2,4,8,16,32 where the rule that takes us to the next term in the sequence is to “multiply by 2”. We can say that the second term of the sequence is u2 = 2 and the fifth term is u5 = 16. A series is what we find when we add up all of the terms in a sequence. We call this process the “sum” or “summation” of the sequence. For example the corresponding series: 1 + 2 + 4 + 8 + 16 + 32 takes the value 63. In this section we will only be investigating two types of sequences/series: arithmetic and geometric.

2.13.1 Arithmetic Sequence An arithmetic sequence is a list of numbers such that the difference between each term in the sequence is constant, for example, the following are examples of arithmetic sequences, 2, 5, 8, 11, 14, ... –3, –1, 1, 3, 5, ... 12, 8, 4, 0, –4, ... We call the constant difference between each term the common difference and it is normally denoted with the letter d. In the first sequence d = 3, for the second the common difference is d = 2 and for the final one d = –4; we can add a negative value each time too.

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nth Term of an Arithmetic Sequence If we denote the first term of an arithmetic sequence with a and the common difference as d, we can write down the nth term of the sequence, un via un = a + (n –1)d

Example 12.13.1 Write down the nth term of the following sequence: 3,7,11,15,19,... In this sequence the first term a = 3 and the common difference d = 4. We can therefore write down the nth term of the sequence as

un = 3 + (n – 1) × 4 = 3 + 4(n –1) = 3 + 4n – 4 = 4n –1.

Sum of the First n Terms In many cases we wish to add up all of the terms in a sequence, for example we may want to add the first 50 terms of the following sequence: 4,7,10,13,16,... It would be an arduous task to add all 50 of these terms together. Instead, we can use a result for the sum of the first n terms of a series, Sn: Sn = n 2 (a + L) n   = 2 (2a + (n –1)d)

where a is the first term of the sequence, L is the last term, d is the common difference and n is the number of terms in our sequence. You can use either equation that you prefer, but you will rarely know the last term in a sequence so it is typically best to use the second form of the equation.

Example 12.13.2 Calculate the sum of the first 50 terms of the following sequence 4,7,10,13,16,...,151 For this example we do indeed have the first and last term, i.e. a = 4, L = 151 and n = 50. Therefore the sum of the first 50 terms is given by S50 = 50 2 (4 + 151) = 3,875 We could of course have calculated it using the second equation by writing down the first term a = 4, common difference d = 3 and the number of terms n = 50 to find S50 = 50 2 (2(4) + (50 –1)(3)) = 3,875

2.13.2 Geometric Sequence A geometric sequence is one where each term after the first is found by multiplying the previous term by some fixed, non-zero constant called the common ratio, denoted by r. The following are examples of a geometric sequence,

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3, 12, 48, 192,768, ... 10, 5, 2.5, 1.25, 0.625, ... 4, –8, 16, –32, 64, ... The common ratio of a sequence can be found by dividing the current term by the one before it, i.e. un u2 u3 r = u = u = ... = u (n-1) 1 2 12 The common ratio for the first sequence above is given by r = 3 = 4, for the second 5 1 sequence it is given by r = 10 = 2 = 0.5 and for the final sequence r = -8 4 = –2.

nth Term of a Geometric Sequence If we denote the first term of a geometric sequence with a and the common ratio r, we can write down the nth term of the sequence, un via un = arn –1

Example 12.13.3 Write down the nth term of the following sequence: 2,6,18,54,162,... The first term a = 2 and the common ratio r = 6/2 = 3, therefore the nth term of the sequence is given by un = 2×3n –1 We could use this expression to find the 15th term if we wished, by setting n = 15, i.e. u15 = 2×315 –1 = 2×314 = 9,565,938.

Sum of the First n Terms As with the arithmetic sequence, we can calculate the sum of the first n terms of a geometric sequence. For a given series with first term a and common ratio r, the sum of the first n terms is given by: a(1 –rn) Sn =    1 – r

Example 12.13.4 For the following sequence, calculate the sum of the first 20 terms: 4,6,9,13.5,20.25,... We can see from the sequence that the first term a = 4 and the common ratio r = 6/4 = 1.5. Therefore the sum of the first 15 terms is 4(1 –1.515) S15 = = 3,495.15      1 –1.5

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Sum to Infinity For a geometric sequence, under certain conditions we can sum an infinite number of terms. An infinite series is one that has no last term. The condition on being able to sum an infinite number of terms is that the common ratio should be between –1 and 1, i.e. –1  8) we use the fact that P(r > 8) = 1 – P(r ≤ 8) as the tables only give the cumulative probability, i.e. P(r  ≤  a). To calculate the value of P(r  ≤  8) we are required to look at the table, find μ = 4.5 and r = 8 to find P(r  ≤  8) = 0.959743. Then P(r > 8) = 1 – 0.959743 = 0.040257.

Normal Distribution The normal distribution is a continuous distribution, which is often used in economics. Many naturally occuring quantities follow the normal distribution, for example weight and height. A quantity that follows the normal distribution has the following properties: „„ mean = mode = median. „„ It is symmetric around the mean which is at the centre of the distribution. „„ 50% of the values are greater than the mean and 50% are less than the mean. „„ To calculate probabilities we need to know the mean and standard deviation of the population. The standard deviation is a measure of spread in a data set (see Section 4.4 for more information on how to calculate the standard deviation).

