Instuctor’s Solutions Manual for Statistics for Business and Economics [12 ed.] 9780321836816, 0321836812

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INSTRUCTOR’S SOLUTIONS MANUAL NANCY S. BOUDREAU Bowling Green State University

S TATISTICS FOR B USINESS AND E CONOMICS TWELFTH EDITION

James T. McClave Info Tech, Inc. University of Florida

P. George Benson College of Charleston

Terry Sincich University of South Florida

Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo





























The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson from electronic files supplied by the author. Copyright © 2014, 2011, 2008 Pearson Education, Inc. Publishing as Pearson, 75 Arlington Street, Boston, MA 02116. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-321-83681-6 ISBN-10: 0-321-83681-2

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Chapter 1 Statistics, Data, and Statistical Thinking 1.1

Statistics is a science that deals with the collection, classification, analysis, and interpretation of information or data. It is a meaningful, useful science with a broad, almost limitless scope of applications to business, government, and the physical and social sciences.

1.2

Descriptive statistics utilizes numerical and graphical methods to look for patterns, to summarize, and to present the information in a set of data. Inferential statistics utilizes sample data to make estimates, decisions, predictions, or other generalizations about a larger set of data.

1.3

The four elements of a descriptive statistics problem are: 1. 2. 3. 4.

1.4

The population or sample of interest. This is the collection of all the units upon which the variable is measured. One or more variables that are to be investigated. These are the types of data that are to be collected. Tables, graphs, or numerical summary tools. These are tools used to display the characteristic of the sample or population. Identification of patterns in the data. These are conclusions drawn from what the summary tools revealed about the population or sample.

The five elements of an inferential statistical analysis are: 1. 2. 3. 4.

5.

The population of interest. The population is a set of existing units. One or more variables that are to be investigated. A variable is a characteristic or property of an individual population unit. The sample of population units. A sample is a subset of the units of a population. The inference about the population based on information contained in the sample. A statistical inference is an estimate, prediction, or generalization about a population based on information contained in a sample. A measure of reliability for the inference. The reliability of an inference is how confident one is that the inference is correct.

1.5

The first major method of collecting data is from a published source. These data have already been collected by someone else and are available in a published source. The second method of collecting data is from a designed experiment. These data are collected by a researcher who exerts strict control over the experimental units in a study. These data are measured directly from the experimental units. The final method of collecting data is observational. These data are collected directly from experimental units by simply observing the experimental units in their natural environment and recording the values of the desired characteristics. The most common type of observational study is a survey.

1.6

Quantitative data are measurements that are recorded on a meaningful numerical scale. Qualitative data are measurements that are not numerical in nature; they can only be classified into one of a group of categories.

1.7

A population is a set of existing units such as people, objects, transactions, or events. A variable is a characteristic or property of an individual population unit such as height of a person, time of a reflex, amount of a transaction, etc.

1 Copyright © 2014 Pearson Education, Inc.

2

Chapter 1

1.8

A population is a set of existing units such as people, objects, transactions, or events. A sample is a subset of the units of a population.

1.9

A representative sample is a sample that exhibits characteristics similar to those possessed by the target population. A representative sample is essential if inferential statistics is to be applied. If a sample does not possess the same characteristics as the target population, then any inferences made using the sample will be unreliable.

1.10

An inference without a measure of reliability is nothing more than a guess. A measure of reliability separates statistical inference from fortune telling or guessing. Reliability gives a measure of how confident one is that the inference is correct.

1.11

A population is a set of existing units such as people, objects, transactions, or events. A process is a series of actions or operations that transform inputs to outputs. A process produces or generates output over time. Examples of processes are assembly lines, oil refineries, and stock prices.

1.12

Statistical thinking involves applying rational thought processes to critically assess data and inferences made from the data. It involves not taking all data and inferences presented at face value, but rather making sure the inferences and data are valid.

1.13

The data consisting of the classifications A, B, C, and D are qualitative. These data are nominal and thus are qualitative. After the data are input as 1, 2, 3, and 4, they are still nominal and thus qualitative. The only differences between the two data sets are the names of the categories. The numbers associated with the four groups are meaningless.

1.14

Answers will vary. First, number the elements of the population from 1 to 200,000. Using MINITAB, generate 10 numbers on the interval from 1 to 200,000, eliminating any duplicates. The 10 numbers selected for the random sample are: 135075 89127 189226 83899 112367 191496 110021 44853 42091 198461 Elements with the above numbers are selected for the sample.

1.15

a.

The experimental unit for this study is a single-family residential property in Arlington, Texas.

b.

The variables measured are the sale price and the Zillow estimated value. Both of these variables are quantitative.

c.

If these 2,045 properties were all the properties sold in Arlington, Texas in the past 6 months, then this would be considered the population.

d.

If these 2,045 properties represent a sample, then the population would be all the single-family residential properties sold in the last 6 months in Arlington, Texas.

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Statistics, Data, and Statistical Thinking

1.16

1.17

1.18

1.19

3

e.

No. The real estate market across the United States varies greatly. The prices of single-family residential properties in this small area are probably not representative of all properties across the United States.

a.

The experimental unit for this study is an NFL quarterback.

b.

The variables measured in this study include draft position, NFL winning ratio, and QB production score. Since the draft position was put into 3 categories, it is a qualitative variable. The NFL winning ratio and the QB production score are both quantitative.

c.

Since we want to project the performance of future NFL QBs, this would be an application of inferential statistics.

a.

The population of interest is all citizens of the United States.

b.

The variable of interest is the view of each citizen as to whether the president is doing a good or bad job. It is qualitative.

c.

The sample is the 2000 individuals selected for the poll.

d.

The inference of interest is to estimate the proportion of all U.S. citizens who believe the president is doing a good job.

e.

The method of data collection is a survey.

f.

It is not very likely that the sample will be representative of the population of all citizens of the United States. By selecting phone numbers at random, the sample will be limited to only those people who have telephones. Also, many people share the same phone number, so each person would not have an equal chance of being contacted. Another possible problem is the time of day the calls are made. If the calls are made in the evening, those people who work in the evening would not be represented.

a.

High school GPA is a number usually between 0.0 and 4.0. Therefore, it is quantitative.

b.

Honors/awards would have responses that name things. Therefore, it would be qualitative.

c.

The scores on the SAT's are numbers between 200 and 800. Therefore, it is quantitative.

d.

Gender is either male or female. Therefore, it is qualitative.

e.

Parent's income is a number: $25,000, $45,000, etc. Therefore, it is quantitative.

f.

Age is a number: 17, 18, etc. Therefore, it is quantitative.

I.

Qualitative; the possible responses are "yes" or "no," which are non-numerical.

II.

Quantitative; age is measured on a numerical scale, such as 15, 32, etc.

III.

Qualitative; the possible responses are “yes” or “no,” which are non-numerical.

IV.

Qualitative; the possible responses are "laser printer" or "another type of printer," which are nonnumerical.

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1.20

Chapter 1 V.

Qualitative; the speeds can be classified as "slower," "unchanged," or "faster," which are nonnumerical.

VI.

Quantitative; the number of people in a household who have used Windows 95 at least once is measured on a numerical scale, such as 0, 1, 2, etc.

a.

For question 1, the data collected would be qualitative. The possible response would be “yes” or “no”. For question 2, the data collected would be quantitative. The responses would be numbers such as 0, 1, 2, etc. For question 3, the data collected would be qualitative. The possible responses would be “yes” or “no”.

1.21

1.22

1.23

b.

The data collected from the 1,066 adults would be a sample. These adults would only be a part of all adults in the United States.

a.

Whether the data collected on the chief executive officers at the 500 largest U. S. companies is a population or a sample depends on what one is interested in. If one is only interested in the information from the CEO’s of the 500 largest U.S. companies, then these data form a population. If one is interested in the information on CEO’s from all U.S. firms, then these data would form a sample.

b.

1.

The industry type of the CEO’s company is a qualitative variable. The industry type is a name.

2.

The CEO’s total compensation is a meaningful number. Thus, it is a quantitative variable.

3.

The CEO’s total compensation over the previous five years is a meaningful number. Thus, it is a quantitative variable.

4.

The number of company stock shares (millions) held is a meaningful number. Thus, it is a quantitative variable.

5.

The CEO’s age is a meaningful number. Thus, it is a quantitative variable.

6.

The CEO’s efficiency rating is a meaningful number. Thus, it is a quantitative variable.

a.

The population of interest is the status of computer crime at all United States businesses and government agencies.

b.

The method of data collection was a survey. Since not all of those who were sent a survey responded, the sample was self-selected. The results are probably not representative of the population. Usually, those who respond to surveys have very strong opinions, either positive or negative.

c.

The variable of interest is whether or not the firm or agency had unauthorized use of its computer systems during the year. Since the response would be either yes or no, the variable would be qualitative.

d.

If the sample was representative, we could infer that approximately 41% of all U. S. corporations and government agencies experienced unauthorized use of their computer systems during the year.

Since the data collected consist of the entire population, this would represent a descriptive study. Flaherty used the data to help describe the condition of the U.S. Treasury in 1861.

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Statistics, Data, and Statistical Thinking

5

1.24

This study would be an example of inferential statistics. The researchers collected data over 2 years. Using this information, the researchers are projecting or making inferences about what will happen in the future.

1.25

a.

The population of interest is all individuals who earned MBA degrees since January 1990.

b.

The method of data collection was a survey.

c.

This is probably not a representative sample. The sample was self-selected. Not all of those who were selected for the study responded to all four surveys. Those who did respond to all 4 surveys probably have very strong opinions, either positive or negative, which may not be representative of all of those in the population.

a.

The population of interest is all CPA firms.

b.

A survey was used to collect the data.

c.

This sample was probably not representative. Not all of those selected to be in the sample responded. In fact, only 992 of the 23,500 people who were sent the survey responded. Generally, those who do respond to surveys have very strong opinions, either positive or negative. These may not be the opinions of all CPA firms.

d.

Since the sample may not be representative, the inferences drawn in the study may not be valid.

a.

Length of maximum span can take on values such as 15 feet, 50 feet, 75 feet, etc. Therefore, it is quantitative.

b.

The number of vehicle lanes can take on values such as 2, 4, etc. Therefore, it is quantitative.

c.

The answer to this item is "yes" or "no," which is not numeric. Therefore, it is qualitative.

d.

Average daily traffic could take on values such as 150 vehicles, 3,579 vehicles, 53,295 vehicles, etc. Therefore, it is quantitative.

e.

Condition can take on values "good," "fair," or "poor," which are not numeric. Therefore, it is qualitative.

f.

The length of the bypass or detour could take on values such as 1 mile, 4 miles, etc. Therefore, it is quantitative.

g.

Route type can take on values "interstate," U.S.," "state," "county," or "city," which are not numeric. Therefore, it is qualitative.

a.

The variable of interest to the researchers is the rating of highway bridges.

b.

Since the rating of a bridge can be categorized as one of three possible values, it is qualitative.

c.

The data set analyzed is a population since all highway bridges in the U.S. were categorized.

d.

The data were collected observationally. Each bridge was observed in its natural setting.

1.26

1.27

1.28

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6 1.29

1.30

1.31

Chapter 1 a.

The process being studied is the distribution of pipes, valves, and fittings to the refining, chemical, and petrochemical industries by the Wallace Company of Houston.

b.

The variables of interest are the speed of the deliveries, the accuracy of the invoices, and the quality of the packaging of the products.

c.

The sampling plan was to monitor a subset of current customers by sending out a questionnaire twice a year and asking the customers to rate the speed of the deliveries, the accuracy of the invoices, and the quality of the packaging minutes. The sample is the total numbers of questionnaires received.

d.

The Wallace Company's immediate interest is learning about the delivery process of its distribution of pipes, valves, and fittings. To do this, it is measuring the speed of deliveries, the accuracy of the invoices, and the quality of its packaging from the sample of its customers to make an inference about the delivery process to all customers. In particular, it might use the mean speed of its deliveries to the sampled customers to estimate the mean speed of its deliveries to all its customers. It might use the mean accuracy of its invoices from the sampled customers to estimate the mean accuracy of its invoices of all its customers. It might use the mean rating of the quality of its packaging from the sampled customers to estimate the mean rating of the quality of its packaging of all its customers.

e.

Several factors might affect the reliability of the inferences. One factor is the set of customers selected to receive the survey. If this set is not representative of all the customers, the wrong inferences could be made. Also, the set of customers returning the surveys may not be representative of all its customers. Again, this could influence the reliability of the inferences made.

a.

The population of interest would be the set of all students. The sample of interest would be the students participating in the experiment. The variable measured in this study is whether the student would spend money on repairing a very old car or not.

b.

The data-collection method used was a designed experiment. The students participating in the experiment were randomly assigned to one of three emotional states and then asked a question.

c.

The researcher could estimate the proportion of all students in each of the three emotional states who would spend money to repair a very old car.

d.

One factor that might affect the reliability of the inference drawn is whether the students in the experiment were representative of all students. It is stated that the sample was made up of volunteer students. Chances are that these volunteer students were not representative of all students. In addition, if these students were all from the same school, they probably would not be representative of the population of students either.

a.

The population of interest would be all accounting alumni of a large southwestern university.

b.

Age would produce quantitative data – the responses would be numbers. Gender would produce qualitative data – the responses would be ‘male’ or ‘female’. Level of education would produce qualitative data – the responses could be categories such college degree, master’s degree, or PhD degree. Income would produce quantitative data – the responses would be numbers. Job satisfaction score would produce quantitative data. We would assume that a satisfaction score would be a number, where the higher the number, the higher the job satisfaction. Machiavellian rating score would produce quantitative data. We would assume that a rating score Copyright © 2014 Pearson Education, Inc.

Statistics, Data, and Statistical Thinking

7

would be a number, where the higher the score, the higher the Machiavellian traits. c.

The sample is the 198 people who returned the useable questionnaires.

d.

The data collection method used was a survey.

e.

The inference made by the researcher is that Machiavellian behavior is not required to achieve success in the accounting profession.

f.

Generally, those who respond to surveys are those with strong feelings (in either direction) toward the subject matter. Those who do not have strong feelings for the subject matter tend not to answer surveys. Those who did not respond might be those who are not real happy with their jobs or those who are not real unhappy with their jobs. Thus, we might have no idea what type of scores these people would have on the Machiavellian rating score.

1.32

a.

Give each stock in the NYSE-Composite Transactions table of the Wall Street Journal a number (1 to m). Using a random number table or a computer program, select n different numbers on the interval from 1 to m. The stocks with the same numbers as the n chosen numbers will be selected for the sample.

1.33

a.

The experimental units for this study are engaged couples who used a particular website.

b.

There are two variables of interest – the price of the engagement ring and the level of appreciation. Price of the engagement ring is a quantitative variable because it is measured on a numerical scale. Level of appreciation is a qualitative variable. There are 7 different categories for this variable that are then assigned numbers.

c.

The population of interest would be all engaged couples.

d.

No, the sample is probably not representative. Only engaged couples who used a particular web site were eligible to be in the sample. Then, only those with “average” American names were invited to be in the sample.

e.

Answers will vary. First, we will number the individuals from 1 to 50. Using MINITAB, 25 random numbers were generated on the interval from 1 to 50. The random numbers are: 1, 4, 5, 8, 12, 13, 17, 18, 19, 20, 22, 26, 27, 30, 31, 33, 34, 35, 38, 39, 40, 42, 43, 46, 49 The individuals who were assigned the numbers corresponding to the above numbers would be assigned to one role and the remaining individuals would be assigned to the other role.

1.34

Answers will vary. Using MINITAB, the 5 seven-digit phone numbers generated with area code 373 were: 373-639-0598 373-411-9164 373-502-7699 373-782-2719 373-930-3231

1.35

a.

Some possible questions are: 1.

In your opinion, why has the banking industry consolidated in the past few years? Check all that apply. a.

Too many small banks with not enough capital. Copyright © 2014 Pearson Education, Inc.

8

Chapter 1 b. c. d. e. f. 2.

A result of the Savings and Loan scandals. To eliminate duplicated resources in the upper management positions. To provide more efficient service to the customers. To provide a more complete list of financial opportunities for the customers. Other. Please list.

Using a scale from 1 to 5, where 1 means strongly disagree and 5 means strongly agree, indicate your agreement to the following statement: "The trend of consolidation in the banking industry will continue in the next five years." 1 strongly disagree

1.36

1.37

1.38

2 disagree

3 no opinion

4 agree

5 strongly agree

b.

The population of interest is the set of all bank presidents in the United States.

c.

It would be extremely difficult and costly to obtain information from all bank presidents. Thus, it would be more efficient to sample just 200 bank presidents. However, by sending the questionnaires to only 200 bank presidents, one risks getting the results from a sample which is not representative of the population. The sample must be chosen in such a way that the results will be representative of the entire population of bank presidents in order to be of any use.

a.

The process being studied is the process of filling beverage cans with soft drink at CCSB's Wakefield plant.

b.

The variable of interest is the amount of carbon dioxide added to each can of beverage.

c.

The sampling plan was to monitor five filled cans every 15 minutes. The sample is the total number of cans selected.

d.

The company's immediate interest is learning about the process of filling beverage cans with soft drink at CCSB's Wakefield plant. To do this, they are measuring the amount of carbon dioxide added to a can of beverage to make an inference about the process of filling beverage cans. In particular, they might use the mean amount of carbon dioxide added to the sampled cans of beverage to estimate the mean amount of carbon dioxide added to all the cans on the process line.

e.

The technician would then be dealing with a population. The cans of beverage have already been processed. He/she is now interested in the outputs.

a.

The population of interest is the set of all people in the United States over 14 years of age.

b.

The variable being measured is the employment status of each person. This variable is qualitative. Each person is either employed or not.

c.

The problem of interest to the Census Bureau is inferential. Based on the information contained in the sample, the Census Bureau wants to estimate the percentage of all people in the labor force who are unemployed.

Suppose we want to select 900 intersections by numbering the intersections from 1 to 500,000. We would then use a random number table or a random number generator from a software program to select 900 distinct intersection points. These would then be the sampled markets. Now, suppose we want to select the 900 intersections by selecting a row from the 500 and a column from the 1,000. We would first number the rows from 1 to 500 and number the columns from 1 to 1,000. Using a random number generator, we would generate a sample of 900 from the 500 rows. Obviously, many rows will be selected more than once. At the same time, we use a random number generator to select 900 Copyright © 2014 Pearson Education, Inc.

Statistics, Data, and Statistical Thinking

9

columns from the 1,000 columns. Again, some of the columns could be selected more than once. Placing these two sets of random numbers side-by-side, we would use the row-column combinations to select the intersections. For example, suppose the first row selected was 453 and the first column selected was 731. The first intersection selected would be row 453, column 731. This process would be continued until 900 unique intersections were selected. 1.39

Answers will vary. a.

The results as stated indicate that by eating oat bran, one can improve his/her health. However, the only way to get the stated benefit is to eat only oat bran with limited results. People may change their eating habits expecting an outcome that is almost impossible.

b.

To investigate the impact of domestic violence on birth defects, one would need to collect data on all kinds of birth defects and whether the mother suffered any domestic violence or not during her pregnancy. One could use an observational study survey to collect the data.

c.

Very few people are always happy with the way they are. However, many people are happy with themselves most of the time. One might want to ask a series of questions to measure self-esteem rather than just one. One question might ask what percent of the time the high school girl is happy with the way she is.

d.

The results of the study are probably misleading because of the fact that if someone relied on a limited number of foods to feed her children it does not imply that the children are hungry. In addition, one might cut the size of a meal because the children were overweight, not because there was not enough food. One might get better information about the proportion of hungry American children by actually recording what a large, representative sample of children eat in a week.

e.

A leading question gives information that seems to be true, but may not be complete. Based on the incomplete information, the respondent may come to a different decision than if the information was not provided.

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Chapter 2 Methods for Describing Sets of Data 2.1

First, we find the frequency of the grade A. The sum of the frequencies for all five grades must be 200. Therefore, subtract the sum of the frequencies of the other four grades from 200. The frequency for grade A is: 200  (36 + 90 + 30 + 28) = 200  184 = 16 To find the relative frequency for each grade, divide the frequency by the total sample size, 200. The relative frequency for the grade B is 36/200 = .18. The rest of the relative frequencies are found in a similar manner and appear in the table: Grade on Statistics Exam A: 90 100 B: 80  89 C: 65  79 D: 50  64 F: Below 50 Total

2.2

a.

Relative Frequency .08 .18 .45 .15 .14 1.00

To find the frequency for each class, count the number of times each letter occurs. The frequencies for the three classes are: Class X Y Z Total

b.

Frequency 16 36 90 30 28 200

Frequency 8 9 3 20

The relative frequency for each class is found by dividing the frequency by the total sample size. The relative frequency for the class X is 8/20 = .40. The relative frequency for the class Y is 9/20 = .45. The relative frequency for the class Z is 3/20 = .15. Class X Y Z Total

Frequency 8 9 3 20

Relative Frequency .40 .45 .15 1.00

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Methods for Describing Sets of Data c.

11

The frequency bar chart is:

9 8

Frequency

7 6 5 4 3 2 1 0 X

d.

Y C la s s

Z

The pie chart for the frequency distribution is: Pie Chart of Class Category X Z 15.0%

Y Z

X 40.0%

Y 45.0%

2.3

a.

The type of graph is a bar graph.

b.

The variable measured for each of the robots is type of robotic limbs.

c.

From the graph, the design used the most is the “legs only” design.

d.

The relative frequencies are computed by dividing the frequencies by the total sample size. The total sample size is n = 106. The relative frequencies for each of the categories are: Type of Limbs None Both Legs ONLY Wheels ONLY Total

Frequency 15 8 63 20 106

Relative Frequency 15/106 = .142 8 / 106 = .075 63/106 = .594 20/106 = .189 1.000

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12

Chapter 2 e.

Using MINITAB, the Pareto diagram is:

.60

Relative Frequency

.50

.40 .30 .20 .10 0 Legs

Wheels

None

Both

Type Percent within all data.

2.4

a.

From the pie chart, 50.4% or .504 of the sampled adults living in the U.S. use the internet and pay to download music. From the data, 506 out of 1,003 adults or 506/1,003 = .504 of sampled adults in the U.S. use the internet and pay to download music. These two results agree.

b.

Using MINITAB, a pie chart of the data is: Pie Chart of Download-Music Category Pay No Pay

No Pay 33.0%

Pay 67.0%

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Methods for Describing Sets of Data 2.5

13

Using MINITAB, the Pareto diagram for the data is: Chart of Tenants 50

Percent

40

30

20

10

0 Small

SmallStandard

Large Tenants

Major

Anchor

Percent within all data.

Most of the tenants in UK shopping malls are small or small standard. They account for approximately 84% of all tenants ([711 + 819]/1,821 = .84). Very few (less than 1%) of the tenants are anchors. 2.6

a.

The relative frequency for each response category is found by dividing the frequency by the total sample size. The relative frequency for the category “Insurance Companies” is 869/2119 = .410. The rest of the relative frequencies are found in a similar manner and are reported in the table. Most responsible for rising health-care costs Insurance companies Pharmaceutical companies Government Hospitals Physicians Other Not at all sure TOTAL

Number responding 869 339 338 127 85 128 233 2,119

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Relative Frequencies 869/2119 = .410 339/2119 = .160 338/2119 = .160 127/2119 = .060 85/2119 = .040 128/2119 = .060 233/2119 = .110 1.000

14

Chapter 2 b.

Using MINITAB, the relative frequency bar chart is: Chart of Category 40

Count

30

20

10

0 Insurance Co

c.

Pharm

Government Hospitals Physicians Category

Other

Not sure

O ther

Phy sicians

Using MINITAB, the Pareto diagram is: Chart of Category

Relative Frequency

.40

.30

.20

.10

0 Insurance Co. Gov ernment

Pharm

Not sure

Hospitals

Category

Most American adults in the sample (41%) believe that the Insurance companies are the most responsible for the rising costs of health care. The next highest categories are Government and Pharmaceutical companies with about 16% each. Only 4% of American adults in the sample believe physicians are the most responsible for the rising health care costs. 2.7

a.

Since the variable measured is manufacturer, the data type is qualitative.

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Methods for Describing Sets of Data b.

15

Using MINITAB, a frequency bar chart for the data is: Number Shipped 120000

Frequency

100000 80000 60000 40000

Pax Tech

Provenco

SZZT

Toshiba

Urmet

Toshiba

Pax Tech

Glintt

EIntelligent

Urmet

Omron

KwangWoo

EIntelligent

Glintt

Fujian Landi

Bitel

0

CyberNet

20000

Manufacturer

Using MINITAB, the Pareto diagram is: Number Shipped 120000 100000 Frequency

80000 60000 40000

Bitel

CyberNet

Provenco

Omron

KwangWoo

0

SZZT

20000

Fujian Landi

c.

Manufacturer

Most PIN pads shipped in 2007 were manufactured by either Fujian Landi or SZZT Electronics. These two categories make up (119,000 + 67,300)/334,039= 186,300/334,039 = .558 of all PIN pads shipped in 2007. Urmet shipped the fewest number of PIN pads among these 12 manufacturers.

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16 2.8

Chapter 2 Using MINITAB, the bar graphs of the 2 waves is:

Sch

NoWorkGrad

NoWorkBusSch

WorkMBA

Sch

2

NoWorkGrad

WorkMBA

NoWorkBusSch

1

90 80 70 60 50 40 30 20 10 0

WorkNoMBA

Percent

WorkNoMBA

Chart of Job Status

Job Status Panel variable: Wave; Percent within all data.

In wave 1, most of those taking the GMAT were working (2657/3244 =.819) and none had MBA’s. About 20% were not working but were in either a 4-year institution or other graduate school ([36 + 551]/3244 = .181). In wave 2, almost all were now working ([1787 + 1372]/3244 = .974). Of those working, more than half had MBA’s (1787/[1787 + 1372] = .566). Of those not working, most were in another graduate school. 2.9

Using MINITAB, the pie chart is: Pie Chart of Percent vs Blog/Forum C ategory C ompany Employ ees Third Party Not Identified

Not Identified 15.4%

Company 38.5% Third Party 11.5%

Employ ees 34.6%

Companies and Employees represent (38.5 + 34.6 = 73.1) slightly more than 73% of the entities creating blogs/forums. Third parties are the least common entity.

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data 2.10

17

Using MINITAB, a bar chart of the data is:

Chart of INDUSTRY 14 12 Count

10 8 6 4 2 Aerospace & Defense Banking Business Services & Capital Goods Chemicals Conglomerates Construction Consumer Durables Diversified Financia Drugs & biotechnolog Food Drink & tobacco Health care equipmen Hotels, Restaurants Household & personal Insurance Materials Media Oil & Gas Operations Retailing Semiconductors Software & Services Technology Hardware Telecommunications s Transportation Utilities

0

INDUSTRY

Industries with the highest frequencies include Oil & Gas Operations, Retailing, Drugs & biotechnologies, and Health care equipment. Industries with the smallest frequencies include Business Services, Construction, Banking, and Consumer Durables. 2.11

a.

Using MINITAB, a pie chart of the data is: Pie Chart of PREVUSE Category NEVER USED USED 28.8%

NEVER 71.2%

From the chart, 71.2% or .712 of the sampled physicians have never used ethics consultation. Copyright © 2014 Pearson Education, Inc.

18

Chapter 2 b.

Using MINITAB, a pie chart of the data is: Pie Chart of FUTUREUSE C ategory NO YES

NO 19.5%

YES 80.5%

From the chart, 19.5% or .195 of the sampled physicians state that they will not use the services in the future. c.

Using MINITAB, the side-by-side pie charts are: Pie Chart of PREVUSE MED

SURG

C ategory NEVER USED

USED 27.9%

USED 29.3%

NEVER 70.7%

NEVER 72.1%

Panel variable: SPEC

The proportion of medical practitioners who have never used ethics consultation is .707. The proportion of surgical practitioners who have never used ethics consultation is .721. These two proportions are almost the same.

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data d.

19

Using MINITAB, the side-by-side pie charts are: Pie Chart of FUTUREUSE MED

SURG

Category NO YES

NO 17.3%

NO 23.3%

YES 76.7%

YES 82.7%

Panel variable: SPEC

The proportion of medical practitioners who will not use ethics consultation in the future is .173. The proportion of surgical practitioners who will not use ethics consultation in the future is .233. The proportion of surgical practitioners who will not use ethics consultation in the future is greater than that of the medical practitioners. Using MINITAB, the side-by-side bar graphs are: Chart of Acquisitions No 1980

Yes 1990 100 75 50

Percent

2.12

25 0

2000 100 75 50 25 0 No

Yes

Acquisitions Panel variable: Year; Percent within all data.

In 1980, very few firms had acquisitions 18 / 1,963  .009  . By 1990, the proportion of firms having acquisitions increased to 350 / 2,197  .159 . By 2000, the proportion of firms having acquisitions increased to 748 / 2,778  .269 .

Copyright © 2014 Pearson Education, Inc.

20 2.13

Chapter 2 Using MINITAB, the side-by-side bar graphs are: Chart of Dive Left

Middle

Ahead

Right

Behind

80 60

Percent

40 20 0

Tied

80 60 40 20 0 Left

Middle

Right

Dive Panel variable: Situation; Percent within all data.

From the graphs, it appears that if the team is either tied or ahead, the goal-keepers tend to dive either right or left with equal probability, with very few diving in the middle. However, if the team is behind, then the majority of goal-keepers tend to dive right (71%). 2.14

Using MINITAB, a pie chart of the data is: Pie Chart of Measure

Big Shows 20.0%

Total visitors 26.7%

Category Big Shows Funds Raised Members Paying visitors Total visitors

Funds Raised 23.3% Paying visitors 16.7%

Members 13.3%

Since the sizes of the slices are close to each other, it appears that the researcher is correct. There is a large amount of variation within the museum community with regard to performance measurement and evaluation.

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data 2.15

21

a.

The variable measured by Performark is the length of time it took for each advertiser to respond back.

b.

The pie chart is: Pie Chart of Response Time C ategory > 120 days 13-59 day s 60-120 days Never responded

> 120 days 12.0%

Never responded 21.0%

13-59 days 33.0%

60-120 day s 34.0%

Twenty-one percent or .21 17,000  3,570 of the advertisers never respond to the sales lead.

d.

The information from the pie chart does not indicate how effective the "bingo cards" are. It just indicates how long it takes advertisers to respond, if at all.

a.

Using MINITAB, the side-by-side graphs are:

Chart of Frequency vs Stars 5 Content

4

3

2

1

Exposure 16 12 8

Frequency

2.16

c.

4 Faculty

Opportunity

0

16 12 8 4 0 5

4

3

2

1

Stars Panel variable: Criteria

From these graphs, one can see that very few of the top 30 MBA programs got 5-stars in any criteria. In addition, about the same number of programs got 4 stars in each of the 4 criteria. The biggest difference in ratings among the 4 criteria was in the number of programs receiving 3-stars. More programs received 3-stars in Course Content than in any of the other criteria. Consequently, fewer programs received 2-stars in Course Content than in any of the other criteria.

Copyright © 2014 Pearson Education, Inc.

b.

Since this chart lists the rankings of only the top 30 MBA programs in the world, it is reasonable that none of these best programs would be rated as 1-star on any criteria.

a.

Using MINITAB, bar charts for the 3 variables are: Chart of Well Class 120 100

Count

80 60 40 20 0 Private

Public Well Class

Chart of Aquifer 200

150

Count

2.17

Chapter 2

100

50

0 Bedrock

Unconsolidated Aquifer

Chart of Detection 160 140 120 100 Count

22

80 60 40 20 0 Below Limit

Detect Detection

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data b.

23

Using MINITAB, the side-by-side bar chart is: Chart of Detection Below Limit Private

Detect Public

80 70

Percent

60 50 40 30 20 10 0 Below Limit

Detect Detection

Panel variable: Well Class; Percent within all data.

c.