3

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The normal distribution has the following shape: Mean = mode = median

50%

94

50%

Business Analytics in Context Total area = 1 or 100%

For the normal distribution, 68% of the data lies within one standard deviation of the mean as illustrated below:

68% 94

Business Analytics in Context σ=-3 σ=-2 σ=-1

σ=0

σ=1

σ=2

σ=3

It can be shown that 95% of the data lies within two standard deviations of the mean:

95% 94

Business Analytics in Context σ=-3 σ=-2 σ=-1

σ=0

σ=1

σ=2

σ=3

and finally 99.7% of the data lies within three standard deviations of the mean:

99.7%

σ=-3 σ=-2 σ=-1

σ=0

σ=1

σ=2

σ=3

Example 3.3.5 Your annual income is one standard deviation above the population average. What percentage of people earn less than you? We know that 68% of the data lies within one standard deviation of the mean, therefore 68/2 = 34% of the data lies between the mean and one standard deviation. Using the fact that the normal distribution is symmetrical around the mean, i.e. 50% of the data lies

125 Probability 119 Probability

below the mean we find the total percentage of people who earn less than you is:

50% + 34% = 84%

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This may be easier to see with a diagram of the distribution

1 50%

σ=-3 σ=-2 σ=-1

34%

σ=0

σ=1

σ=2

σ=3

Total area = 50% + 34% = 84%

3 3

Example 3.3.6 Your yearly wages are two standard deviation below the population average. What percentage of people earn more than you do? We know that 95% of the data lies within two standard deviations of the mean, therefore 95/2 = 47.5% of the data lies between the mean and two standard deviations. Using the fact that the normal distribution is symmetrical around the mean, i.e. 50% of the data lies below the mean we see that approximately 50% – 47.5% = 2.5% of the population earn less than you do. We have been asked to calculate the percentage of people who earn more than yourself. As the total area under the curve represents 100%, the percentage of people 95 Introduction to Customer Service who earn more than you is given by:

100% – 2.5% = 97.5%

1 47.5% 50% 2.5%

σ=-3 σ=-2 σ=-1

σ=0

σ=1

σ=2

σ=3

Total area to the right = 100% – 2.5% = 97.5%

Example 3.3.7 It is known that 95% of postgraduates in business earn between £25,000 and £32,000. Assuming the data is normally distributed, calculate the mean and standard deviation. We know that the mean will be half way between £25,000 and £32,000: 25,000 + 32,000 Mean = = 28,500 2

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We know that 95% is two standard deviations away from the mean, therefore the difference between the two values is a total of four standard deviations so: 32,000 – 25,000 σ= = 1,750 4

3.3.2 Standard Scores The number of standard deviations from the mean is also known as the standard score or the z-score. To calculate the z-score we use the following formula: x–μ z= σ

Example 3.3.8 Using the data from example 3.3.7 calculate the z-score of someone who earns £33,750. In order to find the z-score, we calculate the difference between the value and the mean and divide by the standard deviation, x – μ 33,750 – 28,500 z= = =3 σ 1,750 We can therefore say that £33,750 is three standard deviations above from the mean.

Example 3.3.9 Using the data from example 3.3.7 calculate the z-score of someone who earns £21,500. We once again calculate the difference between the given value and the mean and divide by the standard deviation, x – μ 21,500 – 28,500 z= = = –4 σ 1,750 We can therefore say that £21,500 is four standard deviations below the mean (below as it is negative). This process of calculating the z-score is a method of standardising the data. By standardising the data we take any normal distribution with mean μ and standard deviation σ 97 Introduction to Customer Service and convert it to the standard normal distribution which has a mean of zero and standard deviation of one. General Normal Distribution

70

80

09

100

110

120

Standardised Normal Distribution

130

z=-3

z=-2

z=-1

z=0

z=1

z=2

z=3

1

Probability

127

What is the purpose of standardising? Standardising the data allows us to use the normal distribution tables, regardless of the actual mean and standard deviation values.

Example 3.3.10 The following are the results of an exam (marks out of 70).

32, 30, 29, 13, 51, 12, 45, 39, 50, 42, 51, 30, 18

Given that the mean score is 34 and the standard deviation is σ = 13.3 what would be your recommendations to the professor marking the paper? As a large majority of the students scored less than half marks, the professor will be required to fail a large number of students. If we assume that the students failed the test due to the test being hard, rather than the students not working hard enough, we can recommend that the professor changes the requirements for passing the exam. If we calculate the z-score for each value we find:

3

-0.15,-0.30,-0.38,-1.58,1.28,-1.65,0.83,0.38,1.20,0.60,1.28,-0.30,-1.20 We can recommend to the professor to change the pass requirement so that everyone who is less than one standard deviation below the mean will still pass the test. Therefore out of the 13 students tested only three of them will fail the test.

Example 3.3.11 Your factory produces bags of sweets with an average bag size of 300g and a standard deviation of 10g. Assuming that the weight of sweets follow a normal distribution, calculate the probability that a bag of sweets weigh less than 317g. We know that the average bag size is 300g with a standard deviation of 10g. To calculate the probability that a bag weighs less than 317g we are required to calculate the z-value, x – μ 317 – 30 z= = σ 10

= 1.7

As the z-score isn’t one of the standard ones we have already looked at, i.e. z = ±1, ±2, ±3, we are required to use the normal distribution standard tables which are given overleaf.

Introduction to Customer Service

The table gives the probability P(z < z0), i.e. the area to the left of our value, i.e. the area given by: z = z0

P(z < z0)

z=-3

z=-2

z=-1

z=0

z=1

z=2

z=3

99

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To calculate the probability P(z < 1.7) we go to the table where the left hand column gives the first decimal place and the row along the top is for the second. Giving P(z < 1.7) = 0.9554. Therefore the probability of the bag of sweets weighing less than 317g is 0.9554.

Example 3.3.12 The average salary of someone in the UK is £28,800 with a standard deviation of £5,500. Assuming it follows a normal distribution, calculate the probabilty of someone chosen at random earning more than £43,000. To find the probability of earning more than £43,000 we first calculate the z-score associated with this value, i.e. x – μ 43,000 – 28,800 z= = ≈ 2.58 σ 5,500 To find the probability of earning more than £43,000 we must calculate the probability of P(z > 2.58). As the table only gives the value to the left, i.e. P(z 2.58) = 1– P(z < 2.58). To calculate the probability P(z < 2.58) we look at the first column to find 2.5 and read across to find 2.58 and P(z < 2.58) = 0.9951. Therefore the probability P(z > 2.58) = 1 – 0.9951 = 0.0049 and the probability of earning more that £43,000 is 0.0049.