Using MINITAB, the side-by-side bar chart is: Chart of Detection Below Limit Bedrock

Detect Unconsoli

70 60

Percent

50 40 30 20 10 0 Below Limit

Detect Detection

Panel variable: Aquifer; Percent within all data.

d.

From the bar charts in parts a-c, one can infer that most aquifers are bedrock and most levels of MTBE were below the limit ( 2 / 3) . Also the percentages of public wells verses private wells are relatively close. Approximately 80% of private wells are not contaminated, while only about 60% of public wells are not contaminated. The percentage of contaminated wells is about the same for both types of aquifers ( 30%) .

Copyright © 2014 Pearson Education, Inc.

24 2.18

Chapter 2 Using MINITAB, the relative frequency histogram is:

.25

Relative Frequency

.20

.15

.10

.05

0 0

2.5

4.5

6.5

8.5 Class

10.5

12.5

14.5

16.5

To find the number of measurements for each measurement class, multiply the relative frequency by the total number of observations, n = 500. The frequency table is: Measurement Class Relative Frequency .10 .5  2.5 .15 2.5  4.5 .25 4.5  6.5 .20 6.5  8.5 .05 8.5  10.5 .10 10.5  12.5 .10 12.5  14.5 .05 14.5  16.5

Frequency 500(.10) = 50 500(.15) = 75 500(.25) = 125 500(.20) = 100 500(.05) = 25 500(.10) = 50 500(.10) = 50 500(.05) = 25 500

Using MINITAB, the frequency histogram is: 140 120 100 Frequency

2.19

.5

80 60 40 20 0 0

.5

2.5

4.5

6.5

8.5 Class

10.5

12.5

14.6

16.5

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data 2.20

a.

The original data set has 1 + 3 + 5 + 7 + 4 + 3 = 23 observations.

b.

For the bottom row of the stem-and-leaf display:

25

The stem is 0. The leaves are 0, 1, 2. Assuming that the data are up to two digits, rounded off to the nearest whole number, the numbers in the original data set are 0, 1, and 2.

2.21

2.22

2.23

c.

Again, assuming that the data are up to two digits, rounded off to the nearest whole number, the dot plot corresponding to all the data points is:

a.

This is a frequency histogram because the number of observations is graphed for each interval rather than the relative frequency.

b.

There are 14 measurement classes.

c.

There are 49 measurements in the data set.

a.

The measurement class 10 – 20 has the highest proportion of respondents.

b.

The approximate proportion of the 144 organizations that reported a percentage monetary loss from malicious insider actions less than 20% is .30 + .38 = .68.

c.

The approximate proportion of the 144 organizations that reported a percentage monetary loss from malicious insider actions greater than 60% is .07 + .03 + .04 + .05 = .19.

d.

The approximate proportion of the 144 organizations that reported a percentage monetary loss from malicious insider actions between 20% and 30% is .11. Therefore about .11(144) = 15.84 or 16 of the 144 organizations reported a percentage monetary loss from malicious insider actions between 20% and 30%.

a.

Since the label on the vertical axis is Percent, this is a relative frequency histogram. We can divide the percents by 100% to get the relative frequencies.

b.

Summing the percents represented by all of the bars above 100, we get approximately 12%.

Copyright © 2014 Pearson Education, Inc.

2.24

Chapter 2 a.

Using MINITAB, the stem-and-leaf display and histogram are: Stem-and-Leaf Display: Score Stem-and-leaf of Score Leaf Unit = 1.0 1 1 2 3 3 4 4 5 7 11 17 24 41 62 (37) 87 27

6 7 7 7 7 7 8 8 8 8 8 9 9 9 9 9 10

N

= 186

9 3 4 8 3 44 6667 888999 0001111 22222222222333333 444444555555555555555 6666666666666667777777777777777777777 888888888888888888888888888899999999999999999999999999999999 000000000000000000000000000 Histogram of Score

60 50

Frequency

26

40 30

20 10

0 72

b.

c.

76

80

84 Score

88

92

96

100

From the stem-and-leaf display, there are only 7 observations with sanitation scores less than 86. The proportion of ships with accepted sanitation standards is (186  7) / 186  179 / 186  .962 . The score of 69 is highlighted in the stem-and-leaf display.

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data 2.25

a.

Using MINITAB, a dot plot of the data is: Dotplot of Acquisitions

0

2.26

a.

240

360 480 Acquisitions

600

720

840

By looking at the dot plot, one can conclude that the years 1996-2000 had the highest number of firms with at least one acquisition. The lowest number of acquisitions in that time frame (748) is almost 100 higher than the highest value from the remaining years. Using MINITAB, a histogram of the current values of the 32 NFL teams is: Histogram of Value ($mil) 14 12 10 Frequency

b.

120

8 6 4 2 0 750

900

1050

1200 1350 Value ($mil)

1500

1650

1800

Copyright © 2014 Pearson Education, Inc.

27

Chapter 2 b.

Using MINITAB, a histogram of the 1-year change in current value for the 32 NFL teams is: Histogram of Chang1Yr (%) 10

Frequency

8

6

4

2

0 -4

c.

-2

0

2 4 Chang1Yr (%)

6

8

10

Using MINITAB, a histogram of the debt-to-value ratios for the 32 NFL teams is: Histogram of Debt/Value (%) 20

15 Frequency

28

10

5

0 0

16

32 Debt/Value (%)

48

64

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data d.

29

Using MINITAB, a histogram of the annual revenues for the 32 NFL teams is: Histogram of Revenue ($mil) 16 14

Frequency

12 10 8 6 4 2 0 225

e.

250

275

300 325 Revenue ($mil)

350

375

400

Using MINITAB, a histogram of the operating incomes for the 32 NFL teams is: Histogram of Income ($mil) 10

Frequency

8

6

4

2

0 0

f.

20

40 60 Income ($mil)

80

100

120

For all of the histograms, there is 1 team that has a very high score. The Dallas Cowboys have the largest values for current value, annual revenues, and operating income. However, the New York Giants have the highest 1-year change, while the New York Jets have the highest debt-to-value ratio. All of the graphs except the one showing the 1-Yr Value Changes are skewed to the right.

Copyright © 2014 Pearson Education, Inc.

a.

Using MINITAB, the frequency histograms for 2011 and 2010 SAT mathematics scores are: His togr a m of M ATH 2 0 1 1 , M A TH 2 0 1 0 480

520

M A T H2011

Frequency

2.27

Chapter 2

560

600

M A T H2010

14

14

12

12

10

10

8

8

6

6

4

4

2

2

0

0 480

520

560

600

It appears that the scores have not changed very much at all. The graphs are very similar. b.

Using MINITAB, the frequency histograms for 2011 and 2001 SAT mathematics scores are: His togr am of M ATH2 0 1 1 , M ATH2 0 0 1 480 M A T H2011

510

540

570

600

M A T H2001

14

12

12 10 10 Frequency

30

8 8 6 6 4

4

2

2 0

0 480

520

560

600

It appears that the scores have shifted to the right. The scores in 2011 appear to be somewhat better than the scores in 2011.

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data c.

31

Using MINITAB, the frequency histogram of the differences is: H is togr a m of D iffM a th 16 14

Frequency

12 10 8 6 4 2 0 -32

-16

0 DiffM a t h

16

32

From this graph of the differences, we can see that there are more observations to the right of 0 than to the left of 0. This indicates that, in general, the scores have improved since 2001. d.

2.28

From the graph, the largest improvement score is in the neighborhood of 32. The actual largest score is 32 and it is associated with Michigan.

Using MINITAB, the two dot plots are: Dotplot of Arrive, Depart

Arrive Depart 108

120

132

144

156

168

Data

Yes. Most of the numbers of items arriving at the work center per hour are in the 135 to 165 area. Most of the numbers of items departing the work center per hour are in the 110 to 140 area. Because the number of items arriving is larger than the number of items departing, there will probably be some sort of bottleneck.

Copyright © 2014 Pearson Education, Inc.

32 2.29

Chapter 2 Using MINITAB, the stem-and-leaf display is: Stem-and-Leaf Display: Dioxide Stem-and-leaf of Dioxide Leaf Unit = 0.10

5 7 (2) 7 7 5 5 4 4

0 0 1 1 2 2 3 3 4

N

= 16

12234 55 34 44 3 0000

The highlighted values are values that correspond to water specimens that contain oil. There is a tendency for crude oil to be present in water with lower levels of dioxide as 6 of the lowest 8 specimens with the lowest levels of dioxide contain oil. 2.30

Yes, we would agree with the statement that honey may be the preferable treatment for the cough and sleep difficulty associated with childhood upper respiratory tract infection. For those receiving the honey dosage, 14 of the 35 children (or 40%) had improvement scores of 12 or higher. For those receiving the DM dosage, only 9 of the 33 (or 24%) children had improvement scores of 12 or higher. For those receiving no dosage, only 2 of the 37 children (or 5%) had improvement scores of 12 or higher. In addition, the median improvement score for those receiving the honey dosage was 11, the median for those receiving the DM dosage was 9 and the median for those receiving no dosage was 7.

2.31

Using MINITAB, the relative frequency histograms of the years in practice for the two groups of doctors are: Histogram of YRSPRAC 0.0 NO

25

7.5

15.0

22.5

30.0

37.5

YES

Percent

20 15

10 5 0 0.0

7.5

15.0

22.5

30.0

37.5 YRSPRAC

Panel variable: FUTUREUSE

The researchers hypothesized that older, more experienced physicians will be less likely to use ethics consultation in the future. From the histograms, approximately 38% of the doctors that said “no” have more than 20 years of experience. Only about 19% of the doctors that said “yes” had more than 20 years of experience. This supports the researchers’ assertion.

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data a.

Using MINITAB, the stem-and-leaf display is as follows, where the stems are the units place and the leaves are the decimal places:

Stem-and-Leaf Display: Time Stem-and-leaf of Time Leaf Unit = 0.10

(26) 23 15 9 4 2 2 1 1 1

1 2 3 4 5 6 7 8 9 10

N

= 49

00001122222344444445555679 11446799 002899 11125 24 8

1

b.

A little more than half (26/49 = .53) of all companies spent less than 2 months in bankruptcy. Only two of the 49 companies spent more than 6 months in bankruptcy. It appears that, in general, the length of time in bankruptcy for firms using "prepacks" is less than that of firms not using prepacks."

c.

A dot diagram will be used to compare the time in bankruptcy for the three types of "prepack" firms: Dotplot of Time vs Votes

Votes

2.32

33

Joint None Prepack 1.2

2.4

3.6

4.8

6.0

7.2

8.4

9.6

Time

d.

The highlighted times in part a correspond to companies that were reorganized through a leverage buyout. There does not appear to be any pattern to these points. They appear to be scattered about evenly throughout the distribution of all times.

Copyright © 2014 Pearson Education, Inc.

34 2.33

Chapter 2 Using MINITAB, the histogram of the data is: Histogram of INTTIME 60

50

Frequency

40

30

20

10

0 0

75

150

225 300 INTTIME

375

450

525

This histogram looks very similar to the one shown in the problem. Thus, there appears that there was minimal or no collaboration or collusion from within the company. We could conclude that the phishing attack against the organization was not an inside job. 2.34

Using MINITAB, the stem-and-leaf display for the data is:

Stem-and-Leaf Display: Time Stem-and-leaf of Time Leaf Unit = 1.0 3 7 (7) 11 6 4 2 1

3 4 5 6 7 8 9 10

N

= 25

239 3499 0011469 34458 13 26 5 2

The numbers in bold represent delivery times associated with customers who subsequently did not place additional orders with the firm. Since there were only 2 customers with delivery times of 68 days or longer that placed additional orders, I would say the maximum tolerable delivery time is about 65 to 67 days. Everyone with delivery times less than 67 days placed additional orders. 2.35

 x  3.2  2.5  2.1  3.7  2.8  2.0  16.3  2.717

Assume the data are a sample. The sample mean is:

x

n

6

6

The median is the average of the middle two numbers when the data are arranged in order (since n = 6 is even). The data arranged in order are: 2.0, 2.1, 2.5, 2.8, 3.2, 3.7. The middle two numbers are 2.5 and 2.8. The median is:

2.5  2.8 5.3   2.65 2 2

Copyright © 2014 Pearson Education, Inc.

2.36

Methods for Describing Sets of Data

 x  85  8.5

a.

x

b.

x

400  25 16

c.

x

35  .778 45

d.

x

242  13.44 18

n

35

10

2.37

The mean and median of a symmetric data set are equal to each other. The mean is larger than the median when the data set is skewed to the right. The mean is less than the median when the data set is skewed to the left. Thus, by comparing the mean and median, one can determine whether the data set is symmetric, skewed right, or skewed left.

2.38

The median is the middle number once the data have been arranged in order. If n is even, there is not a single middle number. Thus, to compute the median, we take the average of the middle two numbers. If n is odd, there is a single middle number. The median is this middle number. A data set with five measurements arranged in order is 1, 3, 5, 6, 8. The median is the middle number, which is 5. A data set with six measurements arranged in order is 1, 3, 5, 5, 6, 8. The median is the average of the 5  5 10 middle two numbers which is   5. 2 2

2.39

Assume the data are a sample. The mode is the observation that occurs most frequently. For this sample, the mode is 15, which occurs three times.

 x  18  10  15  13  17  15  12  15  18  16  11  160  14.545

The sample mean is:

x

11

n

11

The median is the middle number when the data are arranged in order. The data arranged in order are: 10, 11, 12, 13, 15, 15, 15, 16, 17, 18, 18. The middle number is the 6th number, which is 15. 2.40

a.

b.

x

 x  7    4  15  2.5

x

 x  2    4  40  3.08

6 6 33 Median =  3 (mean of 3rd and 4th numbers, after ordering) 2 Mode = 3 n

13 13 n Median = 3 (7th number, after ordering) Mode = 3

Copyright © 2014 Pearson Education, Inc.

36

Chapter 2

c.

2.41

2.42

x

 x  51    37  496  49.6

10 10 48  50 Median =  49 (mean of 5th and 6th numbers, after ordering) 2 Mode = 50 n

a.

For a distribution that is skewed to the left, the mean is less than the median.

b.

For a distribution that is skewed to the right, the mean is greater than the median.

c.

For a symmetric distribution, the mean and median are equal.

a.

b.

The mean is  x  9  (.1)  (1.6)  14.6  16.0  7.7  19.9  9.8  3.2  24.8  17.6  10.7  9.1  140.7  10.82 x n 13 13 The average annualized percentage return on investment for 13 randomly selected stock screeners is 10.82. Since the number of observations is odd, the median is the middle number once the data have been arranged in order. The data arranged in order are: -1.6 -.1 3.2 7.7 9.0 9.1 9.8 10.7 14.6 16.0 17.6 19.9 24.8 The middle number is 9.8 which is the median. Half of the annualized percentage returns on investment are below 9.8 and half are above 9.8.

2.43

2.44

a.

The mean amount exported on the printout is 653. This means that the average amount of money per market from exporting sparkling wine was $653,000.

b.

The median amount exported on the printout is 231. Since the median is the middle value, this means that half of the 30 sparkling wine export values were above $231,000 and half of the sparkling wine export values were below $231,000.

c.

The mean 3-year percentage change on the printout is 481. This means that in the last three years, the average change is 481%, which indicates a large increase.

d.

The median 3-year percentage change on the printout is 156. Since the median is the middle value, this means that half, or 15 of the 30 countries’ 3-year percentage change values were above 156% and half, or 15 of the 30 countries’ 3-year percentage change values were below 156%.

a.

The sample mean is:

x n

x

i 1

n

i



1.72  2.50  2.16    1.95 37.62   1.881 20 20

The sample average surface roughness of the 20 observations is 1.881.

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data

b.

37

The median is found as the average of the 10th and 11th observations, once the data have been ordered. The ordered data are: 1.06 1.09 1.19 1.26 1.27 1.40 1.51 1.72 1.95 2.03 2.05 2.13 2.13 2.16 2.24 2.31 2.41 2.50 2.57 2.64

The 10th and 11th observations are 2.03 and 2.05. The median is: 2.03  2.05 4.08   2.04 2 2 The middle surface roughness measurement is 2.04. Half of the sample measurements were less than 2.04 and half were greater than 2.04.

2.45

c.

The data are somewhat skewed to the left. Thus, the median might be a better measure of central tendency than the mean. The few small values in the data tend to make the mean smaller than the median.

a.

The mean is x 

b.

 x  1,680,927  885,182  881, 777    563,967  15,192, 021  759, 601.05 .

n 20 20 The average research expenditures for the top 20 ranked universities is 759,601.05 thousand dollars.

Since the number of observations is even, the median is the average of the middle 2 numbers once the data have been arranged in order. Since the data are already arranged in order, the median is 702,592  688, 225  695, 408.5 . 2 Half of the institutions have a research expenditure less than 695,408.5 thousand dollars and half have research expenditures greater than 695,408.5 thousand dollars.

2.46

2.47

c.

No, the mean from part a would not be a good measure for the center of the distribution for all American universities. The data in part a come from only the top 20 universities. These universities would not be representative of all American universities.

a.

The mean is 67.755. The statement is accurate.

b.

The median is 68.000. The statement is accurate.

c.

The mode is 64. The statement is not accurate. A better statement would be: “The most common reported level of support for corporate sustainability for the 992 senior managers was 64.

d.

Since the mean and median are almost the same, the distribution of the 992 support levels should be fairly symmetric. The histogram in Exercise 2.23 is almost symmetric.

a.

The median is the middle number (18th) once the data have been arranged in order because n = 35 is odd. The honey dosage data arranged in order are: 4,5,6,8,8,8,8,9,9,9,9,10,10,10,10,10,10,11,11,11,11,12,12,12,12,12,12,13,13,14,15,15,15,15,16 The 18th number is the median = 11.

Copyright © 2014 Pearson Education, Inc.

38

Chapter 2

b.

The median is the middle number (17th) once the data have been arranged in order because n = 33 is odd. The DM dosage data arranged in order are: 3,4,4,4,4,4,4,6,6,6,7,7,7,7,7,8,9,9,9,9,9,10,10,10,11,12,12,12,12,12,13,13,15 The 17th number is the median = 9.

c.

The median is the middle number (19th) once the data have been arranged in order because n = 37 is odd. The No dosage data arranged in order are: 0,1,1,1,3,3,4,4,5,5,5,6,6,6,6,7,7,7,7,7,7,7,7,8,8,8,8,8,8,9,9,9,9,10,11,12,12 The 19th number is the median = 7.

2.48

d.

Since the median for the Honey dosage is larger than the other two, it appears that the honey dosage leads to more improvement than the other two treatments.

a.

The mean dioxide level is x 

3.3  0.5  1.3    4.0 29   1.81 . The average dioxide amount is 16 16

1.81. b.

Since the number of observations is even, the median is the average of the middle 2 numbers once the data are arranged in order. The data arranged in order are: 0.1 0.2 0.2 0.3 0.4 0.5 0.5 1.3 1.4 2.4 2.4 3.3 4.0 4.0 4.0 4.0 The median is

1.3  1.4 2.7   1.35 . Half of the dioxide levels are below 1.35 and half are above 2 2

1.35. c.

The mode is the number that occurs the most. For this data set the mode is 4.0. The most frequent level of dioxide is 4.0.

d.

Since the number of observations is even, the median is the average of the middle 2 numbers once the data are arranged in order. The data arranged in order are: 0.1 0.3 1.4 2.4 2.4 3.3 4.0 4.0 4.0 4.0 The median is

e.

2.4  3.3 5.7   2.85 . 2 2

Since the number of observations is even, the median is the average of the middle 2 numbers once the data are arranged in order. The data arranged in order are: 0.2 0.2 0.4 0.5 0.5 1.3 The median is

f.

0.4  0.5 0.9   0.45 . 2 2

The median level of dioxide when crude oil is present is 0.45. The median level of dioxide when crude oil is not present is 2.85. It is apparent that the level of dioxide is much higher when crude oil is not present.

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data

2.49

2.50

39

a.

Skewed to the right. There will be a few people with very high salaries such as the president and football coach.

b.

Skewed to the left. On an easy test, most students will have high scores with only a few low scores.

c.

Skewed to the right. On a difficult test, most students will have low scores with only a few high scores.

d.

Skewed to the right. Most students will have a moderate amount of time studying while a few students might study a long time.

e.

Skewed to the left. Most cars will be relatively new with a few much older.

f.

Skewed to the left. Most students will take the entire time to take the exam while a few might leave early.

a.

The sample means is:

x

 x  3.58  3.48  3.27    1.17  77.07  1.927 40

n

40

The median is found as the 20th and 21st observations, once the data have been ordered. The 20th and 21st observations are 1.75 and 1.76. The median is: 1.75  1.76 3.51   1.755 2 2 The mode is the number that occurs the most and is 1.4, which occurs 3 times. b.

The sample average driving performance index is 1.927. The median driving performance index is 1.755. Half of all driving performance indexes are less than 1.755 and half are higher. The most common driving performance index value is 1.4.

c.

Since the mean is larger than the median, the data are skewed to the right. Using MINITAB, a histogram of the driving performance index values is: Histogram of INDEX 10

Frequency

8

6

4

2

0 1.5

2.51

2.0

2.5 INDEX

3.0

3.5

The mean is 141.31 hours. This means that the average number of semester hours per candidate for the CPA exam is 141.31 hours. The median is 140 hours. This means that 50% of the candidates had more than 140 semester hours of credit and 50% had less than 140 semester hours of credit. Since the mean and median are so close in value, the data are probably not skewed, but close to symmetric. Copyright © 2014 Pearson Education, Inc.

40

2.52

Chapter 2

a.

Using MINITAB, the output is: Descriptive Statistics: YRSPRAC Variable YRSPRAC

N 112

N* 6

Mean 14.598

Minimum 1.000

Median 14.000

Maximum 40.000

Mode 14, 20, 25

N for Mode 9

The mean is 14.598. The average length of time in practice for this sample is 14.598 years. The median is 14. Half of the physicians have been in practice less than 14 years and half have been in practice longer than 14 years. There are 3 modes: 14, 20, and 25. The most frequent years in practice are 14, 20, and 25 years. b.

Using MINITAB, the results are: Descriptive Statistics: YRSPRAC Variable YRSPRAC

FUTUREUSE NO YES

N 21 91

N* 2 4

Mean 16.43 14.176

Minimum 1.00 1.000

Median 18.00 14.000

Maximum 35.00 40.000

Mode 25 14, 20

N for Mode 5 8

The mean for the physicians who would refuse to use ethics consultation in the future is 16.43. The average time in practice for these physicians is 16.43 years. The median is 18. Half of the physicians who would refuse ethics consultation in the future have been in practice less than 18 years and half have been in practice more than 18 years. The mode is 25. The most frequent years in practice for these physicians is 25 years.

2.53

c.

From the results in part b, the mean for the physicians who would use ethics consultation in the future is 14.176. The average time in practice for these physicians is 14.176 years. The median is 14. Half of the physicians who would use ethics consultation in the future have been in practice less than 14 years and half have been in practice more than 14 years. There are 2 modes: 14 and 20. The most frequent years in practice for these physicians are 14 and 20 years.

d.

The results in parts b and c confirm the researchers’ theory. The mean, median and mode of years in practice are larger for the physicians who would refuse to use ethics consultation in the future than those who would use ethics consultation in the future.

For the "Joint exchange offer with prepack" firms, the mean time is 2.6545 months, and the median is 1.5 months. Thus, the average time spent in bankruptcy for "Joint" firms is 2.6545 months, while half of the firms spend 1.5 months or less in bankruptcy. For the "No prefiling vote held" firms, the mean time is 4.2364 months, and the median is 3.2 months. Thus, the average time spent in bankruptcy for "No prefiling vote held" firms is 4.2364 months, while half of the firms spend 3.2 months or less in bankruptcy. For the "Prepack solicitation only" firms, the mean time is 1.8185 months, and the median is 1.4 months. Thus, the average time spent in bankruptcy for "Prepack solicitation only" firms is 1.8185 months, while half of the firms spend 1.4 months or less in bankruptcy. Since the means and medians for the three groups of firms differ quite a bit, it would be unreasonable to use a single number to locate the center of the time in bankruptcy. Three different "centers" should be used.

2.54

a.

The sample mean is:

x n

x

i 1

n

i



5  2  4  ...  3 78   3.90 20 20

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data

41

The sample median is found by finding the average of the 10th and 11th observations once the data are arranged in order. The data arranged in order are: 1 1 1 1 1 2 2 3 3 3 4 4 4 5 5 5 6 7 9 11 The 10th and 11th observations are 3 and 4. The average of these two numbers (median) is: median 

3 4 7   3.5 2 2

The mode is the observation appearing the most. For this data set, the mode is 1, which appears 5 times. b.

Eliminating the largest number which is 11 results in the following: The sample mean is:

x n

x

i

i 1

n



5  2  4  ...  3 67   3.53 19 19

The sample median is found by finding the middle observation once the data are arranged in order. The data arranged in order are: 1 1 1 1 1 2 2 3 3 3 4 4 4 5 5 5 6 7 9 The 10th observation is 3. The median is 3 The mode is the observations appearing the most. For this data set, the mode is 1, which appears 5 times. By dropping the largest number, the mean is reduced from 4.05 to 3.68. The median is reduced from 3.5 to 3. There is no effect on the mode. c.

The data arranged in order are: 1 1 1 1 1 2 2 3 3 3 4 4 4 5 5 5 6 7 9 11 If we drop the lowest 2 and largest 2 observations we are left with: 1 1 1 2 2 3 3 3 4 4 4 5 5 5 6 7 The sample 10% trimmed mean is:

x n

x

i 1

n

i



1  1  2  ...  7 56   3.5 16 16

The advantage of the trimmed mean over the regular mean is that very large and very small numbers that could greatly affect the mean have been eliminated. 2.55

2.56

a.

Due to the "elite" superstars, the salary distribution is skewed to the right. Since this implies that the median is less than the mean, the players' association would want to use the median.

b.

The owners, by the logic of part a, would want to use the mean.

a.

The primary disadvantage of using the range to compare variability of data sets is that the two data sets can have the same range and be vastly different with respect to data variation. Also, the range is greatly affected by extreme measures. Copyright © 2014 Pearson Education, Inc.

42

Chapter 2

b.

The sample variance is the sum of the squared deviations of the observations from the sample mean divided by the sample size minus 1. The population variance is the sum of the squared deviations of the values from the population mean divided by the population size.

c.

The variance of a data set can never be negative. The variance of a sample is the sum of the squared deviations from the mean divided by n  1. The square of any number, positive or negative, is always positive. Thus, the variance will be positive. The variance is usually greater than the standard deviation. However, it is possible for the variance to be smaller than the standard deviation. If the data are between 0 and 1, the variance will be smaller than the standard deviation. For example, suppose the data set is .8, .7, .9, .5, and .3. The sample mean is:

x

 x  .8  .7  .9  .5  .3  3.2  .64 .5

n

5

The sample variance is: s 2 

x

2

 x  n 1

2

n



3.22 13  .232  .058 5 1 4

2.28 

The standard deviation is s  .058  .241 2.57

a.

s2 

b.

a.

b.

n

 x

82 5  2.3  5 1 22 

s  2.3  1.52

 x2 

n 1

2

n



 x 

17 2 7  3.619 7 1

s  3.619  1.9

302 10  7.111 10  1

s  7.111  2.67

63 

x

2

n 1

2

n



 x

154 

Range = 1  (3) = 4 s2 

2.58

n 1

2

Range = 8  (2) = 10 s2 

d.

x

2

Range = 6  0 = 6 s2 

c.

 x 

Range = 4  0 = 4

s2 

s2 

 x2 

n 1

x

2

x

2

n



 x 

n 1

2

n

 x 

n 1

2

n

(6.8) 2 17  1.395 17  1

25.04 

202 10  4.8889  10  1 84 

2



1002 40  3.3333 40  1

380 

s  1.395  1.18

s  4.8889  2.211

s  3.3333  1.826

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data

c.

2.59

a.

s2 

 x  n 1

2

17 2 20  .1868  20  1 18 

n

 x  3  1  10  10  4  28 x

s2 

b.

x

2

 x  28  5.6

s2 

s  .1868  .432 2

 32  12  102  102  42  226

5

n

x

2

 x  n 1

2

n



282 5  69.2  17.3 5 1 4

226 

 x  8 10  32  5  55 x

x

43

x

 x  55  13.75 feet

2

s  17.3  4.1593

 82  102  322  52  1213

4

n

 x2 

 x

n 1

2

n



552 4  456.75  152.25 square feet 4 1 3

1213 

s  152.25  12.339 feet

c.

 x  1  (4)  (3)  1  (4)  (4)  15  x x

s2 

d.

 x  15  2.5

x

s2 

 (1)2  (4)2  (3)2  12  (4)2  (4)2  59

6

n

x

2

 x  n 1

x 

2

2

n



(15) 2 6  21.5  4.3 6 1 5

59 

1 1 1 2 1 4 10       2 5 5 5 5 5 5 5

 x  2  1  .33 ounce 6

n

x

2

24 1 1 1 2 1 4  x 2   5    5    5    5    5    5   25  .96 2

2

2

2

2

2

3

 x  n 1

s  4.3  2.0736

n

2

24 22  .2933  25 6   .0587 square ounce 6 1 5

s  .0587  .2422 ounce

Copyright © 2014 Pearson Education, Inc.

44

2.60

Chapter 2

a.

s2 

b.

n 1

2

n

 x



1992 5  3.7 5 1

7935 

s  3.7  1.92

 x2 

n 1

2

n

 x 



3032 9  1,949.25 9 1

s  1,949.25  44.15

2952 8  1,307.84 8 1

s  1,307.84  36.16

25, 795 

Range = 100  2 = 98 s2 

2.61

x

2

Range = 100  1 = 99 s2 

c.

 x 

Range = 42  37 = 5

x

2

n 1

n

2



20, 033 

This is one possibility for the two data sets. Data Set 1: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 Data Set 2: 0, 0, 1, 1, 2, 2, 3, 3, 9, 9 The two sets of data above have the same range = largest measurement  smallest measurement = 9  0 = 9. The means for the two data sets are:

x1 

x2 

 x  0  1  2  3  4  5  6  7  8  9  45  4.5 n

10

10

n

10

10

 x  0  0  1  1  2  2  3  3  9  9  30  3

The dot diagrams for the two data sets are shown below. Dotplot of x1, x2

x1 0

2

4

x

6

8

6

8

x2 0

2

x

4

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data

2.62

This is one possibility for the two data sets. Data Set 1: 1, 1, 2, 2, 3, 3, 4, 4, 5, 5 Data Set 2: 1, 1, 1, 1, 1, 5, 5, 5, 5, 5

x1 

 x  1  1  2  2  3  3  4  4  5  5  30  3

x2 

 x  1  1  1  1  1  5  5  5  5  5  30  3

n

10

n

10

10

10

Therefore, the two data sets have the same mean. The variances for the two data sets are:

s12 

s22 

x

2

 x 

n 1

x

2

n

 x 

n 1

2



302 10  20  2.2222 9 9

110 

2

n



302 10  40  4.4444 9 9

130 

The dot diagrams for the two data sets are shown below. Dotplot of x1, x2

x1 x 1

2

3 x2

1

2

3

4

5

4

5

x

2.63

a.

s2 

b.

 x 

Range = 3  0 = 3

x

2

n 1

n

2

72 5  1.3  5 1 15 

s  1.3  1.14

After adding 3 to each of the data points, Range = 6  3 = 3 Copyright © 2014 Pearson Education, Inc.