Probability

129

Example 3.3.13 A machine dispenses drinks with an average volume of 330ml with a standard deviation of 5ml. Calculate the probability that the machine will dispense a drink with a volume less than 323ml. Once again we are required to calculate the z-score for this value, i.e. z = 323 – 330 = –1.4 5 As we have a negative z-score we must be careful in how we use the table. The table provided (you may have a different table) only gives values for positive z-scores. When we are dealing with negative z-scores we have to use the symmetric nature of the normal distribution and notice that:

3

P(z < – z0) = P(z > z0) = 1– P(z < z0) Therefore to calculate:

Introduction to Customer Service

P(z < –1.4) = P(z > 1.4) = 1– P(z < 1.4) = 1– 0.9192 = 0.0808

We therefore calculate the probability of the machine dispensing a drink less than 323ml is equal to 0.0808.

Example 3.3.14 It is known that the average income of a graduate in business and economics is £33,000 with a standard deviation of £1,500. Calculate the probability of chosing at random a graduate earning between £34,000 and £37,000. We wish to calculate the probability P(34,000 < x < 37000) which is given by the area below. 34,000

37,000

28,500 30,000 31,500 33,000 34,500 36,000 37,500

As we can see the area we wish to calculate is between £34,000 and £37,000 which is given by P(x = 37,000) – P(x = 34,000). Writing these in the standardised model we write P(z < 2.67) – P(z < 0.67). Using the normal distribution table we find P(z < 2.67) = 0.9962 and P(z < 0.67) = 0.7486 therefore:

P(x = 37,000) – P(x = 34,000) = 0.9962 – 0.7486 = 0.2476

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Confidence Builder Questions: Set 3.3 1. In your restaurant you have found that 2 out of every 5 customers ask for a glass of water with their meal. A random sample of 12 customers is selected. (a) Calculate the probability that exactly 7 customers ask for a glass of water with their meal. (b) Calculate the probability that fewer than 9 people ask for a glass of water with their meal. (c) Calculate the probability that more than 4 people will ask for a glass of water with their meal. 2. The probability of a telesales representative making a sale on a particular customer call is 0.2. Calculate the following: (a) The probability that there are exactly 5 sales in 10 calls. (b) The probability that there are more than 4 sales out of 10 calls. 3. You own a factory that produces components of which 2% are known to be defective. The components are packed into boxes of 20. A box is selected at random. (a) Find the probability that the box contains exactly three defective components. (b) Find the probability that there are at least two defective components in the box. 4. There are on average 12 phone calls per hour to a call centre. Assuming the number of phone calls follow a Poisson distribution calculate the following: (a) The probability that there are fewer than 10 phone calls in an hour. (b) The probability that there are more than 15 phone calls in an hour. (c) The probability that there are fewer than 8 phone calls in half an hour. 5. There are on average 10 visitors to a hospital per hour. Assuming the number of visitors follow a Poisson distribution calculate the following: (a) The probability that there are fewer than 2 visitors in an hour. (b) The probability that there are more than 12 visitors in an hour. (c) The probability that there are more than 7 visitors in half an hour. (d) The probability that there are more than 15 visitors during visiting hours (2 hours in total). 6. Assuming a random variable x follows a normal distribution with mean μ = 12 and σ = 3 re-write the following probabilities in terms of their z-score (a) P(x < 15) (b) P(x > 9) (c) P(x > 14)

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131

7. Assuming a random variable x follows a normal distribution with mean μ = 240 and σ = 20 re-write the following probabilities in terms of their z-score: (a) P(x < 250) (b) P(x > 280) (c) P(x > 230) 8. A drinks machine dispenses drinks with an average volume of 400ml with a standard deviation of 12ml. Assuming the volume of drinks follows a normal distribution calculate the following: (a) The probability that the machine dispenses a drink with a volume less than 380ml. (b) The probability that the machine dispenses a drink with a volume greater than 420ml. (c) The probability that the machine dispenses a drink with a volume between 370ml and 410ml. 9. You manufacture bags of flour with an average weight of 1kg and a standard deviation of 0.04kg. Assuming the weight of flour follows a normal distribution calculate: (a) The probability that the bag weighs more than 1.1kg. (b) The probability that the bag weights less than 0.95kg. (c) The probability that the bag weighs between 0.9kg and 1.08kg.

3.4 Decision Analysis Making a decision in business can be a complicated process requiring an assessment of a variety of intangible factors, such as people’s view and ethics, in addition to implementing the outcomes and recommendations of various analytic models. Decision making is a process that involves an element of risk and uncertainty, however there are many different ways in which we can mitigate this risk.

3.4.1 Payoff Table A payoff table is a method of representing a scenario when there are a range of possible outcomes and responses. The purpose of a payoff table is to illustrate the possible profits and losses of each decision we could make. We use a payoff table when the probability of an event occuring is unknown or difficult to estimate. We can also use a payoff table when we are required to perform a simple best-case/worst-case analysis. There are a number of approaches we can take to analyse a payoff table. „„ Average payoff strategy: For this strategy we assume that each event is equally likely to occur and we take the average of all of the outcomes. „„ Opportunity-loss strategy: For this we consider the lost opportunity, also known as the regret associated with each possible decision.

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„„ Aggressive strategy: This is also known as the optimistic approach. We are required to choose the best payoff/highest risk option. „„ Conservative strategy: This is the opposite to the aggressive strategy and is also known as the pessimistic approach. We are required to choose the smallest payoff/less risky option.