45

46

Chapter 2

s2 

c.

x

2

 x 

2

n

n 1



222 5  1.3 5 1

102 

s  1.3  1.14

After subtracting 4 from each of the data points,

 x 

Range = 1  (4) = 3 s2 

2.64

x

2

2

n

n 1



(13) 2 5  1.3 5 1

39 

s  1.3  1.14

d.

The range, variance, and standard deviation remain the same when any number is added to or subtracted from each measurement in the data set.

a.

The range is the difference between the maximum and minimum values. The range  24.8 –  1.6   26.4 . The units of measurement are percents.

b.

The variance is s2 

 x2 

 x

n 1

n

2



140.7 2 13  2236.41  1522.8069  713.6031  59.4669 13  1 12 12

2236.41 

The units are square percents. c. 2.65

a.

The standard deviation is s  59.4669  7.7115 . The units are percents. The range is the difference between the largest observation and the smallest observation. From the printout, the largest observation is $4,852 thousand and the smallest observation is $70 thousand. The range is:

R  $4,852  $70  $4,882 thousand b.

From the printout, the standard deviation is s = $1,113 thousand.

c.

The variance is the standard deviation squared. The variance is: s 2  1,1132  1, 238, 769 million dollars squared

2.66

a.

The sample variance of the honey dosage group is:

s2 

x

2

 x 

n 1

n

2



3752 35  277.142857  8.1512605 35-1 34

4295-

The standard deviation is: s  8.1512605  2.855

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data

b.

47

The sample variance of the DM dosage group is:

s2 

x

2

 x 

n 1

2

n



2752 33  339.33333  10.604167 33-1 32

2631-

The standard deviation is: s  10.604167  3.256 c.

The sample variance of the control group is:

s2 

x

2

 x 

n 1

n

2



2412 37  311.243243  8.6456456 37-1 36

1881-

The standard deviation is: s  8.6456456  2.940 d.

2.67

2.68

The group with the most variability is the group with the largest standard deviation, which is the DM group. The group with the least variability is the group with the smallest standard deviation, which is the honey group.

a.

The range is 155. The statement is accurate.

b.

The variance is 722.036. The statement is not accurate. A more accurate statement would be: “The variance of the levels of supports for corporate sustainability for the 992 senior managers is 722.036.”

c.

The standard deviation is 26.871. If the units of measure for the two distributions are the same, then the distribution of support levels for the 992 senior managers has less variation than a distribution with a standard deviation of 50. If the units of measure for the second distribution is not known, then we cannot compare the variation in the two distributions by looking at the standard deviations alone.

d.

The standard deviation best describes the variation in the distribution. The range can be greatly affected by extreme measures. The variance is measured in square units, which is hard to interpret. Thus, the standard deviation is the best measure to describe the variation.

a.

Using MINITAB, the results are: Descriptive Statistics: YRSPRAC Variable YRSPRAC

N 112

N* 6

Mean 14.598

StDev 9.161

Variance 83.918

Range 39.000

The range is 39. The difference between the largest years in practice and the smallest years in practice is 39 years. The variance is 83.918 square years. The standard deviation is 9.161 years. b.

Using MINITAB, the results are: Descriptive Statistics: YRSPRAC Variable YRSPRAC

FUTUREUSE NO YES

N 21 91

N* 2 4

Mean 16.43 14.176

StDev 10.05 8.950

Variance 100.96 80.102

Range 34.00 39.000

For the physicians who would refuse to use ethics consultation in the future, the standard deviation is 10.05 years. Copyright © 2014 Pearson Education, Inc.

48

2.69

Chapter 2

c.

For the physicians who would use ethics consultation in the future, the standard deviation is 8.95 years.

d.

The variation in the length of time in practice for the physicians who would refuse to use ethics consultation in the future is greater than that for the physicians who would use ethics consultation in the future.

a.

The range is the largest observation minus the smallest observation or 11 – 1 = 10.

    xi  782 xi2   i  450   n 20  7.6737  The variance is: s 2  i n 1 20  1 2

The standard deviation is: s  s 2  7.6737  2.77 b.

The largest observation is 11. It is deleted from the data set. The new range is: 9 – 1 = 8.     xi  67 2 xi2   i  329   n 19  5.1520  The variance is: s 2  i n 1 19  1 2

The standard deviation is: s  s 2  5.1520  2.27 When the largest observation is deleted, the range, variance and standard deviation decrease. c.

The largest observation is 11 and the smallest is 1. When these two observations are deleted from the data set, the new range is: 9 – 1 = 8.     xi  i   662 xi2  328   n 18  5.0588  The variance is: s 2  i n 1 18  1 2

The standard deviation is: s  s 2  5.0588  2.25 When the largest and smallest observations are deleted, the range, variance and standard deviation decrease. 2.70

a.

A worker's overall time to complete the operation under study is determined by adding the subtasktime averages. Worker A

The average for subtask 1 is: x  The average for subtask 2 is: x 

 x  211  30.14  x  21  3 n

7

n 7 Worker A's overall time is 30.14 + 3 = 33.14.

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data Worker B

The average for subtask 1 is: x  The average for subtask 2 is: x 

49

 x  213  30.43  x  29  4.14 n

7

n 7 Worker B's overall time is 30.43 + 4.14 = 34.57. b.

Worker A

s

x

2

n 1

Worker B

s

 x 

 x2 

n

 x

n 1

2



2

n



2112 7  15.8095  3.98 7 1

6455 

2132 7  .9524  .98 7 1

6487 

c.

The standard deviations represent the amount of variability in the time it takes the worker to complete subtask 1.

d.

Worker A

s

x

2

n 1

Worker B

s

e.

 x 

x

2

n

 x  n 1

2

n



2



212 7  .6667  .82 7 1

67 

292 7  4.4762  2.12 7 1

147 

I would choose workers similar to worker B to perform subtask 1. Worker B has a slightly higher average time on subtask 1 (A: x  30.14 , B: x  30.43 ). However, Worker B has a smaller variability in the time it takes to complete subtask 1 (part b). He or she is more consistent in the time needed to complete the task. I would choose workers similar to Worker A to perform subtask 2. Worker A has a smaller average time on subtask 2 (A: x  3 , B: x  4.14 ). Worker A also has a smaller variability in the time needed to complete subtask 2 (part d).

2.71

a.

The unit of measurement of the variable of interest is dollars (the same as the mean and standard deviation). Based on this, the data are quantitative.

b.

Since no information is given about the shape of the data set, we can only use Chebyshev's Rule. $900 is 2 standard deviations below the mean, and $2100 is 2 standard deviations above the mean. Using Chebyshev's Rule, at least 3/4 of the measurements (or 3/4  200 = 150 measurements) will fall between $900 and $2100.

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50

Chapter 2

$600 is 3 standard deviations below the mean and $2400 is 3 standard deviations above the mean. Using Chebyshev's Rule, at least 8/9 of the measurements (or 8/9  200  178 measurements) will fall between $600 and $2400. $1200 is 1 standard deviation below the mean and $1800 is 1 standard deviation above the mean. Using Chebyshev's Rule, nothing can be said about the number of measurements that will fall between $1200 and $1800. $1500 is equal to the mean and $2100 is 2 standard deviations above the mean. Using Chebyshev's Rule, at least 3/4 of the measurements (or 3/4  200 = 150 measurements) will fall between $900 and $2100. It is possible that all of the 150 measurements will be between $900 and $1500. Thus, nothing can be said about the number of measurements between $1500 and $2100. 2.72

2.73

2.74

Since no information is given about the data set, we can only use Chebyshev's Rule. a.

Nothing can be said about the percentage of measurements which will fall between x  s and x  s .

b.

At least 3/4 or 75% of the measurements will fall between x  2 s and x  2 s .

c.

At least 8/9 or 89% of the measurements will fall between x  3s and x  3s .

According to the Empirical Rule: a.

Approximately 68% of the measurements will be contained in the interval x  s to x  s .

b.

Approximately 95% of the measurements will be contained in the interval x  2 s to x  2 s .

c.

Essentially all the measurements will be contained in the interval x  3s to x  3s .

a.

x

s2 

 x  206  8.24 25

n

x

2

 x  n 1

n

2



2062 25  3.357 25  1

1778 

s  3.357  1.83

b. Number of Measurements in Interval

Interval

c.

Percentage

x  s , or (6.41, 10.07)

18

18 / 25  .72 or 72%

x  2 s , or (4.58, 11.90)

24

24 / 25  .96 or 96%

x  3s , or (2.75, 13.73)

25

25 / 25  1.00 or 100%

The percentages in part b are in agreement with Chebyshev's Rule and agree fairly well with the percentages given by the Empirical Rule.

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Methods for Describing Sets of Data

d.

Range  12  5  7 and s 

51

Range 7   1.75 4 4

The range approximation provides a satisfactory estimate of s  1.83 from part a. 2.75

Using Chebyshev's Rule, at least 8/9 of the measurements will fall within 3 standard deviations of the mean. Thus, the range of the data would be around 6 standard deviations. Using the Empirical Rule, approximately 95% of the observations are within 2 standard deviations of the mean. Thus, the range of the data would be around 4 standard deviations. We would expect the standard deviation to be somewhere between Range/6 and Range/4. For our data, the range  760  135  625 . The

Range 625 Range 625   156.25 .   104.17 and 6 6 4 4

Therefore, I would estimate that the standard deviation of the data set is between 104.17 and 156.25. It would not be feasible to have a standard deviation of 25. If the standard deviation were 25, the data would span 625/25 = 25 standard deviations. This would be extremely unlikely. a.

Using MINITAB, the histogram of the data is: Histogram of Wheels 12

10

8 Frequency

2.76

6

4

2

0 1

2

3

4

5

6

7

8

Wheels

Since the distribution is skewed to the right, it is not mound-shaped and it is not symmetric. b.

Using MINITAB, the results are: Descriptive Statistics: Wheels Variable Wheels

N 28

Mean 3.214

StDev 1.371

Minimum 1.000

Q1 2.000

Median 3.000

Q3 4.000

Maximum 8.000

The mean is 3.214 and the standard deviation is 1.371. c. d.

The interval is: x  2 s  3.214  2(1.371)  3.214  2.742  (0.472, 5.956) . According to Chebyshev’s rule, at least 75% of the observations will fall within 2 standard deviations of the mean.

Copyright © 2014 Pearson Education, Inc.

52

2.77

2.78

Chapter 2

e.

According to the Empirical Rule, approximately 95% of the observations will fall within 2 standard deviations of the mean.

f.

Actually, 26 of the 28 or 26/28 = .929 of the observations fall within the interval. This value is close to the 95% that we would expect with the Empirical Rule.

a.

The interval x  2 s will contain at least 75% of the observations. This interval is x  2 s  3.11  2(.66)  3.11  1.32  (1.79, 4.43) .

b.

No. The value 1.25 does not fall in the interval x  2 s . We know that at least 75% of all observations will fall within 2 standard deviations of the mean. Since 1.25 falls more than 2 standard deviations from the mean, it would not be a likely value to observe.

a.

Using Chebyshev’s Rule, at least 75% of the observations will fall within 2 standard deviations of the mean. x  2 s  4.25  2(12.02)  4.25  24.04  ( 19.79, 28.29) or (0, 28.29) since we cannot have a negative number blogs.

b.

2.79

2.80

We would expect the distribution to be skewed to the right. We know that we cannot have a negative number of blogs/forums. Even 1 standard deviation below the mean is a negative number. We would assume that there are a few very large observations because the standard deviation is so big compared to the mean.

a.

The 2 standard deviation interval around the mean is:

b.

Using Chebyshev’s Theorem, at least ¾ of the observations will fall within 2 standard deviations of the mean. Thus, at least ¾ of first-time candidates for the CPA exam have total credit hours between 105.77 and 176.85.

c.

In order for the above statement to be true, nothing needs to be known about the shape of the distribution of total semester hours.

a.

Since the data are mound-shaped and symmetric, we know from the Empirical Rule that approximately 95% of the observations will fall within 2 standard deviations of the mean. This interval will be: x  2 s  39  2(6)  39  12  (27, 51) .

b.

We know that approximately .05 of the observations will fall outside the range 27 to 51. Since the distribution of scores is symmetric, we know that half of the .05 or .025 will fall above 51.

c.

We know from the Empirical Rule that approximately 99.7% (essentially all) of the observations will fall within 3 standard deviations of the mean. This interval is: x  3s  39  3(6)  39  18  (21, 57) .

x  2 s  141.31  2(17.77)  141.31  35.54  (105.77, 176.85)

x n

2.81

a.

The sample mean is: x 

i 1

n

i



17,800  95.699 186

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Methods for Describing Sets of Data

53

 n    xi  n 2 17,8002 x   i 1  1,707,998   n 186  24.6332  The sample variance is: s 2  i 1 n 1 186  1 2

The standard deviation is: s  s 2  24.6332  4.9632 x  s  95.699  4.963  (90.736, 100.662)

b.

x  2 s  95.699  2(4.963)  95.699  9.926  (85.773, 105.625)

x  3s  95.699  3(4.963)  95.699  14.889  (80.810, 110.558)

c.

There are 166 out of 186 observations in the first interval. This is (166 / 186)  100%  89.2% . There are 179 out of 186 observations in the second interval. This is (179 / 186)  100%  96.2% . There are 182 out of 186 observations in the second interval. This is (182 / 186)  100%  97.8% . The percentages for the first 2 intervals are much larger than we would expect using the Empirical Rule. The Empirical Rule indicates that approximately 68% of the observations will fall within 1 standard deviation of the mean. It also indicates that approximately 95% of the observations will fall within 2 standard deviations of the mean. Chebyshev’s Theorem says that at least ¾ or 75% of the observations will fall within 2 standard deviations of the mean and at least 8/9 or 88.9% of the observations will fall within 3 standard deviations of the mean. It appears that our observed percentages agree with Chebyshev’s Theorem better than the Empirical Rule.

2.82

2.83

a.

The interval is: x  2 s  13.2  2(19.5)  13.2  39  ( 25.8, 52.2) or (0, 52.2) since we cannot have negative number of minutes.

b.

Since this interval contains negative numbers, we know that the distribution cannot be symmetric. One cannot have negative values for time spent on a laptop computer.

c.

Since we know the data are not symmetric, we must use Chebyshev’s Rule. At least ¾ or 75% of the observations will fall between -25.8 and 52.2 or between 0 and 52.2 minutes.

x

The sample mean is: n

x

i 1

n

i



240.9  248.8  215.7    238.0 2347.4   234.74 10 10

The sample variance deviation is:

 n    xi  n 2 2347.42 xi   i 1  551,912.1   883.424 n 10    98.1582 s 2  i 1 9 9 n 1 2

The sample standard deviation is:

s2  98.1582  9.91

The data are fairly symmetric, so we can use the Empirical Rule. We know from the Empirical Rule that almost all of the observations will fall within 3 standard deviations of the mean. This interval would be: x  3s  234.74  3(9.91)  234.74  29.73  (205.01, 264.47)

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54

2.84

Chapter 2

a.

Using MINITAB, the frequency histogram for the time in bankruptcy is: Histogram of TIME 20

Frequency

15

10

5

0 2

4

6 Time in Bankruptcy

8

10

The Empirical Rule is not applicable because the data are not mound shaped. b.

Using MINITAB, the descriptive measures are:

Descriptive Statistics: TIME Variable TIME

N 49

Mean 2.549

StDev 1.828

Minimum 1.000

Q1 1.350

Median 1.700

Q3 3.500

Maximum 10.100

From Chebyshev’s Theorem, we know that at least 75% of the observations will fall within 2 standard deviations of the mean. This interval is: x  2 s  2.549  2(1.828)  2.549  3.656  ( 1.107, 6.205) or (0, 6.205) since we cannot have negative months.

2.85

c.

There are 47 of the 49 observations within this interval. The percentage would be (47 / 49)  100%  95.9% . This agrees with Chebyshev’s Theorem (at least 75%). It also agrees with the Empirical Rule (approximately 95%).

d.

From the above interval we know that about 95% of all firms filing for prepackaged bankruptcy will be in bankruptcy between 0 and 6.2 months. Thus, we would estimate that a firm considering filing for bankruptcy will be in bankruptcy up to 6.2 months.

a.

b.

The interval x  2 s for the flexed arm group is x  2 s  59  3(4)  59  12  (47, 71) . The interval for the extended are group is x  2 s  43  3(2)  43  6  (37, 49) . We know that at least 8/9 or 88.9% of the observations will fall within 3 standard deviations of the mean using Chebyshev’s Rule. Since these 2 intervals barely overlap, the information supports the researchers’ theory. The shoppers from the flexed arm group are more likely to select vice options than the extended arm group. The interval x  2 s for the flexed arm group is x  2 s  59  2(10)  59  20  (39, 79) . The interval for the extended are group is x  2 s  43  2(15)  43  30  (13, 73) . Since these two intervals overlap almost completely, the information does not support the researcher’s theory. There does not appear to be any difference between the two groups.

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Methods for Describing Sets of Data

2.86

55

a.

Yes. The distribution of the buy-side analysts is fairly flat and skewed to the right. The distribution of the sell-side analysts is more mound shaped and is not spread out as far as the buy-side distribution. Since the buy-side distribution is more spread out, the variance of the buy-side distribution will be larger than the variance of the sell-side distribution. Because the buy-side distribution is skewed to the right, the mean will be pulled to the right. Thus, the mean of the buyside distribution will be greater than the mean of the sell-side distribution.

b.

Since the sell-side distribution is fairly mound-shaped, we can use the Empirical Rule. The Empirical Rule says that approximately 95% of the observations will fall within 2 standard deviations of the mean. The interval for the sell-side distribution would be: x  2 s  .05  2(.85)  .05  1.7  ( 1.75, 1.65)

Since the buy-side distribution is skewed to the right, we cannot use the Empirical Rule. Thus, we will use Chebyshev’s Rule. We know that at least (1 – 1/k2) will fall within k standard deviations of the mean. If we choose k  4 , then (1  1/ 42 )  .9375 or 93.75%. This is very close to 95% requested in the problem. The interval for the buy-side distribution to contain at least 93.75% of the observations would be: x  4 s  .85  4(1.93)  .85  7.72  ( 6.87, 8.57) Note: This interval will contain at least 93.75% of the observations. It may contain more than 93.75% of the observations. 2.87

Since we do not know if the distribution of the heights of the trees is mound-shaped, we need to apply Chebyshev's Rule. We know   30 and   3 . Therefore,   3  30  3(3)  30  9  (21, 39) . According to Chebyshev's Rule, at least 8 / 9  .89 of the tree heights on this piece of land fall within this interval and at most 1 / 9  .11 of the tree heights will fall above the interval. However, the buyer will only 1000 purchase the land if at least  .20 of the tree heights are at least 40 feet tall. Therefore, the buyer 5000 should not buy the piece of land.

2.88

a.

Since we do not have any idea of the shape of the distribution of SAT-Math score changes, we must use Chebyshev’s Theorem. We know that at least 8/9 of the observations will fall within 3 standard deviations of the mean. This interval would be: x  3s  19  3(65)  19  195  ( 176, 214)

Thus, for a randomly selected student, we could be pretty sure that this student’s score would be anywhere from 176 points below his/her previous SAT-Math score to 214 points above his/her previous SAT-Math score. b.

Since we do not have any idea of the shape of the distribution of SAT-Verbal score changes, we must use Chebyshev’s Theorem. We know that at least 8/9 of the observations will fall within 3 standard deviations of the mean. This interval would be: x  3s  7  3(49)  7  147  ( 140, 154)

Thus, for a randomly selected student, we could be pretty sure that this student’s score would be anywhere from 140 points below his/her previous SAT-Verbal score to 154 points above his/her previous SAT-Verbal score.

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56

Chapter 2

c.

2.89

A change of 140 points on the SAT-Math would be a little less than 2 standard deviations from the mean. A change of 140 points on the SAT-Verbal would be a little less than 3 standard deviations from the mean. Since the 140 point change for the SAT-Math is not as big a change as the 140 point on the SAT-Verbal, it would be most likely that the score was a SAT-Math score.

We know   25 and   1 . Therefore,   2  25  2(.1)  25  .2  (24.8, 25.2) The machine is shut down for adjustment if the contents of two consecutive bags fall more than 2 standard deviations from the mean (i.e., outside the interval (24.8, 25.2)). Therefore, the machine was shut down yesterday at 11:30 (25.23 and 25.25 are outside the interval) and again at 4:00 (24.71 and 25.31 are outside the interval).

2.90

2.91

a.

z

b.

z

c.

z

d.

z

x  x 40  30   2 (sample) 5 s x

 x



2 standard deviations above the mean.



90  89  .5 (population) 2

.5 standard deviations above the mean.



50  50  0 (population) 5

0 standard deviations above the mean.

x  x 20  30   2.5 (sample) 4 s

2.5 standard deviations below the mean.

Using the definition of a percentile:

a.

Percentile 75th

Percentage Above 25%

Percentage Below 75%

b.

50th

50%

50%

c.

20th

80%

20%

d.

84th

16%

84%

2.92

QL corresponds to the 25th percentile. QM corresponds to the 50th percentile. QU corresponds to the 75th percentile.

2.93

We first compute z-scores for each x value. a.

z

b.

z

c.

d.

x

 x



100  50 2 25

1 4  3 1  x   0  200  2 z 100  z

x







10  5  1.67 3

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Methods for Describing Sets of Data

57

The above z-scores indicate that the x value in part a lies the greatest distance above the mean and the x value of part b lies the greatest distance below the mean. 2.94

Since the element 40 has a z-score of 2 and 90 has a z-score of 3, 2 

40  



and

 2  40      2  40    40  2

3

90  



 3  90      3  90

By substitution, 40  2  3  90  5  50    10 and   40  2(10)  60 . Therefore, the population mean is 60 and the standard deviation is 10. 2.95

The mean score of U.S. eighth-graders on a mathematics assessment test is 283. This is the average score. The 25th percentile is 259. This means that 25% of the U.S. eighth-graders score below 259 on the test and 75% score higher. The 75th percentile is 308. This means that 75% of the U.S. eighth-graders score below 308 on the test and 25% score higher. The 90th percentile is 329. This means that 90% of the U.S. eighthgraders score below 329 on the test and 10% score higher.

2.96

a.

The z-score is z 

b.

Since the data are mound-shaped and symmetric and 39 is the mean, .5 of the sampled drug dealers will have WR scores below 39.

c.

If 5% of the drug dealers have WR scores above 49, then 95% will have WR scores below 49. Thus, 49 will be the 95th percentile.

x  x 30  39   1.5 . A score of 30 is 1.5 standard deviations below the mean. 6 s

2.97

A median starting salary of $41,100 indicates that half of the University of South Florida graduates had starting salaries less than $41,100 and half had starting salaries greater than $41,100. At mid-career, half of the University of South Florida graduates had a salary less than $71,100 and half had salaries greater than $71,100. At mid-career, 90% of the University of South Florida graduates had salaries under $131,000 and 10% had salaries greater than $131,000.

2.98

a.

From Exercise 2.81, x  95.699 and s = 4.963. The z-score for an observation of 74 is: z

x  x 74  95.699   4.37 4.963 s

This z-score indicates that an observation of 74 is 4.37 standard deviations below the mean. Very few observations will be lower than this one. b.

The z-score for an observation of 98 is: z

x  x 92  95.699   0.75 4.963 s

This z-score indicates that an observation of 92 is .75 standard deviations below the mean. This score Copyright © 2014 Pearson Education, Inc.

58

Chapter 2

is not an unusual observation in the data set. 2.99

2.100

2.101

Since the 90th percentile of the study sample in the subdivision was .00372 mg/L, which is less than the USEPA level of .015 mg/L, the water customers in the subdivision are not at risk of drinking water with unhealthy lead levels. x  x 155  67.755   3.25 . This score would not be s 26.871 considered a typical level of support. It is 3.25 standard deviations above the mean. Very few observations would be above this value.

The z-score associated with a score of 155 is z 

a.

The 10th percentile is the score that has at least 10% of the observations less than it. If we arrange the data in order from the smallest to the largest, the 10th percentile score will be the .10(75) = 7.5 or 8th observation. When the data are arranged in order, the 8th observation is 0. Thus, the 10th percentile is 0.

b.

The 95th percentile is the score that has at least 95% of the observations less than it. If we arrange the data in order from the smallest to the largest, the 95th percentile score will be the .95(75) = 71.25 or 72nd observation. When the data are arranged in order, the 72nd observation is 21. Thus, the 95th percentile is 21.

x n

c.

The sample mean is: x 

i 1

n

i



393  5.24 75

    xi  3932 xi2   i  5943   n 75  52.482  The sample variance is: s 2  i 75  1 n 1 2

The standard deviation is: s  s 2  52.482  7.244 The z-score for a county with 48 Superfund sites is: z 

2.102

x  x 48  5.24   5.90 7.244 s

d.

Yes. A score of 48 is almost 6 standard deviations from the mean. We know that for any data set almost all (at least 8/9 using Chebyshev’s Theorem) of the observations are within 3 standard deviations of the mean. To be almost 6 standard deviations from the mean is very unusual.

a.

Since the data are approximately mound-shaped, we can use the Empirical Rule. On the blue exam, the mean is 53% and the standard deviation is 15%. We know that approximately 68% of all students will score within 1 standard deviation of the mean. This interval is: x  s  53  15  (38, 68)

About 95% of all students will score within 2 standard deviations of the mean. This interval is: x  2 s  53  2(15)  53  30  (23, 83) About 99.7% of all students will score within 3 standard deviations of the mean. This interval is: x  3s  53  3(15)  53  45  (8, 98)

b.

Since the data are approximately mound-shaped, we can use the Empirical Rule. Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data

59

On the red exam, the mean is 39% and the standard deviation is 12%. We know that approximately 68% of all students will score within 1 standard deviation of the mean. This interval is: x  s  39  12  (27, 51)

About 95% of all students will score within 2 standard deviations of the mean. This interval is: x  2 s  39  2(12)  39  24  (15, 63)

About 99.7% of all students will score within 3 standard deviations of the mean. This interval is: x  3s  39  3(12)  39  36  (3, 75)

The student would have been more likely to have taken the red exam. For the blue exam, we know that approximately 95% of all scores will be from 23% to 83%. The observed 20% score does not fall in this range. For the red exam, we know that approximately 95% of all scores will be from 15% to 63%. The observed 20% score does fall in this range. Thus, it is more likely that the student would have taken the red exam.

a.

The z-score for Harvard is z = 5.08. This means that Harvard’s productivity score was 5.08 standard deviations above the mean. This is extremely high and extremely unusual.

b.

The z-score for Howard University is z = .85. This means that Howard University’s productivity score was .85 standard deviations below the mean. This is not an unusual z-score.

c.

Yes. Other indicators that the distribution is skewed to the right are the values of the highest and lowest z-scores. The lowest z-score is less than 1 standard deviation below the mean while the highest z-score is 5.08 standard deviations above the mean. Using MINITAB, the histogram of the z-scores is: Histogram of Z-Score 70 60 50 Frequency

2.103

c.

40 30 20 10 0 -1

0

1

2 Z-Score

3

4

5

This histogram does imply that the data are skewed to the right.

Copyright © 2014 Pearson Education, Inc.

60

Chapter 2

2.104

a.

From the problem,   2.7 and   .5 z

x



 z  x    x    z

For z = 2.0, x  2.7  2.0(.5)  3.7 For z = 1.0, x  2.7  1.0(.5)  2.2 For z = .5, x  2.7  .5(.5)  2.95 For z = 2.5, x  2.7  2.5(.5)  1.45 b.

For z = 1.6, x  2.7  1.6(.5)  1.9

c.

If we assume the distribution of GPAs is approximately mound-shaped, we can use the Empirical Rule.

From the Empirical Rule, we know that .025 or 2.5% of the students will have GPAs above 3.7 (with z = 2). Thus, the GPA corresponding to summa cum laude (top 2.5%) will be greater than 3.7 (z > 2). We know that .16 or 16% of the students will have GPAs above 3.2 (z = 1). Thus, the limit on GPAs for cum laude (top 16%) will be greater than 3.2 (z > 1). We must assume the distribution is mound-shaped. 2.105

Not necessarily. Because the distribution is highly skewed to the right, the standard deviation is very large. Remember that the z-score represents the number of standard deviations a score is from the mean. If the standard deviation is very large, then the z-scores for observations somewhat near the mean will appear to be fairly small. If we deleted the schools with the very high productivity scores and recomputed the mean and standard deviation, the standard deviation would be much smaller. Thus, most of the z-scores would be larger because we would be dividing by a much smaller standard deviation. This would imply a bigger spread among the rest of the schools than the original distribution with the few outliers.

2.106

To determine if the measurements are outliers, compute the z-score. a.

b.

z

x  x 65  57   .727 11 s

Since the z-score is less than 3, this would not be an outlier.

x  x 21  57   3.273 Since the z-score is greater than 3 in absolute value, this would be an 11 s outlier. z

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Methods for Describing Sets of Data

c.

d.

2.107

z

61

x  x 72  57   1.364 Since the z-score is less than 3, this would not be an outlier. 11 s

x  x 98  57   3.727 Since the z-score is greater than 3 in absolute value, this would be an 11 s outlier. z

The interquartile range is IQR  QU  QL  85  60  25 . The lower inner fence = QL  1.5( IQR )  60  1.5(25)  22.5 . The upper inner fence = QU  1.5( IQR )  85  1.5(25)  122.5 . The lower outer fence = QL  3( IQR )  60  3(25)  15 . The upper outer fence = QU  3( IQR )  85  3(25)  160 . With only this information, the box plot would look something like the following:

*

──────────── ──────────────────│ + │────── ────────────

─┼────┼────┼────┼────┼────┼────┼────┼────┼────┼────┼─── 10 20 30 40 50 60 70 80 90 100 110

The whiskers extend to the inner fences unless no data points are that small or that large. The upper inner fence is 122.5. However, the largest data point is 100, so the whisker stops at 100. The lower inner fence is 22.5. The smallest data point is 18, so the whisker extends to 22.5. Since 18 is between the inner and outer fences, it is designated with a *. We do not know if there is any more than one data point below 22.5, so we cannot be sure that the box plot is entirely correct. 2.108

a.

Median is approximately 4.

b.

QL is approximately 3 (Lower Quartile) QU is approximately 6 (Upper Quartile)

c.

IQR  QU  QL  6  3  3

d.

The data set is skewed to the right since the right whisker is longer than the left, there is one outlier, and there are two potential outliers.

e.

50% of the measurements are to the right of the median and 75% are to the left of the upper quartile.

f.

The upper inner fence is QU  1.5( IQR )  6  1.5(3)  10.5 . The upper outer fence is QU  3( IQR )  6  3(3)  15 . Thus, there are two suspect outliers, 12 and 13. There is one highly suspect outlier, 16.

Copyright © 2014 Pearson Education, Inc.

a.

Using MINITAB, the box plot for sample A is given below. Boxplot of Sample A 200

175

Sample A

2.109

Chapter 2

150

125

100

Using MINITAB, the box plot for sample B is given below. Boxplot of Sample B 210 200 190

Sample B

62

180 170 160 150 140

b.

In sample A, the measurement 84 is an outlier. This measurement falls outside the lower outer fence. Lower outer fence = Lower hinge 3( IQR )  150  3(172  150)  150  3(22)  84 Lower inner fence = Lower hinge 1.5( IQR )  150  1.5(22)  117 Upper inner fence = Upper hinge 1.5( IQR )  172  1.5(22)  205 In addition, 100 may be an outlier. It lies outside the inner fence. In sample B, 140 and 206 may be outliers. The point 140 lies outside the inner fence while the point 206 lies right at the inner fence. Lower outer fence = Lower hinge 3( IQR )  168  3(184  169)  168  3(15)  123 Lower inner fence = Lower hinge 1.5( IQR )  168  1.5(15)  145.5 Upper inner fence = Upper hinge 1.5( IQR )  184  1.5(15)  206.5

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Methods for Describing Sets of Data

2.110

2.111

2.112

a.

The approximate 25th percentile PASI score before treatment is 10. The approximate median before treatment is 15. The approximate 75th percentile PASI score before treatment is 28.

b.

The approximate 25th percentile PASI score after treatment is 3. The approximate median after treatment is 5. The approximate 75th percentile PASI score after treatment is 7.5.

c.

Since the 75th percentile after treatment is lower than the 25th percentile before treatment, it appears that the ichthyotherapy is effective in treating psoriasis.

a.

The average expenditure per full-time employee is $6,563. The median expenditure per employee is $6,232. Half of all expenditures per employee were less than $6,232 and half were greater than $6,232. The lower quartile is $5,309. Twenty-five percent of all expenditures per employee were below $5,309. The upper quartile is $7,216. Seventy-five percent of all expenditures per employee were below $7,216.

b.

IQR  QU  QL  $7, 216  $5, 309  $1, 907 .

c.

The interquartile range goes from the 25th percentile to the 75th percentile. Thus, .5  .75  .25 of the 1,751 army hospitals have expenses between $5,309 and $7,216.

a.

From the printout, x  52.334 and s = 9.224. The highest salary is 75 (thousand). The z-score is z 

x  x 75  52.334   2.46 9.224 s

Therefore, the highest salary is 2.46 standard deviations above the mean. The lowest salary is 35.0 (thousand). The z-score is z 

x  x 35.0  52.334   1.88 9.224 s

Therefore, the lowest salary is 1.88 standard deviations below the mean. The mean salary offer is 52.33 (thousand). The z-score is z 

x  x 52.33  52.334  0 9.224 s

The z-score for the mean salary offer is 0 standard deviations from the mean. No, the highest salary offer is not unusually high. For any distribution, at least 8/9 of the salaries should have z-scores between 3 and 3. A z-score of 2.46 would not be that unusual.

2.113

63

b.

Since no salaries are outside the inner fences, none of them are suspect or highly suspect outliers.

a.

The z-score is: z 

x  x 160  141.31   1.05 17.77 s

Since the z-score is not large, it is not considered an outlier. Copyright © 2014 Pearson Education, Inc.

64

Chapter 2

b.

Z-scores with values greater than 3 in absolute value are considered outliers. An observation with a z-score of 3 would have the value: z

xx x  141.31 3  3(17.77)  x  141.31  53.31  x  141.31  x  194.62 17.77 s

An observation with a z-score of 3 would have the value: z

xx x  141.31  3   3(17.77)  x  141.31  53.31  x  141.31  x  88.00 17.77 s

Thus any observation of semester hours that is greater than or equal to 194.62 or less than or equal to 88 would be considered an outlier. 2.114

From Exercise 2.100, x  67.755 and s  26.87 . Using MINITAB, a boxplot of the data is: Boxplot of Support 160 140 120

Support

100 80 60 40 20 0

From the boxplot, the support level of 155 would be an outlier. From Exercise 2.100, we found the z-score x  x 155  67.755 associated with a score of 155 as z    3.25 . Since this z-score is greater than 3, the 26.871 s observation 155 is considered an outlier. a.

Using MINITAB, the boxplots for each type of firm are: Boxplot of TIME vs VOTES 10

8

TIME

2.115

6

4

2

0 Joint

None VOTES

Prepack

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Methods for Describing Sets of Data

2.116

65

b.

The median bankruptcy time for Joint firms is about 1.5. The median bankruptcy time for None firms is about 3.2. The median bankruptcy time for Prepack firms is about 1.4.

c.

The range of the "Prepack" firms is less than the other two, while the range of the "None" firms is the largest. The interquartile range of the "Prepack" firms is less than the other two, while the interquartile range of the "Joint" firms is larger than the other two.

d.

No. The interquartile range for the "Prepack" firms is the smallest which corresponds to the smallest standard deviation. However, the second smallest interquartile range corresponds to the "None" firms. The second smallest standard deviation corresponds to the "Joint" firms.

e.

Yes. There is evidence of two outliers in the "Prepack" firms. These are indicated by the two *'s. There is also evidence of two outliers in the "None" firms. These are indicated by the two *'s.

a.

From Exercise 2.101, x  5.24 , s 2  52.482 , and s  7.244 . We will use 3 standard deviations from the mean as the cutoff for outliers. Z-scores with values greater than 3 in absolute value are considered outliers. An observation with a z-score of 3 would have the value: z

xx x  5.24 3  3(7.244)  x  5.24  21.732  x  5.24  x  26.972 7.244 s

An observation with a z-score of -3 would have the value: xx x  5.24 z  3   3(7.244)  x  5.24  21.732  x  5.24  x  16.492 7.244 s Thus, any observation that is greater than 26.972 or less than -16.492 would be considered an outlier. In this data set there would be 1 outlier: 48.

x n

b.

Deleting the observation 48, the sample mean is: x 

i 1

n

i



345  4.66 74

    xi  3452 xi2   i  3639   n 74  27.8158  The sample variance is: s 2  i 74  1 n 1 2

The standard deviation is: s  s 2  27.8158  5.274 The mean has decreased from 5.24 to 4.66, while the standard deviation decreased from 7.244 to 5.274.

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2.117

Chapter 2

a.

Using MINITAB, the boxplot is: Boxplot of Score 100

95

90 Score

66

85

80

75

70

From the boxplot, there appears to be 10 outliers: 69, 73, 74, 78, 83, 84, 84, 86, 86, and 86.

b.

From Exercise 2.81, x  95.699 and s = 4.963. Since the data are skewed to the left, we will consider observations more than 2 standard deviations from the mean to be outliers. An observation with a z-score of 2 would have the value: z

xx x  95.699 2  2(4.963)  x  95.699  9.926  x  95.699  x  105.625 4.963 s

An observation with a z-score of -2 would have the value: z

xx x  95.699  2   2(4.963)  x  95.699  9.926  x  95.699  x  85.773 4.963 s

Observations greater than 105.625 or less than 85.773 would be considered outliers. Using this criterion, the following observations would be outliers: 69, 73, 74, 78, 83, 84, and 84. c.

No, these methods do not agree exactly. Using the boxplot, 10 observations were identified as outliers. Using the z-score method, only 7 observations were identified as outliers. However, the 3 additional points that were not identified as outliers using the z-score method were very close to the cutoff value.

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Methods for Describing Sets of Data

2.118

a.

67

Using MINITAB, the box plot is: Boxplot of TIME 70 60 50

TIME

40 30 20 10 0

The median is about 18. The data appear to be skewed to the right since there are 3 suspect outliers to the right and none to the left. The variability of the data is fairly small because the IQR is fairly small, approximately 26  10 = 16. b.

The customers associated with the suspected outliers are customers 268, 269, and 264.

c.

In order to find the z-scores, we must first find the mean and standard deviation.

x

 x  815  20.375 n

s2 

40

x

2

 x 

n 1

n

2

2

24129  815 40  192.90705  40  1

s  192.90705  13.89

The z-scores associated with the suspected outliers are: Customer 268 z 

49  20.375  2.06 13.89

Customer 269 z 

50  20.375  2.13 13.89

Customer 264 z 

64  20.375  3.14 13.89

All the z-scores are greater than 2. These are unusual values. 2.119

From the stem-and-leaf display in Exercise 2.34, the data are fairly mound-shaped, but skewed somewhat to the right. The sample mean is x 

 x  1493  59.72 . n

25

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68

Chapter 2

The sample variance is s 2 

x

2

 x  n 1

n

2



14932 25  321.7933 . 25  1

96,885 

The sample standard deviation is s  321.7933  17.9386 . The z-score associated with the largest value is z  Since

x  x 102  59.72   2.36 . 17.9386 s

the data are not extremely skewed to the right, this observation is probably not an outlier.

The observations associated with the one-time customers are 5 of the largest 7 observations. Thus, repeat customers tend to have shorter delivery times than one-time customers. 2.120

For Perturbed Intrinsics, but no Perturbed Projections:

x

2

n

x

i 1

i

n

 n    xi  n  i 1  2 8.12  x 15.63   i n 5  2.508  .627  s 2  i 1 n 1 5 1 4



8.1  1.62 5

The z-score corresponding to a value of 4.5 is z 

s  s 2  .627  .792

x  x 4.5  1.62   3.63 .792 s

Since this z-score is greater than 3, we would consider this an outlier for perturbed intrinsics, but no perturbed projections. For Perturbed Projections, but no Perturbed Intrinsics:

x

2

n

x

i 1

n

i

 n    xi  n 2 125.82 xi   i 1  3350.1   n 5  184.972  46.243  s 2  i 1 5 1 4 n 1



125.8  25.16 5

s  s 2  46.243  6.800 The z-score corresponding to a value of 4.5 is z 

x  x 4.5  25.16   3.038 6.800 s

Since this z-score is less than -3, we would consider this an outlier for perturbed projections, but no perturbed intrinsics. Since the z-score corresponding to 4.5 for the perturbed projections, but no perturbed intrinsics is smaller in absolute value than that for perturbed intrinsics, but no perturbed projections, it is more likely that the that the type of camera perturbation is perturbed projections, but no perturbed intrinsics.

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Methods for Describing Sets of Data

2.121

69

Using MINITAB, the scatterplot is: Scatterplot of Var2 vs Var1 18 16 14

Var2

12 10 8 6 4 2 0 0

1

2

3

4

5

Var1

2.122

Using MINITAB, a scatterplot of the data is: Scatterplot of Var2 vs Var1 14 12

Var2

10 8 6 4 2 0 0

2

4

6

8

Var1

2.123.

From the scatterplot of the data, it appears that as the number of punishments increases, the average payoff decreases. Thus, there appears to be a negative linear relationship between punishment use and average payoff. This supports the researchers conclusion that “winners” don’t punish”.

Copyright © 2014 Pearson Education, Inc.

70

2.124

Chapter 2

Using MINITAB, the scatterplot of the data is: Scatterplot of Catch vs Search 7000

Catch

6000

5000

4000

3000 15

20

25 Search

30

35

There is an apparent negative linear trend between the search frequency and the total catch. As the search frequency increases, the total catch tends to decrease. Using MINITAB, a scattergram of the data is: Scatterplot of SLUGPCT vs ELEVATION 0.625 0.600 0.575 SLUGPCT

2.125

0.550 0.525 0.500 0.475 0.450 0

1000

2000

3000 ELEVATION

4000

5000

6000

If we include the observation from Denver, then we would say there might be a linear relationship between slugging percentage and elevation. If we eliminated the observation from Denver, it appears that there might not be a relationship between slugging percentage and elevation.

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data

2.126

71

Using MINITAB, the scatterplot of the data is: Scatterplot of MATH2011 vs MATH2001 625 600

MATH2011

575 550 525 500 475 450 460

480

500

520

540 560 MATH2001

580

600

620

There appears to be a positive linear trend between the Math SAT scores in 2001 and the Math SAT scores in 2011. As the 2001 Math SAT scores increase, the 2011 Math SAT scores also tend to increase. a.

Using MINITAB, a scatterplot of JIF and cost is: Scatterplot of JIF vs Cost 3.5 3.0 2.5

JIF

2.0 1.5 1.0 0.5 0.0 0

200

400

600

800

1000 Cost

1200

1400

1600

1800

There is a slight negative linear trend to the data. As cost increases, JIF tends to decrease. b.

Using MINITAB, a scatterplot of the number of cities and cost is: Scatterplot of Cites vs Cost 800 700 600 500 Cites

2.127

400 300 200 100 0 0

200

400

600

800

1000 Cost

1200

1400

1600

1800

Copyright © 2014 Pearson Education, Inc.

72

Chapter 2

There is a moderate positive trend to the data. As cost increases, the number of cities tends to increase. c.

Using MINITAB, a scatterplot of RPI and cost is: Scatterplot of RPI vs Cost

4

RPI

3

2

1

0 0

200

400

600

800

1000 Cost

1200

1400

1600

1800

There is a slight positive trend to the data. As cost increases, RPI tends to increase. Using MINITAB, the scatterplot of the data is: Scatterplot of Mass vs Time 7 6 5 4 M ass

2.128

3 2 1 0 0

10

20

30 T ime

40

50

60

There is evidence to indicate that the mass of the spill tends to diminish as time increases. As time is getting larger, the mass is decreasing.

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Methods for Describing Sets of Data

2.129

a.

73

Using MINITAB, a scatterplot of the data is: Scatterplot of Year2 vs Y ear1 55

50

Year2

45

40

35

30

20

30

40 Year1

50

60

There is a moderate positive trend to the data. As the scores for Year1 increase, the scores for Year2 also tend to increase. b.

Using MINITAB, the scattergram of the data is: Scatterplot of Value ($mil) vs Income ($mil) 2000 1800 1600 Value ($mil)

2.130

From the graph, two agencies that had greater than expected PARS evaluation scores for Year2 were USAID and State.

1400 1200 1000 800 600 0

20

40

60 Income ($mil)

80

100

120

There is a moderate positive trend to the data. As operating income increases, the 2011 value also tends to increase. Since the trend is moderate, we would recommend that an NFL executive use operating income to predict a team’s current value.

Copyright © 2014 Pearson Education, Inc.

a.

Using MINITAB, the scatterplot of the data is: Scatterplot of YRSPRAC vs EDHRS 40

30 YRSPRAC

2.131

Chapter 2

20

10

0 0

200

400

600

800

1000

EDHRS

There does not appear to be much of a relationship between the years of experience and the amount of exposure to ethics in medical school. b.

Using MINITAB, a boxplot of the amount of exposure to ethics in medical school is: Boxplot of EDHRS 1000

800

600 EDHRS

74

400

200

0

The one data point that is an extreme outlier is the value of 1000.

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data

c.

75

After removing this data point, the scatterplot of the data is: Scatterplot of YRSPRAC vs EDHRS 40

YRSPRAC

30

20

10

0 0

10

20

30

40 50 EDHRS

60

70

80

90

With the data point removed, there now appears to be a negative trend to the data. As the amount of exposure to ethics in medical school increases, the years of experience decreases. 2.132

Using MINITAB, a scatterplot of the data is: Scatterplot of ACCURACY vs DISTANCE 75 70

ACCURACY

65

60 55 50 45 280

290

300 DISTANCE

310

320

Yes, his concern is a valid one. From the scatterplot, there appears to be a fairly strong negative relationship between accuracy and driving distance. As driving distance increases, the driving accuracy tend to decrease. 2.133

One way the bar graph can mislead the viewer is that the vertical axis has been cut off. Instead of starting at 0, the vertical axis starts at 12. Another way the bar graph can mislead the viewer is that as the bars get taller, the widths of the bars also increase.

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76

Chapter 2

2.134

a.

Using MINITAB, the time series plot is: Time Series Plot of Deaths 900 800 700

Deaths

600 500 400 300 200 100 0 2003

2004

2005

2006

Index

2.135

b.

The time series plot is misleading because the information for 2006 is incomplete – it is based on only 2 months while all of the rest of the years are based on 12 months.

c.

In order to construct a plot that accurately reflects the trend in American casualties from the Iraq War, we would want complete data for 2006 and information for the years 2007 through 2011.

a.

The graph might be misleading because the scales on the vertical axes are different. The left vertical axis ranges from 0 to $120 million. The right vertical axis ranges from 0 to $20 billion.

b.

Using MINITAB, the redrawn graph is: Time Series Plot of Craigslist, NewspaperAds 18000

Variable C raigslist NewspaperA ds

16000 14000

Data

12000 10000 8000 6000 4000 2000 0 2003

2004

2005

2006 Index

2007

2008

2009

Although the amount of revenue produced by Craigslist has increased dramatically from 2003 to 2009, it is still much smaller than the revenue produced by newspaper ad sales. 1.136

a.

This graph is misleading because it looks like as the days are increasing, the number of barrels collected per day are also increasing. However, the bars are the cumulative number of barrels collected. The cumulative value can never decrease.

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data

b.

77

Using MINITAB, the graph of the daily collection of oil is: Chart of Barrells 2500

Barrells

2000

1500

1000

500

0 May-16

May-17

May-18

May-19 May-20 Day

May-21

May-22

May-23

From this graph, it shows that there has not been a steady improvement in the suctioning process. There was an increase for 3 days, then a leveling off for 3 days, then a decrease. 2.137

The relative frequency histogram is: Histogram of Class

Relative frequency

.20

.15

.10

.05

0 1.125

2.625

4.125 5.625 Measurement Class

7.125

8.625

2.138

The mean is sensitive to extreme values in a data set. Therefore, the median is preferred to the mean when a data set is skewed in one direction or the other.

2.139

a.

z

b.

z

c

z

d.

z

x

 x

 x

 x





50  60  1 10

z

70  60 1 10

z

80  60 2 10



50  50 0 5

z

70  50 4 5

z

80  50 6 5



50  40 1 10

z

70  40 3 10

z

80  40 4 10



50  40  .1 100

z

70  40  .3 100

z

80  40  .4 100

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78

Chapter 2

2.140

2.141

a.

If we assume that the data are about mound-shaped, then any observation with a z-score greater than 3 in absolute value would be considered an outlier. From Exercise 2.139, the z-score corresponding to 50 is 1, the z-score corresponding to 70 is 1, and the z-score corresponding to 80 is 2. Since none of these z-scores is greater than 3 in absolute value, none would be considered outliers.

b.

From Exercise 2.139, the z-score corresponding to 50 is 2, the z-score corresponding to 70 is 2, and the z-score corresponding to 80 is 4. Since the z-score corresponding to 80 is greater than 3, 80 would be considered an outlier.

c.

From Exercise 2.139, the z-score corresponding to 50 is 1, the z-score corresponding to 70 is 3, and the z-score corresponding to 80 is 4. Since the z-scores corresponding to 70 and 80 are greater than or equal to 3, 70 and 80 would be considered outliers.

d.

From Exercise 2.139, the z-score corresponding to 50 is .1, the z-score corresponding to 70 is .3, and the z-score corresponding to 80 is .4. Since none of these z-scores is greater than 3 in absolute value, none would be considered outliers.

a.

 x  13  1  10  3  3  30 x

b.

 x  25  6.25

x s2 

4

n

x

2

n 1

x n



49 7 7

s2 

x

 x  3  3  3  3  12 x

 x  12  3 n

4

s2 

n

2

n 1

n 1

x

2

302 5  108  27 5 1 4

288 



 x 

 x 

x

2

s  27  5.20

 132  62  62  02  241

2

x

2

 132  12  102  32  32  288

 x 

2

 x  1  0  1  10  11  11  15  49 x

d.

s2 

5

n

 x  13  6  6  0  25 x

c.

 x  30  6

x

2

2

n

x

2

n





2

252 4  84.75  28.25 4 1 3

241 

 12  02  12  102  112  112  152  569 .

492 7  226  37.67 7 1 6

569 

s  37.67  6.14

 32  32  32  32  36

 x 

n 1

n

2



s  28.25  5.32

122 4  0 0 4 1 3

36 

Copyright © 2014 Pearson Education, Inc.

s 0 0

2.142

a.

 x  4  6  6  5  6  7  34 x

b.

Methods for Describing Sets of Data

 x  34  5.67

s2 

6

n

x

x

2

2

 4 2  6 2  6 2  52  6 2  7 2  198

 x 

n 1

2

n



x

x n



9  $1.5 6

s2 

x

 x 

n 1

n

342 6  5.3333  1.0667 6 1 5

198 

 x  1  4  (3)  0  (3)  (6)  9  x 2

79

2

s  1.067  1.03

 ( 1) 2  4 2  ( 3) 2  0 2  ( 3) 2  ( 6) 2  71

2



(9) 2 6  57.5  11.5 dollars squared 6 1 5

71 

s  11.5  $3.39

c.

 x  5  5  5  5  16  2.0625 3

x

s2 

4

2

1

1

 x  2.0625  .4125%

3  4  2 1  1                  1.2039  5   5   5   5   16  2

2

2

2

2

2

5

n

x

x

2

 x  n 1

n

2



2.06252 .3531 5   .0883% squared 5 1 4

1.2039 

s  .0883  .30%

d.

(a)

Range = 7  4 = 3

(b)

Range = $4  ($-6) = $10

(c)

Range =

4 1 64 5 59 % %  % %  %  .7375% 5 16 80 80 80

2.143

The range is found by taking the largest measurement in the data set and subtracting the smallest measurement. Therefore, it only uses two measurements from the whole data set. The standard deviation uses every measurement in the data set. Therefore, it takes every measurement into account—not just two. The range is affected by extreme values more than the standard deviation.

2.144



range 20  5 4 4

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80

Chapter 2

2.145

Using MINITAB, the scatterplot is: Scatterplot of Var2 vs Var1 30

Var2

25

20

15

10

100

a.

300 Var1

Management System Cause Category Engineering & Design Procedures & Practices Management & Oversight Training & Communication TOTAL

b.

400

500

To find relative frequencies, we divide the frequencies of each category by the total number of incidents. The relative frequencies of the number of incidents for each of the cause categories are: Number of Incidents

Relative Frequencies

27 24 22 10 83

27 / 83 = .325 24 / 83 = .289 22 / 83 = .265 10 / 83 = .120 1

The Pareto diagram is: Management Systen Cause Category 35 30 25 P er cent

2.146

200

20 15 10 5 0 E ng&D es

c.

P roc&P ract M gmt&O v er C ategor y

Trn&C omm

The category with the highest relative frequency of incidents is Engineering and Design. The category with the lowest relative frequency of incidents is Training and Communication. Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data

2.147

a.

81

The relative frequency for each response category is found by dividing the frequency by the total sample size. The relative frequency for the category “Global Marketing” is 235/2863 = .082. The rest of the relative frequencies are found in a similar manner and are reported in the table. Area Global Marketing Sales Management Buyer Behavior Relationships Innovation Marketing Strategy Channels/Distribution Marketing Research Services TOTAL

Number 235 494 478 498 398 280 213 131 136 2,863

Relative Frequencies 235/2863 = .082 494/2863 = .173 478/2863 = .167 498/2863 = .174 398/2863 = .139 280/2863 = .098 213/2863 = .074 131/2863 = .046 136/2863 = .048 1.00

Relationships and sales management had the most articles published with 17.4% and 17.3%, respectively. Not far behind was Buyer Behavior with 16.7%. Of the rest of the areas, only innovation had more than 10%. b.

Using MINITAB, the pie chart of the data is: Pie Chart of Number vs Area Serv ices Mark eting research 4.8% 4.6%

Global Mark eting 8.2%

C hannells/Distribution 7.4% Sales Management 17.3%

Mark eting Strategy 9.8%

Inov ation 13.9%

C ategory Global Mark eting Sales Management Buy er Behavior Relationships Inovation Mark eting Strategy C hannells/Distribution Mark eting research Serv ices

Buy er Behav ior 16.7%

Relationships 17.4%

The slice for Marketing Research is smaller than the slice for Sales Management because there were fewer articles on Marketing Research than for Sales Management. 2.148

a.

The data are time series data because the numbers of bankruptcies were collected over a period of 10 months.

Copyright © 2014 Pearson Education, Inc.

82

Chapter 2

b.

Using MINITAB, the time series plot is: Time Series Plot of Bankrupties 120000 100000

Bankrupties

80000 60000 40000 20000 0

0 Jan

c. 2.149

Feb

Mar

Apr

May Jun Month

Jul

Aug

Sep

Oct

There is a generally increasing trend in the number of bankruptcies as the months increase.

Using MINITAB, the pie chart is: Pie Chart of F vs DrivStar 2 4, 4.1%

5 18, 18.4%

3 17, 17.3%

C ategory 2 3 4 5

4 59, 60.2%

60% of cars have 4-star rating and only 4% have 2-star ratings. 2.150

a.

The average driver’s severity of head injury in head-on collisions is 603.7.

b.

Since the mean and median are close in value, the data should be fairly symmetric. Thus, we can use the Empirical Rule. We know that about 95% of all observations will fall within 2 standard deviations of the mean. This interval is x  2 s  603.7  2(185.4)  603.7  370.8  (232.9, 974.5) Most of the head-injury ratings will fall between 232.9 and 974.5.

c.

x  x 408  603.7   1.06 185.4 s Since the absolute value is not very big, this is not an unusual value to observe.

The z-score would be: z 

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data

2.151

a.

83

Using MINITAB, a Pareto diagram for the data is: Chart Defects 70 60

Frequency

50 40 30 20 10 0 Body

Accessories

Electrical Defect

Transmission

Engine

The most frequently observed defect is a body defect. b.

Using MINITAB, a Pareto diagram for the Body Defect data is: Chart of Body Defects 30

Frequency

25 20 15 10 5 0 Paint

Dents

Upolstery Body Defect

Windshield

Chrome

Most body defects are either paint or dents. These two categories account for  30  25 / 70  55 / 70  .786 of all body defects. Since these two categories account for so much of the body defects, it would seem appropriate to target these two types of body defects for special attention. 2.152

a.

The data collection method was a survey.

b.

Since the data were 4 different categories, the variable is qualitative.

Copyright © 2014 Pearson Education, Inc.

84

Chapter 2

c.

Using MINITAB, a pie chart of the data is: Pie Chart of Made USA Category < 50% 100% 50-74% 75-99%

< 50% 4, 3.8% 75-99% 20, 18.9%

50-74% 18, 17.0%

100% 64, 60.4%

About 60% of those surveyed believe that “Made in USA” means 100% US labor and materials. 2.153

a.

From the information given, we have x  375 and s = 25. From Chebyshev's Rule, we know that at least three-fourths of the measurements are within the interval: x  2 s , or (325, 425) Thus, at most one-fourth of the measurements exceed 425. In other words, more than 425 vehicles used the intersection on at most 25% of the days.

b.

According to the Empirical Rule, approximately 95% of the measurements are within the interval: x  2 s , or (325, 425)

This leaves approximately 5% of the measurements to lie outside the interval. Because of the symmetry of a mound-shaped distribution, approximately 2.5% of these will lie below 325, and the remaining 2.5% will lie above 425. Thus, on approximately 2.5% of the days, more than 425 vehicles used the intersection. 2.154

The percentile ranking of the age of 25 years would be 100%  75% = 25%. Thus, an age of 25 would correspond to the 25th percentile.

2.155

a.

Using MINITAB, the stem-and-leaf display is: Stem-and-Leaf of PENALTY Leaf Unit = 10 (28) 10 5 5 4 3 3 3 3 2 1

b.

0 1 2 3 4 5 6 7 8 9 10

N = 38

0011111222222223333334444899 00239 0 0

5 3 0

See the highlighted leaves in part a. Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data

c.

2.156

85

Most of the penalties imposed for Clean Air Act violations are relatively small compared to the penalties imposed for other violations. All but two of the penalties for Clean Air Act violations are below the median penalty imposed.

Using MINITAB, the pie charts are: Color F (82, 26.6%) E (44, 14.3%)

D (16, 5.2%)

I (40, 13.0%)

G (65, 21.1%)

H (61, 19.8%)

VS1 (81, 26.3%)

IF

(44, 14.3%)

VS2 (53, 17.2%)

VVS2 (78, 25.3%)

VVS1 (52, 16.9%)

Clarity

The F color occurs the most often with 26.6%. The clarity that occurs the most is VS1 with 26.3%. The D color occurs the least often with 5.2%. The clarity that occurs the least is IF with 14.3%. a.

Using MINITAB, the relative frequency histogram is: Histogram of CARAT .20

Relative frequency

2.157

.15

.10

.05

0 0.30

0.45

0.60 CARAT

0.75

0.90

1.05

Copyright © 2014 Pearson Education, Inc.

Chapter 2

b.

Using MINITAB, the relative frequency histogram for the GIA group is: Histogram for GIA .14

Relative frequency

.12 .10 .08 .06 .04 .02 0 0.30

0.45

0.60 CARAT

0.75

0.90

1.05

Using MINITAB, the relative frequency histograms for the HRD and IGI groups are: Histogram for HRD .40

Relative frequency

c.

.30

.20

.10

0 0.5

0.6

0.7

0.8 CARAT

0.9

1.0

1.1

Histogram for IGI .35 .30

Relative frequency

86

.25 .20 .15 .10 .05 0 0.2

0.4

0.6 CARAT

0.8

1.0

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data

d.

The HRD group does not assess any diamonds less than .5 carats and almost 40% of the diamonds they assess are 1.0 carat or higher. The IGI group does not assess very many diamonds over .5 carats and more than half are .3 carats or less. More than half of the diamonds assessed by the GIA group are more than .5 carats, but the sizes are less than those of the HRD group.

x n

The sample mean is: x 

e.

i 1

i

n



194.32  .631 308

The average number of carats for the 308 diamonds is .631. f.

The median is the average of the middle two observations once they have been ordered. The 154th and 155th observations are .62 and .62. The average of these two observations is .62. Half of the diamonds weigh less than .62 carats and half weigh more.

g

The mode is 1.0. This observation occurred 32 times.

h.

Since the mean and median are close in value, either could be a good descriptor of central tendency.

i.

From Chebyshev’s Theorem, we know that at least ¾ or 75% of all observations will fall within 2 standard deviations of the mean. From part e, x  .63 .     xi  194.322 xi2   i  146.19   n 308  .0768 square carats  The variance is: s 2  i 308  1 n 1 2

The standard deviation is: s  s 2  .0768  .277 carats This interval is: x  2 s  .631  2(.277)  .631  .554  (.077, 1.185) Using MINITAB, the scatterplot is: Scatterplot of PRICE vs CARAT 18000 16000 14000 12000 PRICE

2.158

87

10000 8000 6000 4000 2000 0 0.2

0.3

0.4

0.5

0.6 0.7 CARAT

0.8

0.9

1.0

1.1

As the number of carats increases the price of the diamond tends to increase. There appears to be an upward trend.

Copyright © 2014 Pearson Education, Inc.

2.159

Chapter 2

a.

Using MINITAB, a bar graph of the data is: Chart of Cause 12

10

8 Count

88

6

4

2

0 Collision

Fire

Grounding Cause

HullFail

Unknown

Fire and grounding are the two most likely causes of puncture. b.

Using MINITAB, the descriptive statistics are: Descriptive Statistics: Spillage Variable Spillage

N 42

Mean 66.19

StDev 56.05

Minimum 25.00

Q1 32.00

Median 43.00

Q3 77.50

Maximum 257.00

The mean spillage amount is 66.19 thousand metric tons, while the median is 43.00. Since the median is so much smaller than the mean, it indicates that the data are skewed to the right. The standard deviation is 56.05. Again, since this value is so close to the value of the mean, it indicates that the data are skewed to the right. Since the data are skewed to the right, we cannot use the Empirical Rule to describe the data. Chebyshev’s Rule can be used. Using Chebyshev’s Rule, we know that at least 8/9 of the observations will fall within 3 standard deviations of the mean. x  3s  66.19  3(56.05)  66.19  168.15  ( 101.96, 234.34) or (0, 234.34) since we cannot have negative spillage.

Thus, at least 8/9 of all oil spills will be between 0 and 234.34 thousand metric tons.

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data

2.160

Using MINITAB, a pie chart of the data is: Pie Chart of Defectt Category False True

True 49, 9.8%

False 449, 90.2%

A response of ‘true’ means the software contained defective code. Thus, only 9.8% of the modules contained defective software code. 2.161

a.

Since no information is given about the distribution of the velocities of the Winchester bullets, we can only use Chebyshev's Rule to describe the data. We know that at least 3/4 of the velocities will fall within the interval: x  2 s  936  2(10)  936  20  (916, 956)

Also, at least 8/9 of the velocities will fall within the interval: x  3s  936  3(10)  936  30  (906, 966)

b.

Since a velocity of 1,000 is much larger than the largest value in the second interval in part a, it is very unlikely that the bullet was manufactured by Winchester.

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89

90

Chapter 2

2.162

a.