Example 3.4.1 You have decided to purchase some new equipment for your factory. You can choose three different pieces of equipment (A, B, C). We call these the decision options as these are the ones we can choose. You have created your payoff table of the income (in thousands of pounds) you will receive depending on which type of contact you secure with one of your clients. There are two possible contracts, I or II, and only one of them can be agreed. These are also known as states of nature. The payoff table is given below: Equipment A Equipment B Equipment C

Contract I 120 130 80

Contact II 130 140 150

Which equipment should you consider purchasing and why? If we consider an average pay off strategy we simply average each of the possible payoffs and then select the one with the best average payoff (for maximisation problems) or the lowest average payoff (for minimisation problems). If you are unsure about calculating an average, view section 4.3. By taking the average of each decision option we find: Equipment A Equipment B Equipment C

Contract I 120 130 80

Contact II 130 140 150

Average 125 135 115

As we are looking to maximise our profit we can see that the best option to choose is equipment B as it gives us the highest average payoff of £135,000. If we consider an opportunity-loss strategy we should calculate the regret for each outcome. We typically want to minimise the maximum regret of the decision that we take. You may also see this called minimax regret. To calculate the regret we consider the difference between the maximum payoff of each state of nature and the payoff for that particular decision. Calculating the regret for each state of nature we find: Equipment A Equipment B Equipment C

Contract I

Regret Rs1

Contact II

Regret Rs2

120 130 80

10 0 50

130 140 150

20 10 0

A regret of zero means that it is the best option in the column. A regret of 50 as seen in column Rs1 means that it is 50 away from the best option in that column. We should then consider the maximum regret for each possible decision to find:

Probability

Contract I Equipment A Equipment B Equipment C

Contact II

Regret Rs1 10 0 50

120 130 80

130 140 150

Regret Rs2 20 10 0

133

Maximum Regret 20 10 50

From the table we can see that option B will minimise our maximum regret. If we consider the optimistic approach (or sometimes known as the maximax payoff) we simply choose the best possible payoff. To calculate this we identify the maximum for each possible decision, i.e. Contract I 120 130 80

Equipment A Equipment B Equipment C

Contact II 130 140 150

Maximum Payoff 130 140 150

We can see from the table that the maximum payoff occurs with equipment C giving a payoff of £150,000. The pessimistic approach (or sometimes known as the maximin payoff) requires us to identify the minimum payoff for each state of nature. We then make our decision based on the option for which the minimum payoff is maximised. Contract I 120 130 80

Equipment A Equipment B Equipment C

Contact II 130 140 150

Minimum Payoff 120 130 80

The decision that maximises the minimum payoff would be equipment B with £130,000.

3.4.2 Decision Analysis Using Probabilities When we have an estimate of the probabilities of various states of nature we can decide the optimal decision, based on the expected values of the payoffs. We return to Example 3.4.1, but this time you have estimated the probability of getting Contract I to be 70% and the probability of 30% of getting Contract II. To calculate the expected outcome we employ the following formula:





n

Σ p × contract

ExpectedValue = EV =

i

i

i=1

where pi is the probability of that outcome occuring and n is the total number of states of nature. Calculating the expected value for each outcome we find: Equipment A Equipment B Equipment C

Contract I 120 130 80

Contact II 130 140 150

Expected Value = 0.7 × 120 + 0.3 × 130 = 123 = 0.7 × 130 + 0.3 × 140 = 133 = 0.7 × 80 + 0.3 × 150 = 101

We can therefore see that the option with the highest expected value is equipment B with an expected value of £133,000.

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4.4.3 Decision Tree 128

Instead of using a payoff table we can instead use a decision tree. We represent chance nodes with circles and decision nodes with squares and we use a process called backward induction to calculate the result. Representing the payoff table in Example 3.4.1 with a decision tree we find:

Business Analytics in Context

P(S1) = 0.7

120

Equipment A P(S2) = 0.3 P(S1) = 0.7 Start

130 130

Equipment B P(S2) = 0.3

140

P(S1) = 0.7

80

P(S2) = 0.3

150

Equipment C 128

Business Analytics in Context

We can calculate the expected values for each decision and we fill in the circles to find: P(S1) = 0.7 Equipment A

122 P(S2) = 0.3 P(S1) = 0.7

Start

Equipment B

130 130

133 P(S2) = 0.3

Equipment C

120

140

P(S1) = 0.7

80

P(S2) = 0.3

150

101

Once we have calculated the expected values for each decision we look at the highest value and fill in the square box, i.e.

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Business Analytics in Context Probability

P(S1) = 0.7 Equipment A

120

122 P(S2) = 0.3 P(S1) = 0.7

133

135

Equipment B

130 130

133 P(S2) = 0.3

Equipment C

140

P(S1) = 0.7

80

P(S2) = 0.3

150

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101

Confidence Builder Questions: Set 3.4 1. The following payoff table shows the profit associated with a decision analysis problem with two decision alternatives and three states of nature. State 1

State 2

State 3

Decision 1

150

280

240

Decision 2

240

230

220

(a) If you know nothing about the probabilities of each of the three states occuring, what would be the recommended decision using the minimax regret approach? (b) Which decision would you recommend based on the average pay-off approach? (c) If you know that probability of state 1 occuring is 0.7, state 2 occuring is 0.1 and state 3 occuring is 0.2, use the expected value approach to determine the optimal decision. 2. You own an ice cream store and you are looking to expand your business. The payoff table shows the profit associated with expanding your current store or opening a new store based on three states of the economy (figures are profit, in thousands of pounds). Growing Economy

Stagnant Economy

Recession

Expand Current Store

30

10

-20

Open New Store

40

15

-40

(a) If you know nothing about the probabilities of the state of the economy, what would be the recommended decision using the minimax regret approach? (b) Which decision would you recommend based on the average pay-off approach? (c) If you know that probability of a growing economy is 0.4, stagnant economy is 0.5 and a recession is 0.1, use the expected value approach to dermine the optimal decision.

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3. You are considering buying 1000 shares in a company which is currently valued at £4 per share. You have followed the history of the company and have estimated that the shares will either increase in value by 30% over the next year or decrease by 40% over the same period. You believe that over the next year the probability of a 30% rise is 0.55. Your alternative action is to leave the money in the bank and earn £50 from the interest. Which approach would you recommend taking?