First, we must compute the total processing times by adding the processing times of the three departments. The total processing times are as follows: Request

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Total Processing Time 13.3 5.7 7.6 20.0* 6.1 1.8 13.5 13.0 15.6 10.9 8.7 14.9 3.4 13.6 14.6 14.4

Request

17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

Total Processing Time 19.4* 4.7 9.4 30.2 14.9 10.7 36.2* 6.5 10.4 3.3 8.0 6.9 17.2* 10.2 16.0 11.5

Request

33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

Total Processing Time 23.4* 14.2 14.3 24.0* 6.1 7.4 17.7* 15.4 16.4 9.5 8.1 18.2* 15.3 13.9 19.9* 15.4 14.3* 19.0

The stem-and-leaf displays with the appropriate leaves highlighted are as follows: Stem-and-leaf of Mkt Leaf Unit = 0.10 6 0 7 1 14 2 16 3 22 4 (10) 5 18 6 8 7 4 8 2 9 2 10 1 11

0112446 3 0024699 25 001577 0344556889 0002224799 0038 07 0 0

Stem-and-leaf of Engr Leaf Unit = 0.10 7 14 19 23 (5) 22 19 14 9 9 7 6 5 2 1

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

4466699 3333788 12246 1568 24688 233 01239 22379 66 0 3 023 0 4

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data Stem-and-leaf of Accnt Leaf Unit = 0.10 19 (8) 23 21 19 15 15 13 11 11 11 11 10 9 9 8 8

0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 HI

111111111112 2333444 55556888 00 79 0023 23 78

8 2

91

Stem-and-leaf of Total Leaf Unit = 1.00 1 3 5 11 17 21 (5) 24 14 10 6 5 4

0 0 0 0 0 1 1 1 1 1 2 2 2 HI

1 33 45 666677 888999 0000 33333 4444445555 6677 8999 0 3 44 30, 36

0 4 99, 105, 135, 144, 182, 220, 300

Of the 50 requests, 10 were lost. For each of the three departments, the processing times for the lost requests are scattered throughout the distributions. The processing times for the departments do not appear to be related to whether the request was lost or not. However, the total processing times for the lost requests appear to be clustered towards the high side of the distribution. It appears that if the total processing time could be kept under 17 days, 76% of the data could be maintained, while reducing the number of lost requests to 1. b.

For the Marketing department, if the maximum processing time was set at 6.5 days, 78% of the requests would be processed, while reducing the number of lost requests by 4. For the Engineering department, if the maximum processing time was set at 7.0 days, 72% of the requests would be processed, while reducing the number of lost requests by 5. For the Accounting department, if the maximum processing time was set at 8.5 days, 86% of the requests would be processed, while reducing the number of lost requests by 5.

c.

Using MINITAB, the summary statistics are:

Descriptive Statistics: REQUEST, MARKET, ENGINEER, ACCOUNT Variable MARKET ENGINEER ACCOUNT TOTAL

N Mean 50 4.766 50 5.044 50 3.652 50 13.462

StDev 2.584 3.835 6.256 6.820

Minimum 0.100 0.400 0.100 1.800

Q1 2.825 1.775 0.200 8.075

Median Q3 5.400 6.250 4.500 7.225 0.800 3.725 13.750 16.600

Copyright © 2014 Pearson Education, Inc.

Maximum 11.000 14.400 30.000 36.200

92

Chapter 2

d.

The z-scores corresponding to the maximum time guidelines developed for each department and the total are as follows: Marketing: z 

Engineering: z 

x  x 7.0  5.04   .51 3.84 s

Accounting: z 

x  x 8.5  3.65   .77 6.26 s

Total: z  e.

x  x 6.5  4.77   .67 2.58 s

x  x 17  13.46   .52 6.82 s

To find the maximum processing time corresponding to a z-score of 3, we substitute in the values of z, x , and s into the z formula and solve for x. z

xx  x  x  zs  x  x  zs s

Marketing:

x  4.77  3(2.58)  4.77  7.74  12.51 None of the orders exceed this time.

Engineering:

x  5.04  3(3.84)  5.04  11.52  16.56 None of the orders exceed this time.

These both agree with both the Empirical Rule and Chebyshev's Rule. Accounting:

x  3.65  3(6.26)  3.65  18.78  22.43 One of the orders exceeds this time or 1/50 = .02.

Total:

x  13.46  3(6.82)  13.46  20.46  33.92 One of the orders exceeds this time or 1/50 = .02.

These both agree with Chebyshev's Rule but not the Empirical Rule. Both of these last two distributions are skewed to the right. f.

Marketing:

x  4.77  2(2.58)  4.77  5.16  9.93 Two of the orders exceed this time or 2/50 = .04.

Engineering:

x  5.04  2(3.84)  5.04  7.68  12.72

Two of the orders exceed this time or 2/50 = .04. Accounting:

x  3.65  2(6.26)  3.65  12.52  16.17 Three of the orders exceed this time or 3/50 = .06.

Copyright © 2014 Pearson Education, Inc.

Methods for Describing Sets of Data

93

x  13.46  2(6.82)  13.46  13.64  27.10 Two of the orders exceed this time or 2/50 = .04.

Total:

All of these agree with Chebyshev's Rule but not the Empirical Rule. g.

No observations exceed the guideline of 3 standard deviations for both Marketing and Engineering. One observation exceeds the guideline of 3 standard deviations for both Accounting (#23, time = 30.0 days) and Total (#23, time = 36.2 days). Therefore, only (1/10)  100% of the "lost" quotes have times exceeding at least one of the 3 standard deviation guidelines. Two observations exceed the guideline of 2 standard deviations for both Marketing (#31, time = 11.0 days and #48, time = 10.0 days) and Engineering (#4, time = 13.0 days and #49, time = 14.4 days). Three observations exceed the guideline of 2 standard deviations for Accounting (#20, time = 22.0 days; #23, time = 30.0 days; and #36, time = 18.2 days). Two observations exceed the guideline of 2 standard deviations for Total (#20, time = 30.2 days and #23, time = 36.2 days). Therefore, (7/10)  100% = 70% of the "lost" quotes have times exceeding at least one the 2 standard deviation guidelines. We would recommend the 2 standard deviation guideline since it covers 70% of the lost quotes, while having very few other quotes exceed the guidelines. One reason the plot may be interpreted differently is that no scale is given on the vertical axis. Also, since the plot almost reaches the horizontal axis at 3 years, it is obvious that the bottom of the plot has been cut off. Another important factor omitted is who responded to the survey.

b.

A scale should be added to the vertical axis. Also, that scale should start at 0. Using MINITAB, the time series plot of the data is: Time Series Plot of Acquisitions 900 800 700 600 500 400 300 200 100

2000

1999

1998

1997

1996

1995

1994

1993

1992

1991

1990

1989

1988

1987

1986

1985

1984

1983

1982

0 1981

a.

1980

2.164

a.

Acquisitions

2.163

Year

Copyright © 2014 Pearson Education, Inc.

Chapter 2

b.

To find the percentage of the sampled firms with at least one acquisition, we divide number with acquisitions by the total sampled and then multiply by 100%. For 1980, the percentage of firms with at least on acquisition is (18/1963)*100% = .92%. The rest of the percentages are found in the same manner and are listed in the following table: Year

Number of firms

1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 TOTAL

1,963 2,044 2,029 2,187 2,248 2,238 2,277 2,344 2,279 2,231 2,197 2,261 2,363 2,582 2,775 2,890 3,070 3,099 2,913 2,799 2,778 51,567

Number with Acquisitions 18 115 211 273 317 182 232 258 296 350 350 370 427 532 626 652 751 799 866 750 748 9,123

Percentage with Acquisitions .92% 5.63% 10.40% 12.48% 14.10% 8.13% 10.19% 11.01% 12.99% 15.69% 15.93% 16.36% 18.07% 20.60% 22.56% 22.56% 24.46% 25.78% 29.73% 26.80% 26.93%

Using MINITAB, the time series plot is: Time Series Plot of Percent 30 25 20 Percent

15 10 5

2000

1999

1998

1997

1996

1995

1994

1993

1992

1991

1990

1989

1988

1987

1986

1985

1984

1983

1982

1981

0 1980

94

Year

c.

In this case, both plots are almost the same. In general, the time series plot of the percents would be more informative. By changing the observations to percents, one can compare time periods with different sample sizes on the same basis.

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Methods for Describing Sets of Data 2.165

a.

b.

95

Since the mean is greater than the median, the distribution of the radiation levels is skewed to the right. x  s  10  3  (7, 13) ; x  2 s  10  2(3)  (4, 16) ; x  3s  10  3(3)  (1, 19)

Interval (7, 13) (4, 16) (1, 19)

Chebyshev's At least 0 At least 75% At least 88.9%

Empirical 68% 95% 100%

Since the data are skewed to the right, Chebyshev's Rule is probably more appropriate in this case. c.

The background level is 4. Using Chebyshev's Rule, at least 75% or .75(50)  38 homes are above the background level. Using the Empirical Rule,  97.5% or .975(50)  49 homes are above the background level.

d.

z

x  x 20  10   3.333 3 s

It is unlikely that this new measurement came from the same distribution as the other 50. Using either Chebyshev's Rule or the Empirical Rule, it is very unlikely to see any observations more than 3 standard deviations from the mean.

2.167

a.

Since it is given that the distribution is mound-shaped, we can use the Empirical Rule. We know that 1.84% is 2 standard deviations below the mean. The Empirical Rule states that approximately 95% of the observations will fall within 2 standard deviations of the mean and, consequently, approximately 5% will lie outside that interval. Since a mound-shaped distribution is symmetric, then approximately 2.5% of the day's production of batches will fall below 1.84%.

b.

If the data are actually mound-shaped, it would be extremely unusual (less than 2.5%) to observe a batch with 1.80% zinc phosphide if the true mean is 2.0%. Thus, if we did observe 1.8%, we would conclude that the mean percent of zinc phosphide in today's production is probably less than 2.0%.

a.

Both the height and width of the bars (peanuts) change. Thus, some readers may tend to equate the area of the peanuts with the frequency for each year.

b.

Using MINITAB, the frequency bar chart is: Chart of Peanut 5

4

Peanut

2.166

3

2

1

0 1975

1980

1985

1990

1995

2000

2005

2010

Year

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Chapter 2 a.

Clinic A claims to have a mean weight loss of 15 during the first month and Clinic B claims to have a median weight loss of 10 pounds in the first month. With no other information, I would choose Clinic B. It is very likely that the distributions of weight losses will be skewed to the right – most people lose in the neighborhood of 10 pounds, but a couple might lose much more. If a few people lost much more than 10 pounds, then the mean will be pulled in that direction.

b.

For Clinic A, the median is 10 and the standard deviation is 20. For Clinic B, the mean is 10 and the standard deviation is 5. For Clinic A: The mean is 15 and the median is 10. This would indicate that the data are skewed to the right. Thus, we will have to use Chebyshev’s Rule to describe the distribution of weight losses. x  2 s  15  2(20)  15  40  (  25, 55)

Using Chebyshev’s Rule, we know that at least 75% of all weight losses will be between -25 and 55 pounds. This means that at least 75% of the people will have weight losses of between a loss of 55 pounds to a gain of 25 pounds. This is a very large range. For Clinic B: The mean is 10 and the median is 10. This would indicate that the data are symmetrical. Thus, the Empirical Rule can be used to describe the distribution of weight losses. x  2 s  10  2(5)  10  10  (0, 20)

Using the Empirical Rule, we know that approximately 95% of all weight losses will be between 0 and 20 pounds. This is a much smaller range than in Clinic A. I would still recommend Clinic B. Using Clinic A, a person has the potential to lose a large amount of weight, but also has the potential to gain a relatively large amount of weight. In Clinic B, a person would be very confident that he/she would lose weight. c.

2.169

One would want the clients selected for the samples in each clinic to be representative of all clients in that clinic. One would hope that the clinic would not choose those clients for the sample who lost the most weight just to promote their clinic.

First we make some preliminary calculations. Of the 20 engineers at the time of the layoffs, 14 are 40 or older. Thus, the probability that a randomly selected engineer will be 40 or older is 14/20 = .70. A very high proportion of the engineers is 40 or over. In order to determine if the company is vulnerable to a disparate impact claim, we will first find the median age of all the engineers. Ordering all the ages, we get: 29, 32, 34, 35, 38, 39, 40, 40, 40, 40, 40, 41, 42, 42, 44, 46, 47, 52, 55, 64 The median of all 20 engineers is

40  40 80   40 2 2

Now, we will compute the median age of those engineers who were not laid off. The ages underlined 40  40 80   40 . above correspond to the engineers who were not laid off. The median of these is 2 2

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Methods for Describing Sets of Data

97

The median age of all engineers is the same as the median age of those who were not laid off. The median 40  41 81   40.5 , which is not that much different from the median age of those age of those laid off is 2 2 not laid off. In addition, 70% of all the engineers are 40 or older. Thus, it appears that the company would not be vulnerable to a disparate impact claim. 2.170

Answers will vary. The graph is made to look like the amount of money spent on education has risen dramatically from 1980 to 2000, but the 4th grade reading scores have not increased at all. The graph does not take into account that the number of school children has also increased dramatically in the last 20 years. A better portrayal would be to look at the per capita spending rather than total spending.

2.171

There is evidence to support this claim. The graph peaks at the interval above 1.002. The heights of the bars decrease in order as the intervals get further and further from the peak interval. This is true for all bars except the one above 1.000. This bar is greater than the bar to its right. This would indicate that there are more observations in this interval than one would expect, suggesting that some inspectors might be passing rods with diameters that were barely below the lower specification limit.

Copyright © 2014 Pearson Education, Inc.

Chapter 3 Probability 3.1

a.

Since the probabilities must sum to 1, P ( E3 )  1  P ( E1 )  P ( E 2 )  P ( E 4 )  P ( E5 )  1  .1  .2  .1  .1  .5

b.

3.2

P ( E3 )  1  P ( E1 )  P ( E2 )  P ( E4 )  P ( E5 )  1  P ( E3 )  P ( E2 )  P ( E4 )  P ( E5 )  2 P ( E3 )  1  .1  .2  .1  2 P ( E3 )  .6  P ( E3 )  .3

c.

P ( E3 )  1  P ( E1 )  P ( E 2 )  P ( E 4 )  P ( E5 )  1  .1  .1  .1  .1  .6

a.

This is a Venn Diagram.

b.

If the sample points are equally likely, then P (1)  P (2)  P (3)    P (10) 

1 10

Therefore, 1 1 1 3     .3 10 10 10 10 1 1 2 P ( B )  P (6)  P (7)     .2 10 10 10

P ( A)  P (4)  P (5)  P (6) 

c.

3.3

1 1 3 5     .25 20 20 20 20 3 3 6 P ( B )  P (6)  P (7)     .3 20 20 20 P ( A)  P (4)  P (5)  P (6) 

P( A)  P(1)  P(2)  P(3)  .05  .20  .30  .55 P( B )  P (1)  P (3)  P (5)  .05  .30  .15  .50 P(C )  P(1)  P(2)  P(3)  P(5)  .05  .20  .30  .15  .70

3.4

a.

b.

c.

d.

 9 9! 9  8  7  6  5  4  3  2 1   126    4  4!(9  4)! 4  3  2 1  5  4  3  2 1 7 7! 7  6  5  4  3  2 1   21   2 2!(7 2)! 2 1  5  4  3  2 1     4 4! 4  3  2 1  1   4 4!(4 4)! 4   3  2 1 1  

 5 5! 5  4  3  2 1  1    0  0!(5  0)! 1  5  4  3  2 1 98 Copyright © 2014 Pearson Education, Inc.

Probability

3.5

3.6

99

e.

6 6! 6  5  4  3  2 1  6   5   5!(6  5)! 5  4  3  2  1  1

a.

 N  5 5! 5  4  3  2  1 120    10     n   2  2!(5  2)! 2 1  3  2  1 12

b.

 N  6 6! 6  5  4  3  2  1 720    20    n 3  3!(6 3)! 3  2  1  3  2  1 36    

c.

 N   20  20! 20  19  18    3  2  1 2.432902008  1018    15,504    14  n   5  5!(20  5)! 5  4  3  2  1 15 14  13    3  2  1 1.569209242  10

a.

The tree diagram of the sample points is:

b.

If the dice are fair, then each of the sample points is equally likely. Each would have a probability of 1/36 of occurring. Copyright © 2014 Pearson Education, Inc.

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Chapter 3

c.

There is one sample point in A: (3,3). Thus, P ( A) 

1 . 36

There are 6 sample points in B: (1,6) (2,5) (3,4) (4,3) (5,2) and (6,1). P ( B) 

6 1  . 36 6

There are 18 sample points in C: (1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) 18 1  . (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) and (6,6). Thus, P(C )  36 2 3.7

a.

If we denote the marbles as B1, B2, R1, R2, and R3, then the ten sample points are: (B1, B2) (B1, R1) (B1, R2) (B1, R3) (B2, R1) (B2, R2) (B2, R3) (R1, R2) (R1, R3) (R2, R3)

b.

1 . 10 There are 6 sample points in B: (B1, R1) (B1, R2) (B1, R3) (B2, R1) (B2, R2) (B2, R3).  1  6 3 Thus, P ( B )  6     .  10  10 5  1  3 . There are 3 sample points in C: (R1, R2) (R1, R3) (R2, R3). Thus, P (C )  3     10  10 Each student will obtain slightly different proportions. However, the proportions should be close to

c.

3.8

Each of the sample points would be equally likely. Thus, each would have a probability of 1/10 of occurring. There is one sample point in A: (B1, B2). Thus, P( A) 

P ( A)  1 / 10, P ( B )  6 / 10, and P (C )  3 / 10.

3.9

a.

The sample points of this experiment correspond to each of the 6 possible colors of the M&M’s. Let B r = brown, Y = yellow, R = red, Bl = blue, O = orange, G = green. The six sample points are: Br, Y, R, Bl, O, and G

b.

From the problem, the probabilities of selecting each color are: P(Br) = 0.13, P(Y) = 0.14, P(R) = 0.13, P(Bl) = 0.24, P(O) = 0.2, P(G) = 0.16

c.

The probability that the selected M&M is brown is P(Br) = 0.13

d.

The probability that the selected M&M is red, green or yellow is: P ( R or G or Y )  P ( R )  P (G )  P (Y )  0.13  0.16  0.14  0.43

e. 3.10

P (not Bl )  P ( R )  P (G )  P (Y )  P ( Br )  P (O )  0.13  0.16  0.14  0.13  0.20  0.76

Define the following events: I: {personal illness} F: {family issues} N: {personal needs} E: {entitlement mentality} S: {stress}

Copyright © 2014 Pearson Education, Inc.

Probability

a.

The 5 sample points are: I, F, N, E, S

b.

The probability of each sample points are: P ( I )  0.34, P ( F )  0.22, P ( N )  0.18, P ( E )  0.13, P ( S )  0.13

c.

The probability that the absence is due to something other than “personal illness” (I) is: P (not I )  P ( F )  P ( N )  P ( E )  P ( S )  0.22  0.18  0.13  0.13  0.66

3.11

Define the following events: M: {Nanny who was placed in a job last year is a male} P(M ) 

3.12

a.

24  .0057 4,176

Define the following events: H5: {Hurricane develops from 5th tropical storm} H12: {Hurricane develops before the 12th or higher tropical storm}

b. 3.13

a.

P(H 5 ) 

11  .164 67

P ( H 12 ) 

67  5  .925 67

The 5 sample points are the possible responses of a randomly selected person who participated in Harris Poll: None, 1-2, 3-5, 6-9, 10 or more

b.

The probabilities are: P (none)  0.19, P (1  2)  0.31, P (3  5)  0.26, P (6  9)  0.05, P (10 or more)  0.19

c.

Define the following event: A: {Respondent looks for healthcare information online more than two times per month} P ( A)  P (3  5)  P (6  9)  P (10 or more)  0.25  0.05  0.19  0.50

3.14

a.

Define the following events: A: {Respondent works during summer vacation} B: {Respondent does not work during summer vacation} C: {Respondent unemployed} The sample points are A, B, and C.

b.

Reasonable probabilities are: P ( A)  .46, P ( B )  .35, and P (C )  .19

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102

3.15

Chapter 3

c.

P ( B or C )  P ( B )  P (C )  .35  .19  .54

a.

The international consumer is most likely to use the Certification mark on a label to identify a green product.

b.

Define the following events: A: B: C: D: E: F:

{Certification mark on label} {Packaging} {Reading information about the product} {Advertisement} {Brand website} {Other}

P ( A or B )  P ( A)  P ( B )  .45  .15  .60

3.16

c.

P (C or E )  P (C )  P ( E )  .12  .04  .16

d.

P (not D )  P ( A)  P ( B )  P (C )  P ( E )  P ( F )  .45  .15  .12  .04  .18  .94

a.

Define the following events: A: B: C: D: E:

{Total visitors} {Paying visitors} {Big shows} {Funds raised} {Members}

P( A or D)  P( A)  P( D)  b & c.

8 7 15    .5 30 30 30

A tree diagram with the corresponding probabilities for this problem follows. To compute the probabilities, we have to assume that this sample is representative of all such museums. In addition, we have to assume that each selection of a museum is independent of the second selection. The probability of selecting a particular type of museum is estimated by the number of museums in that category divided by 30. Each sample point consists of two museums. The probabilities of each type of museum in the pair are then multiplied together to find the probability of the sample point. The probabilities are shown in the tree.

Copyright © 2014 Pearson Education, Inc.

Probability

3.17

d.

P ( AA or DD or AD or DA)  P ( AA)  P ( DD )  P ( AD )  P ( DA)  .071  .054  .062  .062  .249

a.

Define the following event: C: {Slaughtered chicken passes inspection with fecal contamination} P (C ) 

1  .01 100

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104

Chapter 3

b.

3.18

306  .0095  .01 32, 075 Yes. The probability of a slaughtered chicken passing inspection with fecal contamination rounded off to 2 decimal places is .01.

Based on the data, P (C ) 

Define the following events: B: {Bitel} C: {Cybernet} F: {Fujian Landi} G: {Glint (Pava Rede)} I: {Intelligent} K: {Kwang Woo} O: {Omron} PT: {Pax Tech} PC: {Provenco Cadmus} S: {SZZT Electronics} T: {Toshiba TEC} U: {Urmet} To compute the probability of each event, we first must sum the number of units shipped by all the manufacturers. The sum is 334,039. P ( B )  13, 500 / 344, 039  .040; P (C )  16, 200 / 344, 039  .048; P ( F )  119, 000 / 344, 039  .356; P (G )  5, 990 / 344, 039  .018; P ( I )  4, 562 / 344, 039  .014; P ( K )  42, 000 / 344, 039  .126; P (O )  20, 000 / 344, 039  .060; P ( PT )  10, 072 / 344, 039  .030; P ( PC )  20, 000 / 344, 039  .060; P ( S )  67, 300 / 344, 039  .201; P (T )  12, 415 / 344, 039  .037; P (U )  3, 000 / 344, 039  .009

a.

P ( F or S )  P ( F )  P ( S )  0.356  0.201  0.557

b.

Define the event: D: {PIN pad is defective} P ( D )  1000 / 334, 039  .003

3.19

a.

The probability that any network is selected on a particular day is 1/8. Therefore, P( ESPN selected on July 11) = 1/8.

b.

The number of ways to select four networks for the weekend days is a combination of 8 networks 8  8! 8  7  6  5  4  3  2 1 taken 4 at a time. The number of ways to do this is      70 .  4  4!(8  4)! 4  3  2  1  4  3  2  1

c.

First, we need to find the number of ways one can choose the 4 networks where ESPN is one of the 4. If ESPN has to be chosen, then the number of ways of doing this is a combination of one thing taken 1 1! 1 one at a time or      1 . The number of ways to select the remaining 3 networks is a  1  1!(1  1)! 1  1

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Probability

105

7 7! 7  6  5  4  3  2 1 combination of 7 things taken 3 at a time or      35 . Thus, the total  3  3!(7  3)! 3  2  1  4  3  2  1 number of ways of selecting 4 networks of which one has to be ESPN is 1(35) = 35.

Finally, the probability of selecting ESPN as one of the 4 networks for the weekend analysis is 35 / 70  .5 . 3.20

a.

Since order does not matter, the number of different bets would be a combination of 8 things taken 2 at 8 

a time. The number of ways would be    2  

b.

3.21

8! 8  7  6  5  4  3  2  1 40,320    28 . 2!(8  2)! 2  1  6  5  4  3  2  1 1440

If all players are of equal ability, then each of the 28 sample points would be equally likely. Each would have a probability of occurring of 1/28. There is only one sample point with values 2 and 7. Thus, the probability of winning with a bet of 2-7 would be 1/28 or .0357.

Since one would be selecting 3 stocks from 15 without replacement, the total number of ways to select the 3 stocks would be a combination of 15 things taken 3 at a time. The number of ways would be  15  15! 15  14  13   3  2  1 1.307674368  1012    455   2874009600  3  3!(15  3)! 3  2  1  12  11  10   3  2  1

3.22

Denote Pu = public, Pr = private, B = bedrocks, U = unconsolidated, BL = below limit, D = detect. a.

The 8 sample points for this experiment in which the well class (public or private), aquifer (bedrocks or unconsolidated) and detectible (below limit or detect) MTBE level of a well are observed are as follows: (Pu, B, BL) (Pu, U, BL)

b.

c.

(Pr, B, BL) (Pu, B, D) (Pr, U, BL) (Pu, U, D)

(Pr, B, D) (Pr, U, D)

P ( Pu , B , BL )  57 / 223  0.256

P ( Pr , B , BL )  81 / 223  0.363

P ( Pu , B , D )  41 / 223  0.184

P ( Pr , B , D )  22 / 223  0.099

P ( Pu , U , BL )  15 / 223  0.067

P ( Pr , U , BL )  0 / 223  0.000

P ( Pu , U , D )  7 / 223  0.031

P ( Pr , U , D )  0 / 223  0.000

Define the following event: D = {Well has a detectible level or MTBE} P ( D )  P ( Pu , B , D )  P ( Pu , U , D )  P ( Pr , B , D )  P ( Pr , U , D )  0.184  0.031  0.099  0  0.314

This means that if one well is chosen at random, the probability that it has a detectible level of MTBE is .314. 3.23

a.

Since we want to maximize the purchase of grill #2, grill #2 must be one of the 3 grills in the display. Thus, we have to pick 2 more grills from the 4 remaining grills. Since order does not matter, the number of different ways to select 2 grill displays from 4 would be a combination of 4 things taken 2 at a time. The number of ways is:  4 4! 4  3  2  1 24   6    2  2!(4  2)! 2  1  2  1 4

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106

Chapter 3

Let Gi represent Grill i. The possibilities are: G1G2G3, G1G2G4, G1G2G5, G2G3G4, G2G3 G5, G2G4G5 b.

c. 3.24

a.

To find reasonable probabilities for the 6 possibilities, we divide the frequencies by the total sample size of 124. The probabilities would be: P (G1G 2 G3 )  35 / 124  .282

P (G1G 2 G 4 )  8 / 124  .065

P (G1G 2 G5 )  42 / 124  .339

P (G 2 G3 G 4 )  4 / 124  .032

P (G 2 G3 G5 )  1 / 124  .008

P (G2 G4 G5 )  34 / 124  .274

P( display contained Grill #1)  P (G1G 2 G3 )  P (G1G2 G 4 )  P (G1G2 G5 )  .282  .065  .339  .686 Let H = Hyundai Elantra, T = Toyota Prius, and S = Subaru Forrester. All possible rankings are as follows, where the first car listed is ranked first, the second car listed is ranked second, and the third car listed is ranked third: H,T,S

b.

H,S,T

S,H,T

S,T,H

T,H,S

T,S,H

If each set of rankings is equally likely, then each has a probability of 1/6. The probability that the Toyota Prius is ranked first  P (T , H , S )  P (T , S , H )  1 / 6  1 / 6  2 / 6  1 / 3

The probability that the Hyundai Elantra is ranked third  P ( S , T , H )  P (T , S , H )  1 / 6  1 / 6  2 / 6  1 / 3 . The probability that the Toyota Prius is ranked first and the Subaru Forrester is ranked second  P (T , S , H )  1 / 6 . 3.25

1 3

1 2  or 1 to 2. 3 3

a.

The odds in favor of an Oxford Shoes win are to 1 

b.

If the odds in favor of Oxford Shoes are 1 to 1, then the probability that Oxford Shoes wins is 1 1  . 11 2

c.

If the odds against Oxford Shoes are 3 to 2, then the odds in favor of Oxford Shoes are 2 to 3. Therefore, the probability that Oxford Shoes wins is

3.26

2 2  . 23 5

First, we need to compute the total number of ways we can select 2 bullets (pair) from 1,837 bullets. This is a combination of 1,837 things taken 2 at a time.  1,837 

The number of pairs is:      2  2!(1,837  2)! 2  1  1835  1834    1 1,837!

1837  1836    1

1837  1836  1,686,366 2

The probability of a false positive is the number of false positives divided by the number of pairs and is: P(false positive) = # false positives / # pairs  693 / 1,686,366  .0004 This probability is very small. There would be only about 4 false positives out of every 10,000. I would have confidence in the FBI’s forensic evidence. Copyright © 2014 Pearson Education, Inc.

Probability

3.27

a.

The number of ways the 5 commissioners can vote is 2(2)(2)(2)(2) = 25 = 32 (Each of the 5 commissioners has 2 choices for his/her vote – For or Against.)

b.

Let F denote a vote ‘For’ and A denote a vote ‘Against’. The 32 sample points would be: FFFFF FFAAF AFFAA AAAAF

107

FFFFA FFFAF FFAFF FAFFF AFFFF FFFAA FFAFA FAFFA AFFFA FAFAF AFFAF FAAFF AFAFF AAFFF FFAAA FAFAA FAAFA FAAAF AFAFA AFAAF AAFFA AAFAF AAAFF FAAAA AFAAA AAFAA AAAFA AAAAA

Each of the sample points should be equally likely. Thus, each would have a probability of 1/32. c.

The sample points that result in a 2-2 split for the other 4 commissioners are: FFAAF FAFAF AFFAF FAAFF AFAFF AAFFF FFAAA FAFAA FAAFA AFFAA AFAFA AAFFA There are 12 sample points.

d.

Let V = event that your vote counts. P (V )  12 / 32  0.375 .

e.

If there are now only 3 commissioners in the bloc, then the total number of ways the bloc can vote is 2(2)(2)  23  8 . The sample points would be: FFF

FFA

FAF

AFF

FAA

AFA

AAF

AAA

The number of sample points where your vote would count is 4: FAF, AFF, FAA, AFA Let W = event that your vote counts in the bloc. P (W )  4 / 8  0.5 . 3.28

3.29

a.

P ( B c )  1  P ( B )  1  .7  .3

b.

P ( Ac )  1  P ( A)  1  .4  .6

c.

P ( A  B )  P ( A)  P ( B )  P ( A  B )  .4  .7  .3  .8

a.

A: {HHH, HHT, HTH, THH, TTH, THT, HTT} B: {HHH, TTH, THT, HTT} A  B : {HHH, HHT, HTH, THH, TTH, THT, HTT} Ac: {TTT} A  B : {HHH, TTH, THT, HTT}

b.

P ( A) 

c.

P ( A  B )  P ( A)  P ( B )  P ( A  B ) 

d.