Probability

137

My notes

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Business Analytics in Context

My notes

Probability

139

My notes

3

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My notes

Probability

141

Applying your Knowledge End of Chapter Questions 1. You have done some market research to determine whether it would be worthwhile launching your new product in a new territory. You have found from your market research that 40% of the people in that territory live in the city centre, 52% live in the suburbs and the rest live in the countryside. From selling products in other territories you know that 30% of the people who live in the city centre will buy your product, 35% of those in the suburbs will do so and 20% of those from the countryside will buy it. From your market research you find there are 2,500,000 people living in the territory.

(a) What is the probability that a randomly chosen person from the territory will buy your product?



(b) How many people in the territory would you expect to buy your product?



You decide to go ahead with introducing your product to the territory. You have two strategies: strategy A will increase the percentage of people who live in the city centre who buy your product from 30% to 40% but has the effect of reducing the sales to those in the suburbs down to 30%. Strategy B has the effect of increasing the sales to those from the suburbs from 35% to 45% whilst reducing the sales to those from the city centre by 5%. The percentage of the purchases from those from the countryside will remain unaffected in both strategies.



(c) Which strategy would produce a greater increase in the number of sales of your product?



Upon further analysis of the market it is shown that there is only 55% chance that the markets will be favourable and the outcomes above will be true. If the market condtions are not favourable it has been shown that strategy A will lead to a decrease in 2% of sales in each catagory and if strategy B is chosen then the status quo remains.



(d) Using the expectation criteria identify which strategy you would recommend.

2. You are the owner of a high end electronics store. To try and increase the number of sales of higher priced goods you have introduced an interest free finance scheme.

(a) You have noticed that within the first year of offering the interest free finance that there was an average of 4 loan repayment defaults per month. Using the Poisson distribution calculate the probability that there are 7 or fewer loan defaults in a month.



(b) Using the Poisson distribution calculate the probability that there are more than 30 loan defaults in a year.

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Applying Business Analytics 1. You are the owner of your own business with 12 employees.

(a) Due to a worldwide pandemic the probability of a member of staff being off sick is p = 0.2. What is the probability that half your team will be off sick at the same time.



(b) Due to the number of staff being off sick you are required to take on 8 temporary staff. The new staff have an average salary of £1,800 per month with a standard deviation of 200. How much extra would you have to spend to cover the wages of the temporary staff for three months.



(c) Assuming the staff wages follow a normal distribution, what is the probability that a temporary member of staff earns between £1,600 and £1,900 per month.



After the pandemic is over you need to restructure your business. You have the choice of laying off some members of staff, retain the orginal 12 members of staff or keep on the tempoaray member of staff. The pay-off table is given below (amounts in 10’s of thousands): Lay off staff Retain original staffing levels Retain original and temporary staff



Growing economy 10 20 40

Stagnant economy 5 −5 −20

Recession −20 −30 −35

(d) Given that the probability that the economy is growing is p = 0.1, the probability that the economy is stagnant is p = 0.2 and the probability that there is a recession is 0.7 u se the expected value approach to determine the optimal decision to make.

Probability

143

My workings

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My workings

Probability

145

My workings

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My workings

Probability

147

Solutions Confidence Builders Set 3.1 1. {(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)} 2. 1/12

3. 1/8

4. 2/25 = 0.08

5. (a) 0.02 (b) 0.98 (c) Probabilility of machine A working = 0.93 and of machine B working = 0.85; therefore machine A is the most reliable. (d) As P(Machine A working) × P(Machine B working) ≠ P(Machine A and Machine B working), i.e. 0.7905 ≠ 0.80 we can say that the two machines are not statistically independent. 6. P(B or Q) = P(B) + P(Q) – P(B ∩ Q) = 7/13 7. P(>2000) = 0.45 × 0.6 + 0.55 × 0.3 = 0.435

Set 3.2 1. (a) 5/44 ≈ 0.11

   (b) 12/35 ≈ 0.34

(c) 9/44 ≈ 0.20

2. 66.7%    3. 0.65

4. 0.73

5. 7/13 ≈ 0.54    6. 0.091

76. 0.43

8. 0.014

Set 3.3 1. (a) 0.1009

(b) 0.9847

(c) 0.5618

2. (a) 0.0264

(b) 0.0328

3. (a) 0.0065

(b) 0.0599

4. (a) 0.2424

(b) 0.1556

(c) 0.7440

5. (a) 0.0005

(b) 0.2084

(c) 0.1334

6. (a) P(z < 1)

(b) P(z > –1)

(c) P(z > 2/3)

7. (a) P(z < 1/2)

(b) P(z > 2)

(c) P(z > –1/2)

8. (a) 0.048

(b) 0.048

(c) 0.792

9. (a) 0.006

(b) 0.106

(c) 0.971

(d) 0.8435

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Set 3.4 1. (a) To minimise the maximum regret we should follow decision 2 as the maximum regret would be 50 compared to 90 for decision 1. (b) The average pay-off for decision 1 is equal to 223.33 compared to 230 for decision 2, therefore we should follow decision 2. (c) E(d1) = 181 and E(d2) = 235, therefore decision 2 would be the better option. 2. (a) To minimise the maximum regret we should expand the current store as the maximum regret would be 10 compared to 20 for opening a new business. (b) The average pay-off for expanding the store is equal to 6.67 compared to 5 for opening a new store, therefore we would recommend expanding the current store. (c) Expected return for expanding the current store is 15 compared to 19.5, therefore based on this criteria the recommended approach would be to open a new store. 3. You would expect to lose £60 therefore the recommended course of action would be to leave the money in the bank.

End of Chapter Questions 1.

(a) 0.318

(b) 795,000

2.