No. P ( A  B ) 

7 8

P(B) 

4 1  8 2

P( A  B) 

7 8

P ( Ac ) 

1 8

7 1 1 7    8 2 2 8

1 which is not 0. 2

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P( A  B) 

4 1  8 2

108

3.30

Chapter 3

The experiment consists of rolling a pair of fair dice. The sample points are: 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6

2, 1 2, 2 2, 3 2, 4 2, 5 2, 6

3, 1 3, 2 3, 3 3, 4 3, 5 3, 6

4, 1 4, 2 4, 3 4, 4 4, 5 4, 6

5, 1 5, 2 5, 3 5, 4 5, 5 5, 6

6, 1 6, 2 6, 3 6, 4 6, 5 6, 6

Since each die is fair, each sample point is equally likely. The probability of each sample point is 1/36. a.

A: {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} B: {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)}

A  B : {(3, 4), (4, 3)} A  B : {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (1, 6), (2, 5), (5, 2), (6, 1)} Ac: {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3), (3, 5), (3, 6), (4, 1), (4, 2), (4, 4), (4, 5), (4, 6), (5, 1), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} b.

 1  6 1 P ( A)  6      36  36 6

 1  11 P ( B )  11     36  36

 1  15 5 P ( A  B )  15      36  36 12

 1  30 5 P ( Ac )  30      36  36 6 1 11 1 6  11  2 15 5      6 36 18 36 36 12

c.

P ( A  B )  P ( A)  P ( B )  P ( A  B ) 

d.

A and B are not mutually exclusive. To be mutually exclusive, P ( A  B ) must be 0. Here, P( A  B) 

3.31

1  1  2 P( A  B)  2      36  36 18

1 . 18 1 1 1 1 1 15 3       5 5 5 20 10 20 4

a.

P ( A)  P ( E1 )  P ( E2 )  P ( E3 )  P ( E5 )  P ( E6 ) 

b.

P ( B )  P ( E2 )  P ( E3 )  P ( E4 )  P ( E7 ) 

c.

P ( A  B )  P ( E1 )  P ( E 2 )  P ( E3 )  P ( E 4 )  P ( E5 )  P ( E 6 )  P ( E 7 ) 

1 1 1 1 13     5 5 20 5 20

1 1 1 1 1 1 1       1 5 5 5 20 20 10 5

d.

P ( A  B )  P ( E2 )  P ( E3 ) 

e.

P ( Ac )  1  P ( A)  1 

1 1 2   5 5 5

3 1  4 4

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Probability

3.32

3.33

13 7  20 20

f.

P( B c )  1  P( B)  1 

g.

P ( A  Ac )  P ( E1 )  P ( E2 )  P ( E3 )  P ( E4 )  P ( E5 )  P ( E6 )  P ( E7 )



109

1 1 1 1 1 1 1       1 5 5 5 20 20 10 5 1 1 5 1    20 5 20 4

h.

P ( Ac  B )  P ( E4 )  P ( E7 ) 

a.

P ( Ac )  P ( E3 )  P ( E6 )  .2  .3  .5

b.

P ( B c )  P ( E1 )  P ( E7 )  .10  .06  .16

c.

P ( Ac  B )  P ( E3 )  P ( E6 )  .2  .3  .5

d.

P ( A  B )  P ( E1 )  P ( E 2 )  P ( E3 )  P ( E 4 )  P ( E5 )  P ( E 6 )  P ( E 7 )  .10  .05  .20  .20  .06  .30  .06  .97

e.

P ( A  B )  P ( E 2 )  P ( E 4 )  P ( E5 )  .05  .20  .06  .31

f.

P ( Ac  B c )  P ( E8 )  .03

g.

No. A and B are mutually exclusive if P ( A  B )  0 . Here, P ( A  B )  .31 .

a.

P ( A)  .50  .10  .05  .65

b.

P ( B )  .10  .07  .50  .05  .72

c.

P (C )  .25

d.

P ( D )  .05  .03  .08

e.

P ( Ac )  .25  .07  .03  .35 (Note: P ( Ac )  1  P ( A)  1  .65  .35 )

f.

P ( A  B )  P ( B )  .10  .07  .50  .05  .72

g.

P( A  C )  0

h.

Two events are mutually exclusive if they have no sample points in common or if the probability of their intersection is 0. P ( A  B )  P ( A)  .50  .10  .05  .65 . Since this is not 0, A and B are not mutually exclusive. P ( A  C )  0 . Since this is 0, A and C are mutually exclusive. P ( A  D )  .05 . Since this is not 0, A and D are not mutually exclusive. P ( B  C )  0 . Since this is 0, B and C are mutually exclusive.

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110

Chapter 3 P ( B  D )  .05 . Since this is not 0, B and D are not mutually exclusive. P (C  D )  0 . Since this is 0, C and D are mutually exclusive.

3.34

3.35

3.36

a.

The outcome "On" and "High" is A  D .

b.

The outcome "Low" or "Medium" is Dc.

a.

The analyst makes an early forecast and is only concerned with accuracy is the event ( A  B ) .

b.

The analyst is not only concerned with accuracy is the event Ac.

c.

The analyst is from a small brokerage firm or makes an early forecast is the event C  B .

d.

The analyst makes a late forecast and is not only concerned with accuracy is the event B c  Ac .

Define the following events: A: {problems with absenteeism} T: {problems with turnover} From the problem, P ( A)  .55, P (T )  .41 , and P ( A  T )  .22 P(problems with either absenteeism or turnover)  P ( A  T )  P ( A)  P (T )  P ( A  T )  .55  .41  .22  .74

3.37

a.

Define the following events: L: {Legs only} W: {Wheels only} B: {Both legs and wheels} N: {Neither legs nor wheels} The sample points are: L, W, B, and N

b.

From the given data: P( L) 

3.38

63  .594 106

P (W ) 

20  .189 106

P(B) 

8  .075 106

c.

P (Wheels)  P (W or B )  P (W )  P ( B )  .189  .075  .264

d.

P (Legs)  P ( L or B )  P ( L )  P ( B )  .594  .075  .669

e.

P (Either legs or wheels)  1  P ( N )  1  .142  .858

P( N ) 

Define the following event: A: {Store violates the NIST scanner accuracy standard} Then P ( Ac )  1  P ( A)  1  52 / 60  8 / 60  .133

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15  .142 106

Probability

3.39

Define the following events: A: {oil structure is active} I: {oil structure is inactive} C: {oil structure is caisson} W: {oil structure is well protector} F: {oil structure is fixed platform} a.

The simple events are all combinations of structure type and activity type. The simple events are: AC, AW, AF, IC, IW, IF

b.

Reasonable probabilities would be the frequency divided by the sample size of 3,400. The probabilities are: P ( AC )  503 / 3, 400  .148

3.40

P ( AW )  225 / 3, 400  .066

P ( AF )  1, 447 / 3, 400  .426

P ( IC )  598 / 3, 400  .176

P ( IW )  177 / 3, 400  .052

P ( IF )  450 / 3, 400  .132

c.

P ( A)  P ( AC )  P ( AW )  P ( AF )  .148  .066  .426  .640

d.

P (W )  P ( AW )  P ( IW )  .066  .052  .118

e.

P ( IC )  .176

f.

P ( I  F )  P ( IC )  P ( IW )  P ( IF )  P ( AF )  .176  .052  .132  .426  .786

g.

P(C c )  1  P(C )  1   P ( AC )  P( IC )   1  .148  .176   1  .324  .676

Define the following events: M: {UK citizen visits MySpace} B: {UK citizen visits Bebo}

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111

112

a.

Chapter 3

The Venn Diagram that illustrates the use of social networking sites in UK is:

M

M∩B

B

1%

3%

4%

3.41

b.

P ( M  B )  P ( M )  P ( B )  P ( M  B )  0.04  0.03  0.01  0.06

c.

P ( M c  B c )  1  P ( M  B )  1  0.06  0.94

First, define the following events: F: P: N: R:

{Fully compensated} {Partially compensated} {Non-compensated} {Left because of retirement}

From the text, we know 127 45 72 7  11  10 28 , P( P)  , P( N )  , and P( R)  P( F )   244 244 244 244 244

3.42

127 244

a.

P( F ) 

b.

P( F  R) 

c.

P( F c )  1  P( F )  1 

d.

P( F  R)  P( F )  P( R)  P( F  R) 

7 244 127 117  244 244 127 28 7 148    244 244 244 244

Define the following events: I: {Invests in Market} N: {No investment} a.

P(I ) 

44, 651  .283 158, 044

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Probability

3.43

31, 943  17, 958  12,145  9, 531 71, 577   .453 158, 044 158, 044

b.

P (IQ  6) 

c.

P ( I  {IQ  6}) 

d.

P ( I  {IQ  6})  P ( I )  P (IQ  6)  P ( I  {IQ  6})  .283  .453  .168  .568

e.

P ( I c )  1  P ( I )  1  .283  .717

f.

Two events are mutually exclusive if the probability of their intersection is 0. 893 P ( I  {IQ  1})   .006 . Since this value is not 0, these two events are not mutually 158, 044 exclusive.

10, 270  6, 698  5,135  4, 464 26, 567   .168 158, 044 158, 044

a.

P  S  A . Products 6 and 7 are contained in this intersection.

b.

P(possess all the desired characteristics)  P( P  S  A)  P(6)  P(7) 

c.

A S P ( A  S )  P (2)  P (3)  P (5)  P (6)  P (7)  P (8)  P (9)  P (10) 

d.

1 1 1 1 1 1 1 1 8 4          10 10 10 10 10 10 10 10 10 5

PS P ( P  S )  P (2)  P (6)  P (7) 

3.44

a.

1 1 1 3    10 10 10 10

Define the following events: G: {Student is assigned to the guilty state} C: {Student chooses the stated option} Then P (G )  57 / 171  .333 .

3.45

b.

P (C )  60 / 171  .351

c.

P (G  C )  45 / 171  .263

d.

P (G  C )  P (G )  P (C )  P (G  C )  .333  .351  .263  .421

Define the following events: M1: {Model 1} M2: {Model 2} a.

P (5) 

85  .531 160

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1 1 1   10 10 5

113

114

3.46

Chapter 3

b.

P (5  0)  P (5)  P (0)  P (5  0)  .531 

c.

P ( M 2  0) 

35  0  .531  .219  .75 160

15  .094 160

Define the following events: A: {Individual tax return is audited by the IRS} B: {Corporation tax return is audited by the IRS}

3.47

1, 581, 394  .0111 142,823,105

a.

P ( A) 

b.

P ( Ac )  1  P ( A)  1  .0111  .9889

c.

P( B) 

d.

P ( B c )  1  P ( B )  1  .0139  .9861

29,803  .0139 2,143,808

Define the following events: A: {Air pressure is over-reported by 4 psi or more} B: {Air pressure is over-reported by 6 psi or more} C: {Air pressure is over-reported by 8 psi or more

3.48

a.

For gas station air pressure gauges that read 35 psi, P ( B )  .09 .

b.

For gas station air pressure gauges that read 55 psi, P (C )  .09 .

c.

For gas station air pressure gauges that read 25 psi, P ( Ac )  1  P ( A)  1  .16  .84 .

d.

No. If air pressure is over-reported by 6 psi or more, then it is also over-reported by 4 psi or more. Thus, these 2 events are not mutually exclusive.

e.

The columns in the table are not mutually exclusive. All events in the last column (% Over-reported by 8 psi or more) are also part of the events in the first and second columns. All events in the second column are also part of the events in the first column. In addition, there is no column for the event ‘Over-reported by less than 4 psi or not over-reported’.

There are a total of 6  6  6  216 possible outcomes from throwing 3 fair dice. To help demonstrate this, suppose the three dice are different colors – red, blue and green. When we roll these dice, we will record the outcome of the red die first, the blue die second, and the green die third. Thus, there are 6 possible outcomes for the first position, 6 for the second, and 6 for the third. This leads to the 216 possible outcomes.

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Probability

115

The Grand Duke argued that the chance of getting a sum of 9 and the chance of getting a sum of 10 should be the same since the number of partitions for 9 and 10 are the same. These partitions are: 9 126 135 144 225 234 333

10 136 145 226 235 244 334

In each case, there are 6 partitions. However, if we take into account the three colors of the dice, then there are various ways to get each partition. For instance, to get a partition of 126, we could get 126, 162, 216, 261, 612, and 621 (again, think of the red die first, the blue die second, and the green die third). However, to get a partition of 333, there is only 1 way. To get a partition of 144, there are 3 ways: 144, 414, and 441. The numbers of ways to get each of the above partitions are: 9 126 135 144 225 234 333

# ways 6 6 3 3 6 _ 1 25

10 136 145 226 235 244 334

# ways 6 6 3 6 3 _3 27

Thus, there are a total of 25 ways to get a sum of 9 and 27 ways to get a sum of 10. The chance of throwing a sum of 9 (25 chances out of 216 possibilities) is less than the chance of throwing a 10 (27 chances out of 216 possibilities). 3.49

3.50

3.51

3.52

a.

P( A | B) 

P ( A  B ) .1   .5 .2 P( B)

b.

P ( B | A) 

P ( A  B ) .1   .25 .4 P ( A)

c.

Events A and B are said to be independent if P ( A | B )  P ( A) . In this case, P ( A | B )  .5 and P ( A)  .4 . Thus, A and B are not independent.

a.

P ( A  B )  P ( A | B ) P ( B )  .6(.2)  .12

b.

P ( B | A) 

a.

If two events are independent, then P ( A  B )  P ( A) P ( B )  .4(.2)  .08 .

b.

If two events are independent, then P ( A | B )  P ( A)  .4 .

c.

P ( A  B )  P ( A)  P ( B )  P ( A  B )  .4  .2  .08  .52

a.

Since A and B are mutually exclusive events, P ( A  B )  P ( A)  P ( B )  .30  .55  .85

P ( A  B ) .12   .3 P ( A) .4

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116

3.53

Chapter 3

b.

Since A and C are mutually exclusive events, P ( A  C )  0

c.

P( A | B) 

d.

Since B and C are mutually exclusive events, P ( B  C )  P ( B )  P (C )  .55  .15  .70

e.

No, B and C cannot be independent events because they are mutually exclusive events.

a.

P ( A)  P ( E1 )  P ( E 2 )  P ( E3 )  .2  .3  .3  .8

P( A  B) 0  0 .55 P( B)

P ( B )  P ( E 2 )  P ( E3 )  P ( E5 )  .3  .3  .1  .7 P ( A  B )  P ( E 2 )  P ( E3 )  .3  .3  .6

b.

P ( E1 | A) 

P ( E 1  A) P ( E 1) .2    .25 P ( A) P ( A) .8

P ( E2 | A) 

P ( E 2  A) P ( E 2) .3    .375 P ( A) P ( A) .8

P ( E3 | A) 

P ( E 3  A) P ( E 3) .3    .375 P ( A) P ( A) .8

The original sample point probabilities are in the proportion .2 to .3 to .3 or 2 to 3 to 3. The conditional probabilities for these sample points are in the proportion .25 to .375 to .375 or 2 to 3 to 3. c.

(1)

P ( B | A)  P ( E 2 | A)  P ( E3 | A)  .375  .375  .75 (from part b)

(2)

P ( B | A) 

P ( A  B ) .6   .75 (from part a) .8 P ( A)

The two methods do yield the same result.

3.54

d.

If A and B are independent events, P ( B | A)  P ( B ) . From part c, P ( B | A)  .75 . From part a, P ( B )  .7 . Since .75  .7 , A and B are not independent events.

a.

If two fair coins are tossed, there are 4 possible outcomes or simple events. They are: E1 = HH

E2 = HT E3 = TH E4 = TT

Event A contains the simple events E1, E2, and E3. Event B contains the simple events E2 and E3.

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Probability

117

A Venn diagram of this would be:

A

B E2 E3

E1

E4

Since the coins are fair, each of the sample points is equally likely. Each would have probabilities of ¼. b.

1 3 P ( A)  3     .75 4 4

P ( A  B )  P ( E2 )P ( E3 ) 

3.55

1 2 1 P ( B )  2      .5 4 4 2

1 1 2 1     .5 4 4 4 2

P ( A  B ) .5  1 .5 P( B)

P ( B | A) 

P ( A  B ) .5   .667 .75 P ( A)

c.

P( A | B) 

a.

P ( A)  P ( E1 )  P ( E3 )  .22  .15  .37

b.

P ( B )  P ( E 2 )  P ( E3 )  P ( E 4 )  .31  .15  .22  .68

c.

P ( A  B )  P ( E3 )  .15

d.

P( A | B) 

e.

P(B  C )  0

f.

P (C | B ) 

g.

For pair A and B: A and B are not independent because P ( A | B )  P ( A) or .2206  .37 .

P ( A  B ) .15   .2206 P( B) .68

P (C  B ) 0  0 .68 P( B)

For pair A and C: P ( A  C )  P ( E1 )  .22 P( A | C ) 

P (C )  P ( E1 )  P ( E5 )  .22  .10  .32

P ( A  C ) .22   .6875 .32 P (C )

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Chapter 3

A and C are not independent because P ( A | C )  P ( A) or .6875  .37 .

For pair B and C: B and C are not independent because P (C | B )  P (C ) or 0  .32 . 3.56

The 36 possible outcomes obtained when tossing two dice are listed below: (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) A: {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)} B: {(3, 6), (4, 5), (5, 4), (5, 6), (6, 3), (6, 5), (6, 6)}

A  B : {(3, 6), (4, 5), (5, 4), (5, 6), (6, 3), (6, 5)} If A and B are independent, then P ( A) P ( B )  P ( A  B ) . P ( A) 

18 1  36 2

P ( A) P ( B ) 

3.57

a.

P( B) 

7 36

P( A  B) 

6 1  36 6

1 7 7 1     P ( A  B ) . Thus, A and B are not independent. 2 36 72 6

P ( A  C )  0  A and C are mutually exclusive. P ( B  C )  0  B and C are mutually exclusive.

b.

P ( A)  P (1)  P (2)  P (3)  .20  .05  .30  .55

P ( B )  P (3)  P (4)  .30  .10  .40

P (C )  P (5)  P (6)  .10  .25  .35

P ( A  B )  P (3)  .30

P( A | B) 

P ( A  B ) .30   .75 .40 P( B)

A and B are independent if P ( A | B )  P ( A) . Since P ( A | B )  .75 and P ( A)  .55 , A and B are not independent.

Since A and C are mutually exclusive, they are not independent. Similarly, since B and C are mutually exclusive, they are not independent. c.

Using the probabilities of sample points, P ( A  B )  P (1)  P (2)  P (3)  P (4)  .20  .05  .30  .10  .65

Using the additive rule, P ( A  B )  P ( A)  P ( B )  P ( A  B )  .55  .40  .30  .65

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Probability

Using the probabilities of sample points, P ( A  C )  P (1)  P (2)  P (3)  P (5)  P (6)  .20  .05  .30  .10  .25  .90

Using the additive rule, P ( A  C )  P ( A)  P (C )  P ( A  C )  .55  .35  0  .90

3.58

3.59

From the Exercise, P ( A)  .15 , P ( B )  .10 , and P ( A  B )  .05 . a.

If events A and B are mutually exclusive then P ( A  B )  0 . For this problem, P ( A  B )  .05 . Therefore, events A and B are not mutually exclusive.

b.

P ( B | A) 

c.

Events A and B are independent if P ( B | A)  P ( B ) . For this exercise, P ( B | A)  .333 and P ( B )  .10 . Since these are not equal, events A and B are not independent.

P ( A  B ) .05   .333 .15 P ( A)

Define the following events: A: {Company is a banking/investment company} B: {Company is based in United States} From the problem, we know that P ( A  B )  P( A | B) 

3.60

4 9  .20 and P ( B )   .45 20 20

P ( A  B ) .20   .444 . P( B) .45

Define the following events: G: {The respondent is assigned to the guilt state} A: {The respondent is assigned to the anger state} C: {The respondent chooses the stated option to repair car}

a.

From Exercise 3.44, we know P (G )  57 / 171  .333 and P (G  C )  45 / 171  .263 P (C | G ) 

b.

P (G  C ) .263   .790 .333 P (G )

From Exercise 3.44, we know P (C )  60 / 171  .351 . Thus, P (C c )  1  .351  .649 P( A | C c ) 

P ( A  C c )  50 / 171  .292

P ( A  C c ) .292   .450 .649 P (C c )

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Chapter 3

c.

Two events C and G are independent if P (C  G )  P (C ) P (G ) . From Exercise 3.44, P (G )  .333 , P (C )  .351 , and P (G  C )  .263 . P (G ) P (C )  .333(.351)  .117  .263  P (G  C ) .

Thus C and G are not independent. 3.61

Define the following events: A: {Internet user has wireless connection via mobile device} B: {Internet user uses Twitter}

From the exercise, P ( A )  .54 and P ( B | A)  .25 . P ( A  B )  P ( B | A) P ( A)  .25(.54)  .135

3.62

Define the following events: A: {Person is victim of identity theft} B: {Theft occurred from unauthorized use of credit card}

From the exercise, P ( A)  .05 and P ( B | A)  .53

3.63

a.

P ( A)  .05

b.

P ( A  B )  P ( B | A) P ( A)  .05(.53)  .0265

Define the following events: F: P: N: R:

{Worker is fully compensated} {Worker is partially compensated} {Worker is non-compensated} {Worker retired}

From the exercise, P ( F )  127 / 244  .520 , P ( P )  45 / 244  .184 , P ( R | F )  7 / 127  .055 , P ( R | P )  11 / 45  .244 , and P ( R | N )  10 / 72  .139 . a.

P ( R | F )  7 / 127  .055

b.

P ( R | N )  10 / 72  .139

c.

The two events are independent if P ( R | F )  P ( R ) . 7  11  10 28   .115 and P ( R | F )  10 / 72  .055 . Since these are not equal, events R 244 244 and F are not independent. P( R) 

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Probability

3.64.

121

Define the following events: A: B: C: D:

{Respondent works during summer vacation} {Respondent does not work during summer vacation} {Respondent unemployed} {Respondent monitors business emails}

From Exercise 3.14: P ( A )  .46 , P ( B )  .35 , P (C )  .19 . From this exercise, P ( D | A)  .35 .

3.65

a.

P ( D | A)  .35

b.

P ( A  D )  P ( D | A) P ( A)  .35(.46)  .161

c.

P ( B  D )  0 (If an employee is not working, then he/she will not monitor business emails.)

Define the following events: I: {Invests in Market} N: {No investment}

a.

10, 270  6, 698  5,135  4, 464 P ( I  {IQ  6}) 26,567 158, 044    .371 P ( I | IQ  6)  31,943  17,958  12,145  9,531 71,577 P (IQ  6) 158, 044

b.

44, 651  26,567 P( I  {IQ  5}) 18, 084 158, 044    .209 . P( I | IQ  5)  158, 044  71,577 86, 467 P(IQ  5) 158, 044

c.

3.66

Yes, it appears that investing in the stock market is dependent on IQ. If investing in the stock market and IQ were independent, then P ( I | IQ  5)  P ( I | IQ  6)  P ( I ) . Since P ( I | IQ  5)  P ( I | IQ  6) , then investing in the stock market and IQ are dependent.

Define the following events: th Ai : {i CEO has bachelor’s degree}

a. b.

P ( A1 ) 

13  .325 40

If the first 4 CEO’s have just bachelor’s degree, then on the next pick there are only 9 left to choose from. Similarly, after picking 4 CEO’s, there are only 36 observations left to choose from. P ( A5 | A1  A2  A3  A4 ) 

9  .25 36

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3.67

Chapter 3

Define the following events: A: {Ambulance can travel to location A under 8 minutes} B: {Ambulance can travel to location B under 8 minutes} C: {Ambulance is busy}

We are given P ( A)  .58 , P ( B )  .42 , and P (C )  .3 .

3.68

a.

P ( A  C c )  P ( A | C c ) P (C c )  .58(1  .3)  .406

b.

P ( B | C c ) P (C c )  .42(1  .3)  .294

If A and B are independent, then P ( A  B )  P ( A) P ( B ) . For this Exercise, P ( A) 

1174  416 1590 1174  89 1263   .883 , P ( B )    .702 , and 1800 1800 1800 1800

P( A  B) 

1174  .652 . 1800

P ( A) P ( B )  .883(.702)  .620  .652  P ( A  B ) . Thus, A and B are not independent.

3.69

Define the following events: A: {Alarm A sounds alarm} B: {Alarm B sounds alarm} I: {Intruder}

a.

From the problem P  A | I   .9, P  B | I   .95, P( A | I c )  .2 and P( B | I c )  .1 .

b.

Since the two systems are operating independently of each other, P ( A  B | I )  P ( A | I ) P ( B | I )  .9(.95)  .855

3.70

c.

P ( A  B | I c )  P ( A | I c ) P ( B | I c )  .2(.1)  .02

d.

P ( A  B | I )  P ( A | I )  P ( B | I )  P ( A  B | I )  .9  .95  .855  .995

a.

Since there are 2 vineyards and 3 years, there are a total of 2(3) = 6 combinations.

b.

Of the 6 combinations, 3 of them are from the Llarga vineyard. Thus, P (Llarga)  3 / 6  .5 .

c.

Of the 6 combinations, 2 of them are Year 3. Thus, P (Year 3)  2 / 6  .333

d.

If the tasters are independent, then the probability that each selects Llarga is P (Llarga) P (Llarga) P (Llarga) P (Llarga)  .5(.5)(.5)(.5)  .0625 .

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Probability

3.71

123

Define the following event: A: {The specimen labeled “red snapper” was really red snapper}

a.

The probability that you are actually served red snapper the next time you order it at a restaurant is P ( A)  1  .77  .23

b.

P(at least one customer is actually served red snapper) = 1 – P(no customer is actually served red snapper)  1  P ( A c  A c  A c  Ac  Ac )  1  P ( A c ) P ( Ac ) P ( A c ) P ( A c ) P ( Ac )  1  .775  1  .271  .729

Note: In order to compute the above probability, we had to assume that the trials or events are independent. This assumption is likely to not be valid. If a restaurant served one customer a look-alike variety, then it probably served the next one a look-a-like variety. 3.72

First, define the following event: A: {CVSA correctly determines the veracity of a suspect} P(A) = .98 (from claim) a.

The event that the CVSA is correct for all four suspects is the event A  A  A  A . P ( A  A  A  A)  .98(.98)(.98)(.98)(.98)  .9224

b.

The event that the CVSA is incorrect for at least one of the four suspects is the event ( A  A  A  A) c . P ( A  A  A  A) c  1  P ( A  A  A  A)  1  .9224  .0776

c. If the CVSA had an accuracy of .98, then the probability of observing 2 incorrect results is less than .0776. Since 2 incorrect results were observed, it was either a rare event or the accuracy of the CVSA is not .98 but something less than .98. 3.73

Define the following events: A: {Patient receives PMI sheet} B: {Patient was hospitalized} P ( A )  .20 ,

3.74

P ( A  B )  .12 ,

P ( B | A) 

P ( A  B ) .12   .60 .20 P ( A)

Define the following events: I: {Leak ignites immediately (jet fire)} D: {Leak has delayed ignition (flash fire)}

From the problem, P ( I )  .01 and P ( D | I c )  .01 The probability of a jet fire or a flash fire  P( I  D)  P( I )  P( D)  P( I  D)  P ( I )  P ( D | I c ) P ( I c )  P ( I  D )  .01  .01(1  .01)  0  .01  .0099  .0199

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Chapter 3

A tree diagram of this problem is: I .01

I .01

D(.01)

.99

Ic

Dc (.99)

3.75

a.

IcD .99(.01)=.0099

IcDc .99(.99)=.9801

If the coin is balanced, then P ( H )  .5 and P (T )  .5 on any trial. Also, we can assume that the results of any coin toss is independent of any other. Thus,

P( H  H  H  H  H  H  H  H  H  H )  P( H ) P( H ) P( H ) P( H ) P( H ) P( H ) P( H ) P( H ) P( H ) P( H )  .5(.5)(.5)(.5)(.5)(.5)(.5)(.5)(.5)(.5)  .510  .0009766 P( H  H  T  T  H  T  T  H  H  H )  P( H ) P( H ) P(T ) P(T ) P( H ) P(T ) P(T ) P( H ) P( H ) P( H )  .5(.5)(.5)(.5)(.5)(.5)(.5)(.5)(.5)(.5)  .510  .0009766 P(T  T  T  T  T  T  T  T  T  T )  P(T ) P(T ) P(T ) P(T ) P(T ) P(T ) P(T ) P(T ) P(T ) P(T )  .5(.5)(.5)(.5)(.5)(.5)(.5)(.5)(.5)(.5)  .510  .0009766 b.

Define the following events: A: {10 coin tosses result in all heads or all tails} B: {10 coin tosses result in mix of heads and tails} P( A)  P( H  H  H  H  H  H  H  H  H  H )  P(T  T  T  T  T  T  T  T  T  T )  .0009766  .0009766  .0019532

c. d.

3.76

P ( B )  1  P ( A)  1  .0019532  .9980468

From the above probabilities, the chances that either all heads or all tails occurred is extremely rare. Thus, if one of these sequences really occurred, it is most likely sequence #2.

Define the following events: A: {Algorithm predicts defects} B: {Module has defects} C: {Algorithm is correct}

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Probability

a.

Accuracy  P (C )  P ( A  B )  P ( Ac  B c ) 

b.

Detection rate  P ( A | B ) 

d bd

c.

False alarm  P ( A | B c ) 

c ac

d a ad   abcd abcd abcd

d cd

d.

Precision  P ( B | A) 

e.

From the SWDEFECTS file the table is: Module has Defects

Algorithm Predicts Defects

False

True

No

400

29

Yes

49

20

Accuracy  P (C )  P ( A  B )  P ( Ac  B c ) 

20  400 420 d a d a      .843 a  b  c  d a  b  c  d a  b  c  d 400  29  49  20 498

The probability that the algorithm is correct is .843. Detection rate  P ( A | B ) 

20 20 d    .408 b  d 29  20 49

The probability that the algorithm predicts a defect given the module is actually defective is .408. False alarm  P ( A | B c ) 

49 49 c    .109 a  c 400  49 449

The probability that the algorithm predicts a defect given the module is not defective is .109. Precision  P ( B | A) 

20 20 d    .290 c  d 49  20 69

The probability that the module is defective given the algorithm predicted a defect is .290. 3.77

a.

P ( B1  A)  P ( A | B1 ) P ( B1 )  .3(.75)  .225

b.

P ( B2  A)  P ( A | B2 ) P ( B2 )  .5(.25)  .125

c.

P ( A)  P ( B1  A)  P ( B2  A)  .225  .125  .35

d.

P ( B1 | A) 

P ( B1  A) .225   .643 .35 P ( A)

e.

P ( B2 | A) 

P ( B2  A) .125   .357 .35 P ( A)

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Chapter 3

First, we find the following probabilities: P ( A  B1 )  P ( A | B1 ) P ( B1 )  .4(.2)  .08 P ( A  B2 )  P ( A | B2 ) P ( B2 )  .25(.15)  .0375 P ( A  B3 )  P ( A | B3 ) P ( B3 )  .6(.65)  .39 P ( A)  P ( A  B1 )  P ( A  B2 )  P ( A  B3 )  .08  .0375  .39  .5075

3.79

a.

P ( B1 | A) 

P ( A  B1 ) .08   .158 .5075 P ( A)

b.

P ( B2 | A) 

P ( A  B2 ) .0375   .074 .5075 P ( A)

c.

P ( B3 | A) 

P ( A  B3 ) .39   .768 P ( A) .5075

If A is independent of B1, B2, and B3, then P ( A | B1 )  P ( A)  .4 . Then P ( B1 | A) 

3.80

P ( A | B1 ) P ( B1 ) .4(.2)   .2 .4 P ( A)

From the information given, P( D)  1 / 80 , P ( D c )  79 / 80 , P ( N | D )  1 / 2 , P ( N c | D )  1 / 2 , P ( N | D c )  1 , and P ( N c | D c )  0 . Using Bayes’ Rule

P( D  N ) P( N | D) P ( D)  P( N ) P ( N | D) P ( D)  P ( N | D c ) P( D c ) 1 1 1 1  1 2 80 160 160      .0063 1 1 79 1 79 1 158 159   1   2 80 80 160 80 160 160

P( D | N ) 

3.81

Define the following events: E: {Expert makes the correct decision} N: {Novice makes the correct decision} M: {Matched condition} E: {Similar distracter condition} E: {Non-similar distracter condition} a.