(a) 0.9489 (b) 0.99635

(c) Strategy B

(d) Strategy B

Applying Business Analytics 1. (a) We assume that the probability is binomially distribution, i.e. with n = 12 and p = 0.2. The probability of half your team being off sick is given by: 12

C6 × 0.26 × (1 − 0.2)6 = 0.016



(b) You are taking on an additional 8 members of staff who each earn £1,800 × 3 = £5,400 for three months. Therefore to cover 8 eight staff you are required to pay an additional 8 × 5,400 = £43,200



(c) We assume the staff earn on average £1,800 with a standard deviation of £200. To calculate the probability that a member of staff earns between £1,600 and £1,900 we must calculate the z-score associated with each value. For £1,600 we have a z-score of: 1,600 − 1,800 = −1 z= 200



and for £1,900 we have a z-score of 1,900 − 1,800 = 1 z= 200 2

Probability



149

Using the z table we find P (z < 12 ) = 0.6915 and P (z < −1) = P (z > 1) = 1 − P (z < 1) = 1 − 0.8413 = 0.1587. We therefore find: P (−1 < z < 12 ) = 0.6915 − 0.1587 = 0.5328

(d) We calculate the expected return on each of the three possibilities. Lay off staff = 0.1 × 10 + 0.2 × 5 + 0.7 × (−20) = −12 Retain original staff levels = 0.1 × 20 + 0.2 × (−5) + 0.7 × (−30) = −20 Retain original and temporary staff = 0.1 × 40 + 0.2 × (−20) + 0.7 × (−35) = −24.5 Therefore the optimum outcome, in financial terms, would be to lay off staff.

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4

Statistics

A strong grasp of statistics is vital if you are to be successful in business. Statistics can be used to analyse data and help you make decisions. In this chapter we will be considering the different ways in which we represent data, different methods of analysing data and finally how we can perform some statistical tests to aid decision making.

4.1 Representing Data There are many different ways of representing data depending on the type of data that you have. In this section we will be looking at bar charts, histograms, line graphs, pie charts and scatter diagrams.

4.1.1 Bar Charts We use bar charts to compare two or more values with a small set of results. When we represent data with a bar chart, the height of the bar shows the frequency of the result occuring.

Example 4.1.1 The average rainfall for a six month period is illustrated in the bar chart below:

Calculate the total rainfall for the whole of the six month period. To calculate the total rainfall for the six months shown we must read off the amount of rainfall each month. By looking at the graph we see that there was 80cm of rainfall in September, 95cm in October, a further 95cm in November, 100cm in December, 95cm in

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January and February was the driest month with 70cm. Therefore the total rainfall is the sum of all these values, i.e. 80 + 95 + 95 + 100 + 95 + 70 = 535 Therefore over the six months there was 535cm of rainfall.

4.1.2 Line Graphs A line graph is used to show trends in data over time. Line graphs are typically used to show a trend over a number of hours, days or months. They are plotted as a series of points which are connected by a straight line. The first and last points of the line do not need to join to the axes.

Example 4.1.2 The following line graph shows the lowest overnight temperature for a week. 1. What was the lowest temperature and on which day did it occur? 2. What was the difference between the highest and lowest temperature?

1. The lowest temperature can be seen to occur on Wednesday and the temperature on that particular evening was −2◦C. 2. To calculate the difference between the highest and lowest temperatures we first need to find the highest temperature (we found the lowest in the previous question). From the line graph we can see that the highest temperature was 7◦ C which occured on Friday. The difference therefore between the highest and lowest temperature is given by 7 − (−2) = 9.

4.1.3 Pie Charts Pie charts use various sized sectors of a circle to represent the data. To interpret a pie chart we need to recall that there are 360◦ in a circle to calculate the size of each segment.

Example 4.1.3 A market research company runs a survey to determine the most popular flavour of crisp. 60 students were asked and the findings are illustrated in the pie chart below. Given that the numbers represent the angle of each sector of the pie chart, how many people prefered salt and vinegar crisps?

Statistics

153

To calculate the number of people who preferred salt and vinegar crisps we need to determine the angle which represents the salt and vinegar flavour. We can see from the pie chart that salt and vinegar was represented by a segment of 60◦. To calculate the number of people that this represents we use the following: angle 60 total number = × number of people in the survey = × 60 = 10 360 360 We can therefore see that there were 10 people who voted for salt and vinegar as their favourite flavour.

4.1.4 Scatter Diagrams Scatter diagrams show the relationship between two sets of data. We can then describe the relationship between the two variables using correlation.

Example 4.1.4 In the scatter diagram below, the grades for recent Maths and English tests are displayed. Describe the relationship (if any) between the grades.

It should be clear from the scatter diagram that those students who scored highly in the Maths test, also scored highly in the English test.

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4.1.5 Histograms Histograms are used when we were are dealing with continuous data. Histograms look very similar to bar charts but we interpret them very differently. When we consider bar charts, all of the bars have the same width and it is the height of the bars that is most significant. For histograms, it is the area that is the most significant aspect to consider. Consider the following data set on the volume of liquid dispensed from a drinks machine and let us construct the associated histogram. Volume (ml)

Frequency

300 ≤ V < 320

6

320 ≤ V < 340

6

340 ≤ V < 380

18

380 ≤ V < 400

10

When we draw a histogram we typically have the different classes on the x-axis and frequency density on the y-axis. To calculate the frequency density we use the formula: Frequency Frequency Density = Class Width Calculating the frequency density for each class we find: Frequency

Class Width

Frequency Density

300 ≤ V < 320

6

20

0.3

320 ≤ V < 340

6

20

0.3

18

40

0.45

10

20

0.5

340 ≤ Vin < Context 380 Business Analytics 380 ≤ V < 400

which we can plot to create the histogram: 0.5

Frequency density

112

Volume (ml)

0.4 0.3 0.2 0.1 0 300

320

340

360

380

400

Volume (ml)

The important characteristic of a histogram compared to a bar chart is that there should be no gaps between the bars. This is because histograms represent continuous data. The other important point to know about histograms is that the area of each bar represents the frequency or the number of data points in that particular class.

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Given the following histogram: 1. How many cars were travelling between 30 − 40mph? 2. How many cars were travelling between 60 − 80mph? Frequency density

0.5 0.4 0.3 0.2 0.1 0 30

40

50

60

70

80

Speed (mph)

1. To calculate the number of cars traveling between 30 and 40mph, we calculate the area of the bar representing this data. Recall, the area of each bar represents how many data points there are in that class. The area of the bar representing 30 to 40 is given by 10 × 0.4 112= 4.Business Analytics Context Therefore, 4 carsinwere travelling between 30 and 40 mph. 2. To calculate the number of cars travelling between 60 and 80mph we need to calculate the following shaded area on the histogram.