P ( E c | M )  1  .9212  .0788

b.

P ( N c | M )  1  .7455  .2545

c.

Since P ( N c | M )  .2545  P ( E c | M )  .0788 , it is more likely that the participant is a Novice.

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Probability

3.82

P ( E1  error ) P (error ) P (error | E1 ) P ( E1 )  P (error | E1 ) P ( E1 )  P (error | E2 ) P ( E2 )  P (error | E3 ) P ( E3 )

P ( E1 | error ) 

a.



.01(.30) .003 .003    .158 .01(.30)  .03(.20)  .02(.50) .003  .006  .01 .019

P ( E2  error ) P (error ) P (error | E2 ) P ( E2 )  P (error | E1 ) P ( E1 )  P(error | E2 ) P ( E2 )  P (error | E3 ) P ( E3 )

P ( E2 | error ) 

b.



.03(.20) .006 .006    .316 .01(.30)  .03(.20)  .02(.50) .003  .006  .01 .019

P ( E3  error ) P (error ) P (error | E3 ) P ( E3 )  P (error | E1 ) P ( E1 )  P (error | E2 ) P ( E2 )  P(error | E3 ) P ( E3 )

P ( E3 | error ) 

c.



3.83

.02(.50) .01 .01    .526 .01(.30)  .03(.20)  .02(.50) .003  .006  .01 .019

d.

If there was a serious error, the probability that the error was made by engineer 3 is .526. This probability is higher than for any of the other engineers. Thus engineer #3 is most likely responsible for the error.

a.

Converting the percentages to probabilities, P (275  300)  .52 , P (305  325)  .39 , and P (330  350)  .09 .

b.

Using Bayes Theorem, P(275  300  CC ) P(CC ) P(CC | 275  300) P(275  300)  P(CC | 275  300) P(275  300)  P(CC | 305  325) P(305  325)  P(CC | 330  350) P(330  350) P(275  300 | CC ) 



3.84

127

.775(.52) .403 .403    .516 .775(.52)  .77(.39)  .86(.09) .403  .3003  .0774 .7807

Define the following events: U: {Athlete uses testosterone} P: {Test is positive} a.

Sensitivity is P ( P | U ) 

50  .5 100

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128

Chapter 3 9  1  .01  .99 900

b.

Specificity is P ( P c | U c )  1 

c.

First, we need to find the probability that an athlete is a user: P (U )  100 / 1000  .1 . Next, we need to find the probability of a positive test: P ( P )  P ( P | U ) P (U )  P ( P | U c ) P (U c )  .5(.1)  .01(.9)  .05  .009  .059

Positive predictive value is P (U | P )  3.85

P (U  P ) P ( P | U ) P (U ) .5(.1)    .847 P( P) P( P) .059

Define the following events: S: {Shale} D: {Dolomite } G: {Gamma ray reading > 60 } From the exercise: P ( D ) 

476 295 34 280  .617 , P ( S )   .383 , P (G | D )   .071 , and P (G | S )   .949 . 771 771 476 295

P ( D  G )  P (G | D ) P ( D )  .071(.617)  .0438 and P (G )  P (G | D ) P ( D )  P (G | S ) P ( S )  .071(.617)  .949(.383)  .0438  .3635  .4073 .

P ( D  G ) .0438   .1075 . Since this probability is so small, we would suggest that the .4073 P (G ) area should not be mined.

Thus, P ( D | G ) 

3.86

Define the following events: D: {Defect in steel casting} H: {NDE detects ‘Hit” or defect in steel casting} From the problem, P( H | D)  .97 , P ( H | D c )  .005 , and P ( D )  .01 . P ( H )  P ( H | D ) P ( D )  P ( H | D c ) P ( D c )  .97(.01)  .005(.99)  .0097  .00495  .01465

P( D | H ) 

3.87

P ( D  H ) P ( H | D ) P ( D ) .97(.01) .0097     .6621 .01465 .01465 P( H ) P( H )

Define the following event: D: {Chip is defective} From the Exercise, P ( S1 )  .15 , P ( S 2 )  .05 , P ( S 3 )  .10 , P ( S 4 )  .20 , P ( S 5 )  .12 , P ( S 6 )  .20 , and P ( S 7 )  .18 . Also, P ( D | S1 )  .001 , P ( D | S 2 )  .0003 , P ( D | S 3 )  .0007 , P ( D | S 4 )  .006 , P ( D | S 5 )  .0002 , P ( D | S 6 )  .0002 , and P ( D | S 7 )  .001 .

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Probability

a. P ( S1 | D ) 

129

We must find the probability of each supplier given a defective chip.

P ( S1  D )  P( D)

P ( D | S1 ) P ( S1 ) P ( D | S1 ) P ( S1 )  P ( D | S 2 ) P ( S 2 )  P ( D | S3 ) P ( S3 )  P ( D | S 4 ) P ( S 4 )  P ( D | S5 ) P ( S5 )  P ( D | S6 ) P( S6 )  P( D | S7 ) P( S7 ) 

.001(.15) .001(.15)  .0003(.05)  .0007(.10)  .006(.20)  .0002(.12)  .0002(.02)  .001(.18)



.00015 .00015   .0893 .00015  .000015  .00007  .0012  .000024  .00004  .00018 .001679 P( S2 | D) 

P ( S 2  D ) P ( D | S 2 ) P ( S 2 ) .0003(.05) .000015     .0089 .001679 .001679 P( D) P( D)

P ( S3 | D ) 

P ( S3  D ) P ( D | S3 ) P ( S3 ) .0007(.10) .00007     .0417 P( D) P( D) .001679 .001679

P(S4 | D) 

P ( S 4  D ) P ( D | S 4 ) P ( S 4 ) .006(.20) .0012     .7147 .001679 .001679 P( D) P( D)

P ( S5 | D ) 

P ( S5  D ) P ( D | S5 ) P ( S5 ) .0002(.12) .000024     .0143 P( D) P( D) .001679 .001679

P ( S6 | D ) 

P ( S6  D ) P ( D | S6 ) P ( S6 ) .0002(.20) .00004     .0238 P( D) P( D) .001679 .001679

P ( S7 | D ) 

P ( S7  D ) P ( D | S7 ) P ( S7 ) .001(.18) .00018     .1072 .001679 .001679 P( D) P( D)

Of these probabilities, .7147 is the largest. This implies that if a failure is observed, supplier number 4 was most likely responsible. b. If the seven suppliers all produce defective chips at the same rate of .0005, then P ( D | S i )  .0005 for all i = 1, 2, 3, … 7 and P ( D )  .0005 . For any supplier i, P ( S i  D )  P ( D | S i ) P ( S i )  .0005 P ( S i ) and P ( Si | D ) 

P ( Si  D ) P ( D | Si ) P ( Si ) .0005 P ( Si )    P ( Si ) .0005 .0005 P( D)

Thus, if a defective is observed, then it most likely came from the supplier with the largest proportion of sales (probability). In this case, the most likely supplier would be either supplier 4 or supplier 6. Both of these have probabilities of .20.

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130

3.88

Chapter 3

Define the following events: A: {Alarm A sounds alarm} B: {Alarm B sounds alarm} I: {Intruder} From the problem: P ( A | I )  .9 , P ( B | I )  .95 , P ( A | I c )  .2 , P ( B | I c )  .1 , and P ( I )  .4

Since the two systems are operating independently of each other, P ( A  B | I )  P ( A | I ) P ( B | I )  .9(.95)  .855 P ( A  B  I )  P ( A  B | I ) P ( I )  .855(.4)  .342

P ( A  B | I c )  P ( A | I c ) P ( B | I c )  .2(.1)  .02 P ( A  B  I c )  P ( A  B | I c ) P ( I c )  .02(.6)  .012

Thus, P ( A  B )  P ( A  B  I )  P ( A  B  I c )  .342  .012  .354 Finally, P ( I | A  B ) 

3.89

a.

b.

P ( A  B  I ) .342   .966 P( A  B) .354

P (T | E )  1 , then P (T | E )  P (T c | E ) . Thus, the probability of more than two bullets given the c P (T | E ) evidence is greater than the probability of two bullets given the evidence. This supports the theory of more than two bullets were used in the assassination of JFK.

If

Using Bayes Theorem, P (T | E ) 

P (T ) P ( E | T ) P (T c ) P ( E | T c ) . and P (T c | E )  c c P (T ) P ( E | T )  P (T ) P ( E | T ) P (T ) P ( E | T )  P (T c ) P ( E | T c )

P(T ) P( E | T ) P(T | E ) P (T ) P( E | T ) P(T ) P( E | T )  P(T c ) P( E | T c )   . Thus, c P(T | E ) P(T c ) P( E | T c ) P (T c ) P( E | T c ) P(T ) P( E | T )  P(T c ) P( E | T c ) 3.90

a.

If the Dow Jones Industrial Average increases, a large New York bank would tend to decrease the prime interest rate. Therefore, the two events are not mutually exclusive since they could occur simultaneously.

b.

The next sale by a PC retailer could not be both a notebook and a desktop computer. Since the two events cannot occur simultaneously, the events are mutually exclusive.

c.

Since both events cannot occur simultaneously, the events are mutually exclusive.

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Probability

3.91

a.

131

The two probability rules for a sample space are that the probability for any sample point is between 0 and 1 and that the sum of the probabilities of all the sample points is 1. For this Exercise, all the probabilities of the sample points are between 0 and 1 and

 P(S )  P(S )  P(S 4

i 1

b.

1

i

2

)  P ( S3 )  P ( S 4 )  .2  .1  .3  .4  1.0

P ( A)  P ( S1 )  P ( S 4 )  .2  .4  .6

3.92

P ( A  B )  P ( A)  P ( B )  P ( A  B )  .7  .5  .4  .8

3.93

a.

If events A and B are mutually exclusive, then P ( A  B )  0 . P( A | B) 

3.94

P( A  B) 0  0 .3 P( B)

b.

No. If events A and B are independent, then P ( A | B )  P ( A) . However, from the Exercise we know P ( A)  .2 and from part a, we know P ( A | B )  0 . Thus, events A and B are not independent.

a.

Because events A and B are independent, we have: P ( A  B )  P ( A) P ( B )  .3(.1)  .03

Thus, P ( A  B )  0 , and the two events cannot be mutually exclusive.

3.95

P ( A  B ) .03   .3 .1 P( B)

P ( B | A) 

P ( A  B ) .03   .1 .3 P ( A)

b.

P( A | B) 

c.

P ( A  B )  P ( A)  P ( B )  P ( A  B )  .3  .1  .03  .37

P ( A  B )  .4 , P ( A | B )  .8

Since P( A | B ) 

.8 

P( A  B) , substitute the given probabilities into the formula and solve for P(B). P( B)

.4 .4  P ( B )   .5 P( B) .8

3.96

The number of ways to select 5 things from 50 is a combination of 50 things taken 5 at a time or  50  50! 50! 50  49  48  47  46  45!    2,118, 760 .   5  5!(50 5)! 5!45! 5  4  3  2  1  45!  

3.97

a.

P( A  B)  0 P ( B  C )  P (2)  .2 P ( A  C )  P (1)  P (2)  P (3)  P (5)  P (6)  .3  .2  .1  .1  .2  .9 P ( A  B  C )  P (1)  P (2)  P (3)  P (4)  P (5)  P (6)  .3  .2  .1  .1  .1  .2  1

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132

Chapter 3 P ( B c )  P (1)  P (3)  P (5)  P (6)  .3  .1  .1  .2  .7 P ( Ac  B )  P (2)  P (4)  .2  .1  .3

P( B | C ) 

.2 .2 P( B  C ) P (2)     .4 P (C ) P (2)  P (5)  P (6) .2  .1  .2 .5

P ( B | A) 

0 P ( B  A)  0 P ( A) P ( A)

b.

Since P ( A  B )  0 , and P ( A) P ( B )  0 , these two would not be equal, implying A and B are not independent. However, A and B are mutually exclusive, since P ( A  B )  0 .

c.

P ( B )  P (2)  P (4)  .2  .1  .3 . But P ( B | C ) , calculated above, is .4. Since these are not equal, B

and C are not independent. Since P ( B  C )  .2 , B and C are not mutually exclusive. 3.98

3.99

a.

6!  6  5  4  3  2 1  720

b.

 10  10! 10  9  8    1   10   9   8  7    1 1 9!(10 9)! 9  

c.

 10  10! 10  9  8    1   10    1 1!(10  1)! 1  9  8    1

d.

6 6! 6 5 4 3 2 1   20   3  3!(6 3)! 3  2  1 3  2  1  

e.

0!  1

Define the following events: E: {Industrial accident caused by faulty Engineering & Design} P: {Industrial accident caused by faulty Procedures & Practices} M: {Industrial accident caused by faulty Management & Oversight} T: {Industrial accident caused by faulty Training & Communication} a.

The sample points for this problem are: E, P, M, and T. Reasonable probabilities are: P ( E )  27 / 83  .3253 , P ( P )  24 / 83  .2892 , P ( M )  22 / 83  .2651 , and P (T )  10 / 83  .1205 .

b.

P ( E )  27 / 83  .3253 . Approximately 32.53% of all industrial accidents are caused by faulty Engineering and Design.

c.

P(Industrial accident caused by something other than procedures & practices)  1  P ( P c )  1  .2892  .7108 . Approximately 71.08% of all industrial accidents are caused by something other than faulty procedures & practices.

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Probability

3.100

a.

133

Define the following events: J: {Raise based on job performance} C: {Raise based on cost of living} U: {Unsure.} The 3 sample points are: J, C, and U

b.

c. 3.101

We will base the probabilities on the proportions of the 10,000 U.S. workers surveyed who responded in each category. Thus, P ( J )  .35 , P (C )  .50 , and P (U )  .15 P(Raise based on either job performance or cost of living)  P ( J )  P (C )  .35  .50  .85

Define the event: B: {Small business owned by non-Hispanic white female} From the problem, P ( B )  .27 The probability that a small business owned by a non-Hispanic white is male-owned is P ( B c )  1  P ( B )  1  .27  .73 .

3.102

Define the following events: C: {Public school building has inadequate plumbing} D: {Public school has plans for repairing building} From the problem, we know P (C )  .25 and P ( D | C )  .38 . P (C  D )  P ( D | C ) P (C )  .38(.25)  .095

3.103

a.

This statement is false. All probabilities are between 0 and 1 inclusive. One cannot have a probability of 4.

b.

If we assume that the probabilities are the same as the percents (changed to proportions), then this is a true statement. P (4 or 5)  P (4)  P (5)  .6020  .1837  .7857

3.104

c.

This statement is true. There were no observations with one star. Thus, P (1)  0 .

d.

This statement is false. P (2)  .0408 and P (5)  .1837 . P (5)  P (2) .

Define the following events: S: {cause of fatal crash is speeding} C: {cause of fatal crash is missing a curve} From the problem, we know P(S) = .3 and P ( S  C )  .12 . P (C | S ) 

P (C  S ) .12   .4 .3 P(S )

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134

3.105

3.106

Chapter 3

a.

B C

b.

Ac

c.

CB

d.

A  Cc

a.

The 5 sample points are: Total population, Agricultural change, Presence of industry, Growth, and Population concentration.

b.

The probabilities are best estimated with the sample proportions. Thus,

P(Total population) = .18 P(Agricultural change) = .05 P(Presence of industry) = .27 P(Growth) = .05 P(Population concentration) = .45 c.

Define the following event:

A: {Factor specified is population-related} P(A) = P(Total population) + P(Growth) + P(Population concentration)  .18  .05  .45  .68 . 3.107

Define the following events:

G: {regularly use the golf course} T: {regularly use the tennis courts} Given: P (G )  .7 and P (T )  .5 The event "uses neither facility" can be written as G c  T c or (G  T ) c . We are given P (G c  T c )  P[(G  T ) c ]  .05 . The complement of the event "uses neither facility" is the event "uses at

least one of the two facilities" which can be written as G  T . P (G  T )  1  P (G  T ) c  1  .05  .95

From the additive rule, P (G  T )  P (G )  P (T )  P (G  T )  .95  .7  .5  P (G  T )  P (G  T )  .25

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Probability

a.

The Venn Diagram is:

G .45

T .25

.25

.05

3.108

b.

P (G  T )  .95 from above.

c.

P (G  T )  .25 from above.

d.

P (G | T ) 

S

P (G  T ) .25   .5 P (T ) .5

Define the following events:

A: {electrical switch monitors quality of power} B: {electrical switch not wired properly} From the problem, P ( A )  .90 and P ( B | A)  .90 . P ( A  B c )  P ( B c | A) P ( A)  (1  .90)(.90)  .09 .

3.109

a.

P ( A) 

1, 465  .684 2,143

b.

P( B) 

265  .124 2,143

c.

No. There is one sample point that they have in common: Plaintiff trial win – reversed, Jury

d.

P ( Ac )  1  P ( A)  1  .684  .316

e.

P( A  B) 

194  71  429  111  731 1, 536   .717 2,143 2,143

f.

P( A  B) 

194  .091 2,143

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135

136

3.110

Chapter 3

Since there are 11 individuals who are willing to serve on the panel, the number of different panels of 5 experts is a combination of 11 things taken 5 at a time or  11  11! 11  10  9  8  7  6  5  4  3  2  1   462    5  5!6! (5  4  3  2  1)(6  5  4  3  2  1)

3.111

Define the following events:

A: {The watch is accurate} N: {The watch is not accurate} Assuming the manufacturer's claim is correct, P ( N )  .05 and P ( A)  1  P ( N )  1  .05  .95

The sample space for the purchase of four of the manufacturer's watches is listed below.

(A, A, A, A) (A, A, A, N) (A, A, N, A) (A, N, A, A) a.

(N, A, A, A) (A, A, N, N) (A, N, A, N) (N, A, A, N)

(A, N, N, A) (N, A, N, A) (N, N, A, A) (A, N, N, N)

(N, A, N, N) (N, N, A, N) (N, N, N, A) (N, N, N, N)

All four watches not being accurate as claimed is the sample point (N, N, N, N). Assuming the watches purchased operate independently and the manufacturer's claim is correct, P  N , N , N , N   P  N  P  N  P  N  P  N   .054  .00000625

b.

The sample points in the sample space that consist of exactly two watches failing to meet the claim are listed below. (A, A, N, N) (N, A, A, N) (A, N, A, N) (N, A, N, A) (A, N, N, A) (N, N, A, A) The probability that exactly two of the four watches fail to meet the claim is the sum of the probabilities of these six sample points. Assuming the watches purchased operate independently and the manufacturer's claim is correct, P ( A, A, N , N )  P ( A) P ( A) P ( N ) P ( N )  .95(.95)(.05)(.05)  .00225625

All six of the sample points will have the same probability. Therefore, the probability that exactly two of the four watches fail to meet the claim when the manufacturer's claim is correct is 6(.00225625)  .0135

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Probability

c.

137

The sample points in the sample space that consist of three of the four watches failing to meet the claim are listed below. (A, N, N, N) (N, N, A, N) (N, A, N, N) (N, N, N, A) The probability that three of the four watches fail to meet the claim is the sum of the probabilities of the four sample points. Assuming the watches purchased operate independently and the manufacturer's claim is correct, P ( A, N , N , N )  P ( A) P ( N ) P ( N ) P ( N )  .95(.05)(.05)(.05)  .00011875

All four of the sample points will have the same probability. Therefore, the probability that three of the four watches fail to meet the claim when the manufacturer's claim is correct is 4(.00011875)  .000475

If this event occurred, we would tend to doubt the validity of the manufacturer's claim since its probability of occurring is so small. d.

All four watches tested failing to meet the claim is the sample point (N, N, N, N). Assuming the watches purchased operate independently and the manufacturer's claim is correct, P ( N , N , N , N )  P ( N ) P ( N ) P ( N ) P ( N )  .05(.05)(.05)(.05)  .00000625

Since the probability of observing this event is so small if the claim is true, we have strong evidence against the validity of the claim. However, we do not have conclusive proof that the claim is false. There is still a chance the event can occur (with probability .00000625) although it is extremely small. 3.112

The possible ways of ranking the blades are: GSW

SGW

WGS

GWS

SWG

WSG

If the consumer had no preference but still ranked the blades, then the 6 possibilities are equally likely. Therefore, each of the 6 possibilities has a probability of 1/6 of occurring. 1 1 2 1    6 6 6 3

a.

P(Ranks G first)  P (GSW )  P (GWS ) 

b.

P(Ranks G last)  P ( SWG )  P (WSG ) 

c.

P(ranks G last and W second)  P ( SWG ) 

d.

P (WGS ) 

1 1 2 1    6 6 6 3 1 6

1 6

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138

3.113

Chapter 3

Define the following events:

A: {Never smoked cigars} B: {Former cigar smoker} C: {Current cigar smoker} D: {Died from cancer} E: {Did not die from cancer}

3.114

3.115

a.

P ( D | A) 

782 782 P ( D  A) 137, 243    .006 121,529 121,529 P ( A) 137, 243

b.

P( D | B) 

91 91 P( D  B) 137, 243    .012 7,848 7,848 P( B) 137, 243

c.

P( D | C ) 

141 141 P( D  C ) 137, 243    .018 7,866 7,866 P (C ) 137, 243

a.

Consecutive tosses of a coin are independent events since what occurs one time would not affect the next outcome.

b.

If the individuals are randomly selected, then what one individual says should not affect what the next person says. They are independent events.

c.

The results in two consecutive at-bats are probably not independent. The player may have faced the same pitcher both times which may affect the outcome.

d.

The amount of gain and loss for two different stocks bought and sold on the same day are probably not independent. The market might be way up or down on a certain day so that all stocks are affected.

e.

The amount of gain or loss for two different stocks that are bought and sold in different time periods are independent. What happens to one stock should not affect what happens to the other.

f.

The prices bid by two different development firms in response to the same building construction proposal would probably not be independent. The same variables would be present for both firms to consider in their bids (materials, labor, etc.).

Define the following events:

A: {Wheelchair user had an injurious fall} B: {Wheelchair user had all five features installed in the home} C: {Wheelchair user had no falls} D: {Wheelchair user had none of the features installed in the home} a.

P ( A) 

48  .157 306

b.

P(B) 

9  .029 306

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Probability

139

89  .291 306

c.

P (C  D ) 

d.

P( A | B) 

2 P( A  B ) 2  306   .222 9 P( B) 9 306

e.

P( A | D) 

20 P ( A  D) 306  20  .183  109 P ( D) 109 306

3.116 Define the following events:

A1: {Paraguay is assigned to Group A} A2: {Ecuador is assigned to Group A} B1: {Paraguay is assigned to Group B} B2: {Sweden or top team in pot 3 is assigned to Group B} D1: {Paraguay is assigned to Group D} D2: {Ecuador is assigned to Group D} If the teams are drawn at random from each pot, the probability that any team is assigned to a group is 1/8. a.

P ( A1 )  1 / 8  .125

b.

P ( A1  A2 )  P ( A1 )  P ( A2 )  P ( A1  A2 )  1 / 8  1 / 8  0  2 / 8  .25

c.

P ( B1  B2 )  P ( B1 ) P ( B2 )  (1 / 8)(2 / 8)  2 / 64  .03125

d.

We can look at this probability by looking at how the slots can be filled. We will just look at how the teams from pot 2 can be put into Groups A, B, C, and D. The order of filling these really does not matter, so we will look at the ways to fill Group C, then Group D, then Group A, then Group B. First, we will find the total number of ways we can fill these 4 slots or Groups where Group C cannot have Paraguay or Ecuador. Since Group C cannot have Paraguay or Ecuador, then there are only 6 ways to fill Group C. There would then be 7 ways to fill Group D, 6 ways to fill Group A and 5 ways to fill Group B. The total ways to fill these 4 Groups without having Paraguay or Ecuador in Group C is 6(7)(6)(5) = 1,260. Now, we will find the number of ways we can fill these 4 Groups where Group C cannot have Paraguay or Ecuador and Group D does have either Paraguay or Ecuador. There will be 6 ways to fill Group C, 2 ways to fill Group D, 6 ways to fill Group A, and 5 ways to fill Group B. The total ways to fill these 4 Groups without having Paraguay or Ecuador in Group C and having either Paraguay or Ecuador in Group D is 6(2)(6)(5) = 360. Thus, the probability that Group C does not have either Paraguay or Ecuador and Group D does have either Paraguay or Ecuador is 360 / 1, 260  2 / 7  .286 . Finally, the probability that Group D does not have either Paraguay or Ecuador is 1  .286  .714 .

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140

3.117

Chapter 3

Define the following events:

S1: {Salesman makes sale on the first visit} S2: {Salesman makes a sale on the second visit} P ( S1 )  .4

P ( S 2 | S1c )  .65

The sample points of the experiment are: S1  S 2c , S1c  S 2 , S1c  S2c

The probability the salesman will make a sale is: P ( S1  S 2c )  P ( S1c  S 2 )  P ( S1 )  P ( S 2 | S1c ) P ( S1c )  .4  .65(1  .4)  .4  .39  .79

3.118

Define the following events:

S: {System shuts down} F1: {Hardware failure} F2: {Software failure} F3: {Power failure} From the Exercise, we know: P ( F1 )  .01 , P ( F2 )  .05 , and P ( F3 )  .02 . Also, P ( S | F1 )  .73 , P ( S | F2 )  .12 , and P ( S | F3 )  .88 .

The probability that the current shutdown is due to a hardware failure is:

P( F1 | S )  

P( F1  S ) P( S | F1 ) P( F1 )  P( S ) P( S | F1 ) P( F1 )  P( S | F2 ) P( F2 )  P( S | F3 ) P( F3 ) .73(.01) .0073 .0073    .2362 .73(.01)  .12(.05)  .88(.02) .0073  .006  .0176 .0309

The probability that the current shutdown is due to a software failure is: P( F2 | S )  

P( F2  S ) P( S | F2 ) P( F2 )  P( S ) P( S | F1 ) P( F1 )  P( S | F2 ) P( F2 )  P( S | F3 ) P( F3 ) .12(.05) .006 .006    .1942 .73(.01)  .12(.05)  .88(.02) .0073  .006  .0176 .0309

The probability that the current shutdown is due to a power failure is: P( F3 | S )  

P( F3  S ) P( S | F3 ) P( F3 )  P( S ) P( S | F1 ) P( F1 )  P( S | F2 ) P( F2 )  P( S | F3 ) P( F3 ) .88(.02) .0176 .0176    .5696 .73(.01)  .12(.05)  .88(.02) .0073  .006  .0176 .0309

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Probability

3.119

a.

Suppose we let the four positions in a sample point represent in order (1) Raise a broad mix of crops, (2) Raise livestock, (3) Use chemicals sparingly, and (4) Use techniques for regenerating the soil, such as crop rotation. A farmer is either likely (L) to engage in an activity or unlikely (U). The possible classifications are: LLLL LLLU LLUL LULL ULLL LLUU LULU LUUL ULLU ULUL UULL LUUU ULUU UULU UUUL UUUU

b.

Since there are 16 classifications or sample points and all are equally likely, then each has a probability of 1/16. P (UUUU ) 

c.

1 16

The probability that a farmer will be classified as likely on at least three criteria is  1  5 . P ( LLLL )  P ( LLLU )  P ( LLUL )  P ( LULL )  P (ULLL )  5     16  16

3.120

Define the following events: C: {Committee judges joint acceptable} I: {Inspector judges joint acceptable} The sample points of this experiment are:

C  I , C  I c , C c  I , Cc  I c a.

The probability the inspector judges the joint to be acceptable is: P ( I )  P (C  I )  P (C c  I ) 

101 23 124    .810 153 153 153

The probability the committee judges the joint to be acceptable is: P (C )  P (C  I )  P (C  I c ) 

b.

101 10 111    .725 153 153 153

The probability that both the committee and the inspector judge the joint to be acceptable is: P (C  I ) 

101  .660 153

The probability that neither judge the joint to be acceptable is: P (C c  I c )  c.

141

The probability the inspector and committee disagree is: P (C  I c )  P (C c  I ) 

10 23 33    .216 153 153 153

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19  .124 153

142

Chapter 3

The probability the inspector and committee agree is: P (C  I )  P (C c  I c ) 

3.121

101 19 120    .784 153 153 153

Define the following events: O1: O2: O3: A:

{Component #1 in System A operates properly} {Component #2 in System A operates properly} {Component #3 in System A operates properly} {System A works properly}

P (O1 )  1  P  O1c   1  .12  .88

a.

P (O2 )  1  P  O2c   1  .09  .91

P (O3 )  1  P  O3c   1  .11  .89

P ( A)  P (O1  O2  O3 )  P (O1 ) P (O2 ) P (O3 )  .88(.91)(.89)  .7127

(since the three components operate independently) b.

P ( Ac )  1  P ( A)  1  .7127  .2873

(see part a) c.

Define the following events: C1: C2: D3: D4: C: D:

{Component 1 in System B works properly} {Component 2 in System B works properly} {Component 3 in System B works properly} {Component 4 in System B works properly} {Subsystem C works properly} {Subsystem D works properly}

The probability a component fails is .1, so the probability a component works properly is1  .1  .9 . Subsystem C works properly if both components 1 and 2 work properly. P (C )  P (C1  C 2 )  P (C1 ) P (C 2 )  .9(.9)  .81

(since the components operate independently) Similarly, P ( D )  P ( D1  D2 )  P ( D1 ) P ( D2 )  .9(.9)  .81 The system operates properly if either subsystem C or D operates properly. The probability that System B operates properly is: P (C  D )  P (C )  P ( D )  P (C  D )  P (C )  P ( D )  P (C ) P ( D )  .81  .81  .81(.81)  .9639

d.

The probability exactly one subsystem fails in System B is: P (C  D c )  P (C c  D )  P (C ) P ( D c )  P (C c ) P ( D )  .81(1  .81)  (1  .81)(.81)  .1539  .1539  .3078

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Probability

e.

143

The probability that System B fails is the probability that both subsystems fail: P (C c  D c )  P (C c ) P ( D c )  (1  .81)(1  .81)  .0361

f.

The system operates correctly 99% of the time means it fails 1% of the time. The probability one subsystem fails is .19. The probability n subsystems fail is .19n. Thus, we must find n such that (.19) n  .01  n  3

3.122 Define the following events: R: {Successful regime change is achieved} M: {Mission is extended to support a weak Iraq government) a.

The probability that a successful regime change is not achieved is P ( R c )  1  P ( R )  1  .7  .3

b.

P ( R | M )  .26

c.

Given that P ( M )  .55 , the probability that the mission is extended and results in a successful regime change is P ( M  R )  P ( R | M ) P ( M )  .26(.55)  .143

3.123

The probability of a false positive is P ( A | B ) .