Frequency density

0.5 0.4 0.3 0.2 0.1 0 30

40

50

60

70

80

Speed (mph)

The area is given by 10 × 0.5 + 10 × 0.5 = 10. Therefore, there were 10 cars travelling between 60 − 80mph.

Confidence Builder Questions: Set 4.1 1. Consider the following line chart showing how the inflation rate has changed in the UK since 1984.

(a) What was the highest rate of inflation since 1984 and in which year did it occur?



(b) What was the lowest rate of inflation before 2015 and in which year did it occur?



(c) What is the difference between the highest and lowest rate of inflation?

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2. Consider the following line chart showing how the Bank of England base rate has changed in the UK since 1998.



(a) What was the highest base rate since 1998?



(b) What was the lowest base rate before 2007?



(c) What is the difference between the highest and lowest base rate?

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3. Consider the following bar chart showing the number of iPhone sales worldwide since its launch in 2007. 250 231.22 211.88

216.76 217.72

200 169.22 150.26

150 125.05 100 72.27 50

0

148

39.99

1.39

11.63

4

20.73



(a) For which year were the number of sales highest?



(b) What were the total number of sales since 2007.



(c) If each iPhone costs on average £250 to manufacture and sells for £700. How much profit has been made since the iPhone was launched?

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4. Consider the following bar chart showing the mobile phone market share in the first quarter of 2019.

35 30

30 28

25

23

20 15

14

10 5

3

2

0 Apple

Samsung

Huawei

Xiami

Alcatel

Other

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(a) If Apple’s market share had grown 10% from the fourth quarter in 2018. What was its market share in the fourth quarter of 2018?



(b) If Huawei and Xiami merged, what would their total market share be?

5. Consider the following pie chart showing information on the blood types of 300 people.



(a) How many people had Type O blood?



(b) How many more people had Type B blood than Type A?

6. You have a fleet of 200 cars. The following pie chart gives a break down of the types of cars in your fleet. Types of Cars of 200 Fleet Vehicles



(a) How many diesel cars are in the fleet?



(b) How many more hybrid vehicles do you have, compared to electric?

7. Consider the following scatter graph illustrating the relationship between advertising costs and customer enquiries. How would you describe the relationship between advertising spending and customer enquiries?

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4 8. Consider the scatter graph showing the relationship between number of units manu- factured and factory overheads. How would you describe the relationship between number of units and factory overheads?

9. You are given the following data of the distribution of salaries in your company. Salary (thousands of£) Frequency

0 − 10

10 − 20

20 − 30

30 − 40

40 − 60

10

15

12

40

15



(a) Represent the data in a histogram.



(b) Estimate how many people earn between £30, 000 and £35, 000.

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10. You are given the following data of the distribution of waiting times in your store. Waiting Times (minutes) Frequency

0−5 20

5 − 10 10 − 20 15

20

20 − 25

25 − 30

30 − 40

10

10

15



(a) Represent the data in a histogram.



(b) Estimate how many people had to wait less than 12 minutes.

4.2 Sampling Methods To understand sampling methods, we need to understand the difference between a population and a sample. „„ Population: The total set of all observations that can be made. The number of elements in a population is denoted by N . For example if we wanted to study the height of adult males, the population would be defined as the set of heights of all males in the world. „„ Sample: A set of observations which have been drawn from some population. The number of elements in a sample is denoted by n. We typically use a sample when it is impractical or impossible to study a whole population. For example if we wanted to study the height of all adult males it would be impossible to measure the height of all males in the world, but we would be able to measure a sample of adult males.

4.2.1 Sampling Methods We use various sampling methods as a way of taking observations from a population. There are two main types of samples: probability and non-probability. When we are performing a probability sampling method, each element of the population has a known chance, which is non-zero, of being chosen. The main benefit of a probability sampling method is that it guarantees that the chosen sample is representative of the population. For a non-probability sampling method we do not know the probability of a particular element being chosen, or even if it is possible for a particular element to be chosen.

4.2.2 Probability Sampling Methods There are several different methods of probability sampling: simple random sampling, stratified sampling, cluster sampling and systematic random sampling.

Simple Random Sampling A simple random sample is when each of the elements in the population has an equal probability of being selected. There are many ways in which a simple random sample can be performed. For example, if each element of a population is assigned a unique number, and the numbers then get drawn out of a bowl.

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Stratified Sampling When we are performing a stratified sample, we divide the population in groups based on some particular characteristic, e.g. gender or ethnicity. Each group is called a strata, and we sample each strata using typically simple random sampling. The number of elements we sample in each group depends on the number of elements in that particular stata. To calculate the number of elements in each strata we can use the equation: strata size × sample size number selected from each strata = total population size

Example 4.2.1 Arnold owns a cafe and wants to sample 50 of his customers about the new coffee beans that he has been trying out. The table below shows how many customers attended Arnold’s cafe during the last week, based on age and gender. 16 - 25

26 - 35

36-45

46-55

56 or older

Female

12

10

24

18

15

Male

8

15

22

19

7

Arnold has chosen to perform a stratified sampling method. How many females aged 46 − 55 should he ask? Stratified sampling requires us to choose the number of elements in each sample. To find out how many females aged between 46 − 55 we use: strata size 18 number selected from each strata = × sample size = × 50 = 6 total population size 150

Cluster Sampling With cluster sampling, each member of a population is assigned to a group, called a cluster. A cluster is then chosen using another probability sampling method (typically simple random sampling). Once a cluster has been chosen, only individuals in that cluster are surveyed.

Systematic Random Sampling When we are performing a systematic random sample we create a list of every member of the population. From this list, we then randomly select a first sample element from the first m elements on the population list. Thereafter, we choose every mth element on the list.

4.2.3 Non-Probability Sampling Methods There are two main non-probability sampling methods: convenience and voluntary.

Convenience Sampling A convenience sample is simply one that is easy to perform, i.e. it is made up of people who are easy to make observations from. For example, you may interview people in your local supermarket as it is close to your business and would be convenient for you.

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Voluntary Sample A voluntary sample is made up of people who choose to take the survey. For example, if you are watching a TV show and have been asked to vote for a winner, this would be classed as a volunteer sample.

Confidence Builder Questions: Set 4.2 1. You wish to investigate the thoughts and opinions of the staff that work in your department. The following descriptions of different sampling methods are given below, which method are they?

(a) You approach the first 20 members of staff that enter the building in the morning.



(b) Members of staff have their ID numbers checked and if their number ends in a 2, 4, 6 or 8 they are interviewed.

2. Constance introduces a new drink to her cafe menu. She wants to sample 20 of her customers about the new drink. The table below shows how many customers ordered the new drink during the last week based on age and gender.



16 - 25

26 - 35

36-45

46-55

56 or older

Female

14

2

20

28

5

Male

18

5

15

11

9

Assuming Constance is performing a stratified sampling method, how many females aged 16 − 25 should she ask?

4.3 Measures of Location Measures of location are a mathematical way of summarising a list of numbers with some “typical” value. The three most common measures are the mean, median and mode.

4.3.1 The Mean The mean is commonly known as the average. The mean is defined as:

Σ

     n

mean =

     

i=1

n

xi

which when we write it in words requires us to add up all of the data points and then divide by the number of points.

Example 4.3.1 Calculate the mean of the following set of data 10, 13, 17, 11, 16, 10, 13, 15, 19, 18, 15, 14. To calculate the mean we are required to add up all of our data points: 10 + 13 + 17 + 11 + 16 + 10 + 13 + 15 + 19 + 18 + 15 + 14 = 171

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then divide this answer by the number of data points, which in this case is 12. We therefore find: 171 mean = = 14.25 12

Example 4.3.2 Calculate the mean of the following set of data 2, 5, 1, 6, 9, 7, 1. To calculate the mean we are required to add up all of our data points to find 2 + 5 + 1 + 6 + 9 + 7 + 1 = 31 we then divide this value by the number of data points, which in this case is 7: 31 mean = = 4.43 7

Calculating the Mean from Frequency Tables When dealing with a large amount of data we generally use a frequency table. The purpose of a frequency table is to summarise the data in a more succinct manner. For example, if we were presented with the following data set: 1, 2, 2, 1, 1, 1, 1, 2, 3, 4, 3, 2, 2, 3, 4, 3, 2, 1, 1, 2, 3, 4, 1, 3, 2, 2, 5, 2, 1, 1, 2, 3, 4, 5, 4, 4, 4, 4, 3, 2, 1, 1, 1, 1, 1, 1, 2, 2, 3, 2, 1, 5, 3, 2, 1, 5, 4, 3, 5, 5, 5, 5, 5, 3, 2, 1, 2, 3, 4, 2, 2 it would be awkward to work with, as there is simply so much data to look at. A frequency table would summarise this by keeping track of how many occurrences of each outcome there are. For example we can see that 1 occurs 19 times, there are 20 occurrences of 2, 13 occurrences of 3, 10 occurrences of 4 and finally 9 occurrences of 5. All of this data can be summarised in the frequency table below. Score 1 2 3 4 5

Frequency 19 20 13 10 9

To calculate the mean when the data is represented in a frequency table we use the equation:

Σ

     n

mean =

     

i= 1

n

fi xi

where xi is the ith value and fi represents how many times the score xi occurred. To calculate the mean we introduce a third column which represents the score × frequency:

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Score

Frequency

Score × Frequency

1

19

1 × 19 = 19

2

20

2 × 20 = 40

3

13

3 × 13 = 39

4

10

4 × 10 = 40

5

9

5 × 9 = 45

The formula requires us to calculate:

n

Σ

fi xi =

5

Σ

fi xi = 19 + 40 + 39 + 40 + 45 = 183

i =1 i =1

and to calculate the mean we divide this value by n, the number of terms. It is important to note that when we are dealing with data as a frequency, n is equal to the sum of all the frequencies, i.e. n = 19 + 20 + 13 + 10 + 9 = 71. We therefore find the mean is equal to 183/71 = 2.58.

Grouped Frequency Tables We have previously seen how we can use frequency tables to summarise a large amount of data. If we have a large amount of data with many different values it might be sensible to ‘group’ the data. Consider the following values: 9, 12, 15, 4, 6, 2, 6, 2, 1, 18, 18, 14, 5, 18, 12, 11, 12, 3, 10, 14, 15, 2, 5, 14, 12, 14, 15, 12, 13, 8, 9, 2, 6, 7, 8, 5, 4, 3, 2, 6, 17, 16, 14, 13, 12, 15, 16, 12, 11, 10, 5, 4, 2, 6, 5, 5, 6, 3, 5, 7, 1 The first step would be to write the numbers in order and then calculate the range of the data. Writing the numbers in order: 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 12, 12, 12, 12, 12, 12, 12, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 17, 18, 18, 18 The range as we will see in the next section is defined as the largest value in the set minus the smallest, i.e. range = largest – smallest We can see from the data that the largest data point is 18 whereas the smallest is 1 which leads to a range of 18 − 1 = 17. To calculate the group size we divide the range by how many groups we would like. For this data we would like 5 groups, but you may want to chose more or fewer groups. As a consequence, we will divide the range by 5, i.e. 17/5 = 3.4 which we shall round up to a whole number, 4. We should then choose a starting value for our groups which should be less than or equal to the smallest value in our data set. For this set of data, 0 would be a sensible starting value. We shall therefore define the following groups, 0 − 4, 4 − 8, 8 − 12, 12 − 16 and 16 − 20. Now we should be careful as there is an overlap on the groups. To avoid this we will define the groups in the following ways 0 ≤ x < 4, 4 ≤ x < 8, 8 ≤ x < 12, 12 ≤ x < 16, 16 ≤ x < 20 which will avoid any overlap.

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Frequency

0≤x