3.124

Define the following events: A1: {Fuse made by line 1} A2: {Fuse made by line 2} D: {Fuse is defective} From the Exercise, we know P ( D | A1 )  .06 and P ( D | A2 )  .025 . Also, P ( A1 )  P ( A2 )  .5 . Two fuses are going to be selected and we need to find the probability that one of the two is defective. We can get one defective fuse out of two by getting a defective on the first and non-defective on the second ( D  D c ) or non-defective on the first and defective on the second ( D c  D ) . The probability of getting one defective out of two fuses given line 1 is: P ( D  D c | A1 )  P ( D c  D | A1 )  P ( D | A1 ) P( D c | A1 )  P( D c | A1 ) P( D | A1 )  .06(1  .06)  (1  .06)(.06)  .06(.94)  .94(.06)  .1128  P(1 D | A1 )

The probability of getting one defective out of two fuses given line 2 is: P ( D  D c | A2 )  P( D c  D | A2 )  P( D | A2 ) P( D c | A2 )  P( D c | A2 ) P( D | A2 )  .025(1  .025)  (1  .025)(.025)  .025(.975)  .975(.025)  .04875  P(1 D | A2 )

The probability of getting one defective out of two fuses is: P (1 D )  P (1 D  A1 )  P (1 D  A2 )  P (1 D | A1 ) P ( A1 )  P (1 D | A2 ) P ( A2 )

 .1128(.5)  .04875(.5)  .0564  .024375  .080775

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144

Chapter 3

Finally, we want to find: P ( A1 | 1 D ) 

3.125

P (1 D  A1 ) .0564   .6982 P (1 D) .080775

Define the following events: A: {Press is correctly adjusted} B: {Press is incorrectly adjusted} D: {part is defective} From the exercise, P ( A )  .90 , P ( D | A)  .05 , and. We also know that event B is the complement of event A. Thus, P ( B )  1  P ( A)  1  .90  .10 . P( B | D) 

3.126

P( B  D) P( D | B) P( B) .50(.10) .05 .05      .526 P( D) P ( D | B ) P ( B )  P ( D | A) P( A) .50(.10)  .05(.90) .05  .045 .095

There are a total of 6  6 = 36 outcomes when rolling 2 dice. If we let the first number in the pair represent the outcome of die number 1 and the second number in the pair represent the outcome of die number 2, then the possible outcomes are: 1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6

If both dice are fair, then each of these outcomes are equally like and have a probability of 1/36. a.

To win on the first roll, a player must roll a 7 or 11. There are 6 ways to roll a 7 and 2 ways to roll an 11. Thus the probability of winning on the first roll is: P (7 or 11) 

b.

To lose on the first roll, a player must roll a 2 or 3. There is 1 way to roll a 2 and 2 ways to roll a 3. Thus the probability of losing on the first roll is: P (2 or 3) 

c.

8  .2222 36

3  .0833 36

If a player rolls a 4 on the first roll, the game will end on the next roll if the player rolls 4 (player wins) or if the player rolls a 7 (player loses). There are 3 ways to roll a 4 and 6 ways to roll a 7. Thus, P (4or 7 on 2 nd roll) 

36 9   .25 . 36 36

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Probability

3.127

145

Define the flowing events: A: {Dealer draws a blackjack} B: {Player draws a blackjack} a.

For the dealer to draw a blackjack, he needs to draw an ace and a face card. There are  4 4! 4  3  2 1   4 ways to draw an ace and    1  1!(4  1)! 1  3  2  1  12  12! 12  11  10    1   12 ways to draw a face card (there are 12 face    1  1!(12  1)! 1  11  10  9   1

cards in the deck). The total number of ways a dealer can draw a blackjack is 4  12 = 48. The total number of ways a dealer can draw 2 cards is  52  52! 52  51  50   1   1326    2  2!(52  2)! 2  1  50  49  48   1

Thus, the probability that the dealer draws a blackjack is P ( A)  b.

48  .0362 1326

In order for the player to win with a blackjack, the player must draw a blackjack and the dealer does not. Using our notation, this is the event B  AC . We need to find the probability that the player draws a blackjack  P( B)  and the probability that the dealer does not draw a blackjack given the

player does  P ( Ac | B )  . Then, the probability that the player wins with a blackjack is P ( Ac | B ) P ( B ) .

The probability that the player draws a blackjack is the same as the probability that the dealer draws a blackjack, which is P ( B )  .0362 . There are 5 scenarios where the dealer will not draw a blackjack given the player does. First, the dealer could draw an ace and not a face card. Next, the dealer could draw a face card and not an ace. Third, the dealer could draw two cards that are not aces or face cards. Fourth, the dealer could draw two aces, and finally, the dealer could draw two face cards. The number of ways the dealer could draw an ace and not a face card given the player draws a blackjack is  3  36  3! 36! 3  2  1 36  35  34    1     3(36)  108     1 1    2  1 1  35  34  33    1 1!(3 1)! 1!(36 1)! 1   

(Note: Given the player has drawn blackjack, there are only 3 aces left and 36 non-face cards.) The number of ways the dealer could draw a face card and not an ace given the player draws a blackjack is  11 36  11! 36! 11  10  9   1 36  35  34   1     11(36)  396     1 1    10  9  8   1 1  35  34  33   1 1!(11 1)! 1!(36 1)! 1   

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146

Chapter 3

The number of ways the dealer could draw neither a face card nor an ace given the player draws a blackjack is  36  36! 36  35  34   1   630   2   2!(36 2)! 2 1  34  33  32    1  

The number of ways the dealer could draw two aces given the player draws a blackjack is 3 3! 3  2 1  3   2  2!(3 2)! 2 1 1  

The number of ways the dealer could draw two face cards given the player draws a blackjack is  11  11! 11  10  9    1   55   2   9  8  7   1 2!(11 2)! 2  

The total number of ways the dealer can draw two cards given the player draws a blackjack is  50  50! 50  49  48   1   1225   2    48  47  46   1 2!(50 2)! 2 1  

The probability that the dealer does not draw a blackjack given the player draws a blackjack is P ( Ac | B ) 

108  396  630  3  55 1192   .9731 1225 1225

Finally, the probability that the player wins with a blackjack is P ( B  Ac )  P ( Ac | B ) P ( B )  .9731(.0362)  .0352

3.128

a.

Define the following events: W: F:

{Player wins the game Go} {Player plays first (black stones)}

P (W  F )  319 / 577  .553

b.

P (W  F | CA)  34 / 34  1

P (W  F | CB )  69 / 79  .873

P (W  F | CC )  66 / 118  .559

P (W  F | BA)  40 / 54  .741

P (W  F | BB )  52 / 95  .547

P (W  F | BC )  27 / 79  .342

P (W  F | AA)  15 / 28  .536

P (W  F | AB )  11 / 51  .216

P (W  F | AC )  5 / 39  .128

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Probability

c.

147

There are three combinations where the player with the black stones (first) is ranked higher than the player with the white stones: CA, CB, and BA. P (W  F | CA  CB  BA)  (34  69  40) / (34  79  54)  143 / 167  .856

d.

There are three combinations where the players are of the same level: CC, BB, and AA. P (W  F | CC  BB  AA)  (66  52  15) / (118  95  28)  133 / 241  .552

3.129

First, we will list all possible sample points for placing a car (C) and 2 goats (G) behind doors #1, #2, and #3. If the first position corresponds to door #1, the second position corresponds to door #2, and the third position corresponds to door #3, the sample space is: (C G G) (G C G) (G G C) Now, suppose you pick door #1. Initially, the probability that you will win the car is 1/3 – only one of the sample points has a car behind door #1. The host will now open a door behind which is a goat. If you pick door #1 in the first sample point (C G G), the host will open either door #2 or door #3. Suppose he opens door #3 (it really does not matter). If you pick door #1 in the second sample point (G C G), the host will open door #3. If you pick door #1 in the third sample point (G G C), the host will open door #2. Now, the new sample space will be: (C G) (G C) (G C) where the first position corresponds to door #1 (the one you chose) and the second position corresponds to the door that was not opened by the host. Now, if you keep door #1, the probability that you win the car is 1/3. However, if you switch to the remaining door, the probability that you win the car is now 2/3. Based on these probabilities, it is to your advantage to switch doors. The above could be repeated by selecting door #2 initially or door #3 initially. In either of these cases, again, the probability of winning the car is 1/3 if you do not switch and 2/3 if you switch. Thus, Marilyn was correct.

3.130

Suppose we define the following event: E: {Error produced when dividing} From the problem, we know that P ( E )  1 / 9, 000, 000, 000 The probability of no error produced when dividing is P ( E c )  1  P ( E )  1  1 / 9,000,000,000  8,999,999,999 / 9,000,000,000  .999999999  1.0000

Suppose we want to find the probability of no errors in 2 divisions (assuming each division is independent): P ( E c  E c )  .999999999(.999999999)  .999999999  1.0000

Thus, in general, the probability of no errors in k divisions would be: c c k k P (  Ec  Ec  Ec   E   )  P ( E )  [8, 999, 999, 999 / 9, 000, 000, 000] k

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148

Chapter 3

Suppose a user ran a program that performed 1 billion divisions. The probability of no errors in these 1 billion divisions would be: P ( E c )1,000,000,000  [8,999,999,999 / 9,000,000,000]1,000,000,000  .8948

Thus, the probability of at least 1 error in 1 billion divisions would be 1  P ( E c )1,000,000,000  1  [8,999,999,999 / 9,000,000,000]1,000,000,000  1  .8948  .1052

Copyright © 2014 Pearson Education, Inc.

Chapter 4 Random Variables and Probability Distributions 4.1

4.2

a.

The number of newspapers sold by New York Times each month can take on a countable number of values. Thus, this is a discrete random variable.

b.

The amount of ink used in printing the Sunday edition of the New York Times can take on an infinite number of different values. Thus, this is a continuous random variable.

c.

The actual number of ounces in a one gallon bottle of laundry detergent can take on an infinite number of different values. Thus, this is a continuous random variable.

d.

The number of defective parts in a shipment of nuts and bolts can take on a countable number of values. Thus, this is a discrete random variable.

e.

The number of people collecting unemployment insurance each month can take on a countable number of values. Thus, this is a discrete random variable.

a.

The closing price of a particular stock on the New York Stock Exchange is discrete. It can take on only a countable number of values.

b.

The number of shares of a particular stock that are traded on a particular day is discrete. It can take on only a countable number of values.

c.

The quarterly earnings of a particular firm is discrete. It can take on only a countable number of values.

d.

The percentage change in yearly earnings between 2011 and 2012 for a particular firm is continuous. It can take on any value in an interval.

e.

The number of new products introduced per year by a firm is discrete. It can take on only a countable number of values.

f.

The time until a pharmaceutical company gains approval from the U.S. Food and Drug Administration to market a new drug is continuous. It can take on any value in an interval of time.

4.3

Since there are only a fixed number of outcomes to the experiment, the random variable, x, the number of stars in the rating, is discrete.

4.4

The number of customers, x, waiting in line can take on values 0, 1, 2, 3, … . Even though the list is never ending, we call this list countable. Thus, the random variable is discrete.

4.5

The variable x, total compensation in 2011 (in $ millions), is reported in whole number dollars. Since there are a countable number of possible outcomes, this variable is discrete.

4.6

A banker might be interested in the number of new accounts opened in a month, or the number of mortgages it currently has, both of which are discrete random variables.

149 Copyright © 2014 Pearson Education, Inc.

150

Chapter 4

4.7

An economist might be interested in the percentage of the work force that is unemployed, or the current inflation rate, both of which are continuous random variables.

4.8

The manager of a hotel might be concerned with the number of employees on duty at a specific time, or the number of vacancies there are on a certain night.

4.9

The manager of a clothing store might be concerned with the number of employees on duty at a specific time of day, or the number of articles of a particular type of clothing that are on hand.

4.10

A stockbroker might be interested in the length of time until the stock market is closed for the day.

4.11

a.

p (22)  .25

b.

P( x  20 or x  24)  P( x  20)  P( x  24)  .15  .20  .35

c.

P( x  23)  P( x  20)  P( x  21)  P( x  22)  P( x  23)  .15  .10  .25  .30  .80

a.

The variable x can take on values 1, 3, 5, 7, and 9.

b.

The value of x that has the highest probability associated with it is 5. It has a probability of .4.

c.

Using MINITAB, the probability distribution of x as a graph is:

4.12

.4

p(x)

.3

.2

.1

0 1

3

4

5 x

6

7

8

9

d.

P( x  7)  .2

e.

P( x  5)  p(5)  p(7)  p(9)  .4  .2  .1  .7

f.

P( x  2)  p(3)  p(5)  p(7)  p(9)  .2  .4  .2  .1  .9

g. 4.13

2

E( x)   xp( x)  1(.1)  3(.2)  5(.4)  7(.2)  9(.1)  .1  .6  2.0  1.4  .9  5.0

 p(x)  1 .

Thus, p(2)  p(3)  p(5)  p(8)  p(10)  1

a.

We know

b.

P( x  2 or x  10)  P( x  2)  P( x  10)  .15  .25  .40

c.

P( x  8)  P( x  2)  P( x  3)  P( x  5)  P( x  8)  .15  .10  .25  .25  .75

 p(5)  1  p(2)  p(3)  p(8)  p(10)  1  .15  .10  .25  .25  .25

Copyright © 2014 Pearson Education, Inc.

Random Variables and Probability Distributions

4.14

4.15

 p( x)  .1 .3  .3  .2  .9  1 .

a.

This is not a valid distribution because

b.

This is a valid distribution because 0  p( x)  1 for all values of x and

c.

This is not a valid distribution because p(4)  .3  0 .

d.

The sum of the probabilities over all possible values of the random variable is p( x)  .15  .15  .45  .35  1.1  1 , so this is not a valid probability distribution.

a.

When a die is tossed, the number of spots observed on the upturned face can be 1, 2, 3, 4, 5, or 6. Since the six sample points are equally likely, each one has a probability of 1/6.

 p(x)  .25  .5  .25  1 .



The probability distribution of x may be summarized in tabular form: x

1

2

3

4

5

6

p(x)

1 6

1 6

1 6

1 6

1 6

1 6

The probability distribution of x may also be presented in graphical form:

p(x)

b.

1/6

0 1

2

3

4

5

6

x

4.16

151

a.

The sample points are (where H = head, T = tail):

x = # heads b.

HHH HHT HTH THH HTT THT TTH TTT 3 2 2 2 1 1 1 0

If each event is equally likely, then P(sample point) 

1 1  k 8

1 1 1 1 3 1 1 1 3 1 p(3)  , p(2)     , p (1)     , and p(0)  8 8 8 8 8 8 8 8 8 8

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152

Chapter 4 c.

Using Minitab, the graph of p(x) is:

.500

p(x)

.375

.250

.125

0 0

1

2

3

x

d. a.

3 1 4 1    8 8 8 2

  E( x)   xp( x)  4(.02)  (3)(.07)  (2)(.10)  (1)(.15)  0(.3)  1(.18)  2(.10)  3(.06)  4(.02)

 .08  .21  .2  .15  0  .18  .2  .18  .08  0

 2  E[( x   )2 ]   ( x   ) 2 p( x)

 ( 4  0) 2 (.02)  ( 3  0) 2 (.07)  ( 2  0) 2 (.10)  ( 1  0) 2 (.15)  (0  0) 2 (.30)  (1  0) 2 (.18)  (2  0) 2 (.10)  (3  0) 2 (.06)  (4  0) 2 (.02)  .32  .63  .4  .15  0  .18  .4  .54  .32  2.94

  2.94  1.715 b.

Using MINITAB, the graph is: Histogram of x .30

.25

.20 p(x)

4.17

P( x  2 or x  3)  p (2)  p(3) 

.15

.10

.05

0 -4

-3

  2

-2

-1

0

1

2

 0

3

4

  2

  2  0  2(1.715)  0  3.430  (3.430, 3.430) Copyright © 2014 Pearson Education, Inc.

Random Variables and Probability Distributions c.

4.18

a.

153

P(3.430  x  3.430)  p (3)  p (2)  p (1)  p (0)  p(1)  p (2)  p (3)  .07  .10  .15  .30  .18  .10  .06  .96

  E( x)   xp( x)

 10(.05)  20(.20)  30(.30)  40(.25)  50(.10)  60(.10)  .5  4  9  10  5  6  34.5

 2  E( x   )2   ( x   )2 p( x)

 (10  34.5)2 (.05)  (20  34.5) 2 (.20)  (30  34.5) 2 (.30)  (40  34.5)2 (.25)  (50  34.5) 2 (.10)  (60  34.5) 2 (.10)  30.0125  42.05  6.075  7.5625  24.025  65.025  174.75

  174.75  13.219 b.

Using MINITAB, the graph is: Histogram of x .30 .25

p(x)

.20 .15 .10 .05 0 10

20

30

40

50

60

x

  2

c.

  34.5

  2

  2  34.5  2(13.219)  34.5  26.438  (8.062, 60.938)

P(8.062  x  60.938)  p(10)  p(20)  p(30)  p(40)  p(50)  p(60)  .05  .20  .30  .25  .10  .10  1.00 4.19

a.

It would seem that the mean of both would be 1 since they both are symmetric distributions centered at 1.

b.

P(x) seems more variable since there appears to be greater probability for the two extreme values of 0 and 2 than there is in the distribution of y.

c.

For x:   E( x) 

 xp( x)  0(.3) 1(.4)  2(.3)  0  .4  .6  1

 2  E[( x   ) 2 ]   ( x   )2 p( x)

 (0  1)2 (.3)  (1  1)2 (.4)  (2  1)2 (.3)  .3  0  .3  .6

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154

Chapter 4 For y:   E( y) 

 yp( y)  0(.1) 1(.8)  2(.1)  0  .8  .2  1

 2  E[( y   )2 ]   ( y   ) 2 p( y )

 (0  1) 2 (.1)  (1  1) 2 (.8)  (2  1) 2 (.1)  .1  0  .1  .2

The variance for x is larger than that for y. 4.20

4.21

a.

The possible values of x are 1, 2, 3, and 4 or more.

b.

P( x  1)  .26

c.

P( x  4)  .25

d.

We cannot compute E(x) because the last value of x has more than one value (4 or more). To find the E(x), each possible value of x can have only one value.

a.

The probability distribution for x is found by converting the Percent column to a probability column by dividing the percents by 100. The probability distribution of x is: x 2 3 4 5

b.

P( x  5)  p(5)  .1837 .

c.

P( x  2)  p(2)  .0408 .

p(x) .0408 .1735 .6020 .1837

  E ( x)   xi p ( xi )  2(.0408) 3(.1735)  4(.6020)  5(.1837) 4

d.

i 1

 .0816  .5205  2.4080  .9185  3.9286  3.93

The average star rating for a car’s drivers-side star rating is 3.93. 4.22

a.

Yes. Relative frequencies are observed values from a sample. Relative frequencies are commonly used to estimate unknown probabilities. In addition, relative frequencies have the same properties as the probabilities in a probability distribution, namely 1. all relative frequencies are greater than or equal to zero 2. the sum of all the relative frequencies is 1

Copyright © 2014 Pearson Education, Inc.

Random Variables and Probability Distributions b.

155

Using MINITAB, the graph of the probability distribution is: .16 .14 .12

p(age)

.10 .08 .06 .04 .02 0 20

c.

22

24

26 age

28

30

32

Let x = age of employee. Then P( x  30)  .13  .15  .12  .40 .

P( x  40)  0 P( x  30)  .02  .04  .05  .07  .04  .02  .07  .02  .11  .07  .51

4.23

d.

P( x  25or x  26)  .02  .07  .09

a.

In order for this to be a valid probability distribution, all probabilities must be between 0 and 1 and the sum of all the probabilities must be 1. For this data, all the probabilities are between 0 and 1. If you sum all of the probabilities, the sum is 1.

b.

P( x  10)  P( x  10)  P( x  11)    P( x  20)

 .02  .02  .02  .02  .01  .01  .01  .01  .01  .005  .005  .14 c.

The mean of x is

  E ( x)   xp( x)  0(.17)  1(.10)  2(.11)    20(.005)  0  .1  .22  .33    .1  4.655

The variance of x is

 2  E ( x   ) 2   ( x   )2 p( x)  (0  4.655) 2 (.17)  (1  4.655) 2 (.1)  (2  4.655) 2 (.11)    (20  4.655)2 (.005)

 3.6837  1.3359  .7754    1.1773  19.8560 d.

From Chebyshev’s Rule, we know that at least .75 of the observations will fall within 2 standard deviations of the mean. The standard deviation is   19.8560  4.456 . The interval is:   2  4.655  2(4.456)  4.655  8.912  (4.257, 13.567) .

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156 4.24

Chapter 4 a.

The probability distribution for x is: Grill Display Combination 1-2-3 1-2-4 1-2-5 2-3-4 2-3-5 2-4-5

b. 4.25

x 6 7 8 9 10 11

p(x) 35 /124  .282 8 /124  .065 42 /124  .339 4 /124  .032 1/124  .008 34 /124  .274

P( x  10)  p(11)  .274

a.

The possible values of x are 0, 2, 3, and 4.

b.

To find the probability distribution of x, we first find the frequency distribution of x. We then divide the frequencies by n  106 to get the probabilities. The probability distribution of x is:

0 35 .3302

x f(x) p(x)

c.

2 58 .5472

3 5 .0472

4 8 .0755

  E( x)   xp( x)  0(.3302)  2(.5472)  3(.0472)  4(.0755)  1.538 . For all social robots, the

average number of legs on the robot is 1.538. 4.26

a.

For this problem, x = sequence number of a Florida tropical storm within a season that develops into a hurricane. Thus, x can take on values 1, 2, 3, … . Since this is a countable number of outcomes, x is a discrete random variable.

b.

The probability distribution of x is found by dividing the number of storms by the total number of storms, n = 67. The probability distribution of x is:

x 1 2 3 4 5 6 7 8

f(x) 4 10 5 6 11 5 5 5

p(x) .0597 .1493 .0746 .0896 .1642 .0746 .0746 .0746

x 9 10 11 12 13 14 15 22

f(x) 4 2 5 1 1 1 1 1

p(x) .0597 .0299 .0746 .0149 .0149 .0149 .0149 .0149

c.

P( x  5)  .1642

d.

P( x  5)  P( x  1)  P( x  2)  P( x  3)  P( x  4)  .0597  .1493  .0746  .0896  .3732

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Random Variables and Probability Distributions e.

157

The expected value is

  E ( x)   xp( x)  1(.0597)  2(.1493)  3(.0746)    22(.0149)  .0597  .2986  .2238    .3278  6.1174

The average sequence number of a Florida tropical storm within a season that develops into a hurricane is 6.1174.

4.27

f.

No, it is not likely. The probability is only .0149.

a.

The random variable x is a discrete random variable because it can take on only values 0, 1, 2, 3, 4, or 5 in this example.

b.

c.

p (0) 

5!(.35)0 (.65)5  0 5  4  3  2 1(1)(.65)5   .655  .1160 0!(5  0)! 1  5  4  3  2 1

p(1) 

5!(.35)1 (.65)51 5  4  3  2 1(.35)1 (.65) 4   5(.35)(.65) 4  .3124 1!(5  1)! 1  4  3  2 1

p(2) 

5!(.35) 2 (.65)5 2 5  4  3  2 1(.35) 2 (.65)3   10(.35) 2 (.65)3  .3364 2!(5  2)! 2 1  3  2 1

p(3) 

5!(.35)3 (.65)53 5  4  3  2 1(.35)3 (.65) 2   10(.35)3 (.65) 2  .1811 3!(5  3)! 3  2 1  2 1

p(4) 

5!(.35) 4 (.65)5 4 5  4  3  2 1(.35) 4 (.65)1   5(.35) 4 (.65)1  .0488 4!(5  4)! 4  3  2 1 1

p(5) 

5!(.35)5 (.65)55 5  4  3  2 1(.35)5 (.65)0   (.35)5  .0053 5!(5  5)! 5  4  3  2 1 1

The two properties of discrete random variables are that 0  p( x)  1 for all x and above, all probabilities are between 0 and 1 and

 p(x)  1 .

 p( x)  .1160  .3124  .3364  .1811  .0488  .0053  1

4.28

d.

P( x  4)  p(4)  p(5)  .0488  .0053  .0541

a.

First, we must find the probability distribution of x. Define the following events: C: {Chicken is contaminated} N: {Chicken is not contaminated} If 3 slaughtered chickens are randomly selected, then the possible outcomes are: CCC, CCN, CNC, NCC, CNN, NCN, NNC, and NNN Each of these outcomes are NOT equally likely since P(C )  1/100  .01 . Copyright © 2014 Pearson Education, Inc.

From

158

Chapter 4

P( N )  1  P(C )  1  .01  .99 . P(CCC )  P(C  C  C )  P(C ) P(C ) P(C )  .01(.01)(.01)  .000001 P(CCN )  P(CNC )  P( NCC )  P(C  C  N )  P(C ) P(C ) P( N )  .01(.01)(.99)  .000099 P(CNN )  P( NCN )  P( NNC )  P(C  N  N )  P(C ) P( N ) P( N )  .01(.99)(.99)  .009801 P( NNN )  P( N  N  N )  P( N ) P( N ) P( N )  .99(.99)(.99)  .970299 The variable x is defined as the number of contaminated chickens in the sample. The value of x for each of the outcomes is: x 3 2 2 2 1 1 1 0

Event CCC CCN CNC NCC CNN NCN NNC NNN

p(x) .000001 .000099 .000099 .000099 .009801 .009801 .009801 .970299

The probability distribution of x is: x 0 1 2 3

b.

p(x) .970299 .029403 .000297 .000001

Using MINITAB, the probability graph for x is: 1

.8

p(x)

.6

.4

.2

0 0

1

2

3

x

4.29

c.

P( x  1)  P( x  0)  P( x  1)  .970299  .029403  .999702

a.

p(1)  .23(.77)11  .23(.77)0  .23 . The probability that one would encounter a contaminated cartridge on the first trial is .23.

b.

p(5)  .23(.77)51  .23(.77)4  .0809 . The probability that one would encounter a the first contaminated cartridge on the fifth trial is .0809. Copyright © 2014 Pearson Education, Inc.

Random Variables and Probability Distributions

c.

4.30

159

P( x  2)  1  P( x  1)  1  P( x  1)  1  .23  .77 . The probability that the first contaminated cartridge is found on the second trial or later is .77.

a.

If the first letters of consumers’ last names are all equally likely, then P( x  i)  1/ 26 for i = 1, 2, …, 26.

b.

The expected value is

  E ( x)   xp( x)  1

 1   1   1   1    2    3      26    13.5  26   26   26   26 

The average number given to a consumer based on his last name is 13.5. c.

4.31

4.32

This probability distribution is probably not realistic. Very few consumers have last names that begin with Q or U. However, many consumers have last names that begin with S and T. One could estimate the true probability distribution of x by taking a random sample of names from a phone book and looking at the relative frequency distribution of the values of x assigned to the sampled names.

a.

 20   100  20  20! 80! 20! 80!    0   3 - 0  0!(20  0)! 3!(80  3)! 0!20! 3!77! 82,160  p (0)      .508 100! 100! 161, 700  100    3!(100  3)! 3!97!  3 

b.

 20   100  20  20! 80! 20! 80!    1   3 - 1  1!(20  1)! 2!(80  2)! 1!19! 2!78! 63, 200  p (1)      .391 100! 100! 161, 700  100    3!(100  3)! 3!97!  3 

c.

 20   100  20  20! 80! 20! 80!    2 3 2 2!(20  2)!1!(80  1)! 2!18!1!79! 15, 200     p (2)      .094 100! 100! 161, 700  100    3!(100  3)! 3!97!  3 

d.

 20   100  20  20! 80! 20!    1 3 3 0 1,140 3!(20  3)! 0!(80  0)! 3!17!     p (3)      .007 100! 100! 161, 700  100    3!(100  3)! 3!97!  3 

a.

E ( x) 

 xp( x)

All x

Firm A: E ( x )  0(.01)  500(.01)  1000(.01)  1500(.02)  2000(.35)  2500(.30)  3000(.25)  3500(.02)  4000(.01)  4500(.01)  5000(.01)  0  5  10  30  700  750  750  70  40  45  50  2450

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160

Chapter 4 Firm B: E ( x )  0(.00)  200(.01)  700(.02)  1200(.02)  1700(.15)  2200(.30)  2700(.30)  3200(.15)  3700(.02)  4200(.02)  4700(.01)  0  2  14  24  255  660  810  480  74  84  47  2450

b.

  2

2 

 (x  )

2

p ( x)

All x

Firm A:  2  (0  2450) 2 (.01)  (500  2450) 2 (.01)    (5000  2450) 2 (.01)  60, 025  38, 025  21, 025  18, 050  70,875  750  75, 625  22, 050  24, 025  42, 025  65, 025  437,500

  437,500  661.44 Firm B:  2  (0  2450) 2 (.00)  (200  2450) 2 (.01)    (4700  2450) 2 (.01)  0  50, 625  61, 250  31, 250  84,375  18, 750  84, 375  31, 250  61, 250  50, 625  492, 500

  492,500  701.78 Firm B faces greater risk of physical damage because it has a higher variance and standard deviation. 4.33

To find the probability distribution of x, we sum the probabilities associated with the same value of x. The probability distribution is: x p(x)

4.34

8.5 .462189

9 .288764

9.5 .141671

10 .069967

10.5 .025236

11 .011657

12 .000518

To determine which group of Finnish citizens has the highest average IQ score, we must find the expected value for each group. To do this, we first find the probability distribution for each group by dividing the frequency for each IQ level in each group by the group total. The probability distributions are: IQ 1 2 3 4 5 6 7 8 9

Invest .020 .030 .045 .120 .190 .230 .150 .115 .100

No Invest .041 .083 .088 .174 .217 .191 .099 .062 .045

For Investors,   E( x)   xp( x)  1(.020)  2(.030)  3(.045)    9(.100)  5.895 Copyright © 2014 Pearson Education, Inc.

Random Variables and Probability Distributions

For Non-investors,   E( x) 

 xp(x)  1(.041)  2(.083)  3(.088)  9(.045)  4.992

161

Thus, the investors had a higher average IQ than the non-investors. 4.35

a.

Let x = the potential flood damages. Since we are assuming if it rains the business will incur damages and if it does not rain the business will not incur any damages, the probability distribution of x is: 0 .7

x p(x) b.

300,000 .3

The expected loss due to flood damage is

E ( x)   xp( x)  0(.7)  300, 000(.3)  0  90, 000  $90, 000 All x

4.36

Let x = winnings in the Florida lottery. The probability distribution for x is: x $1 $6,999,999

p(x) 22,999,999/23,000,000 1/23,000,000

The expected net winnings would be:

  E ( x )   xp ( x)  ( 1)  All x

1  22, 999, 999      6,999,999    $.70 23, 000, 000 23, 000, 000    

The average winnings of all those who play the lottery is $.70. 4.37

a.

Since there are 20 possible outcomes that are all equally likely, the probability of any of the 20 numbers is 1/20. The probability distribution of x is:

P( x  5)  1/ 20  .05 ; P( x  10)  1/ 20  .05 ; etc. x

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

p(x) .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 b.

E( x)   xp( x)  5(.05)  10(.05)  15(.05)  20(.05)  25(.05)  30(.05)  35(.05)  40(.05)  45(.05)  50(.05)  55(.05)  60(.05)  65(.05)  70(.05)  75(.05)  80(.05)  85(.05)  90(.05)  95(.05)  100(.05)  52.5

c.

 2  E ( x   ) 2   ( x   ) 2 p( x)  (5  52.5) 2 (.05)  (10  52.5) 2 (.05)

 (15  52.5) 2 (.05)  (20  52.5) 2 (.05)  (25  52.5) 2 (.05)  (30  52.5) 2 (.05)  (35  52.5) 2 (.05)  (40  52.5) 2 (.05)  (45  52.5) 2 (.05)  (50  52.5) 2 (.05)  (55  52.5) 2 (.05)  (60  52.5) 2 (.05)  (65  52.5) 2 (.05)  (70  52.5) 2 (.05)  (75  52.5) 2 (.05)  (80  52.5) 2 (.05)  (85  52.5) 2 (.05)  (90  52.5) 2 (.05)  (95  52.5) 2 (.05)  (100  52.5) 2 (.05)  831.25

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162

Chapter 4

  831.25  28.83 Since the uniform distribution is not mound-shaped, we will use Chebyshev's theorem to describe the data. We know that at least 8/9 of the observations will fall with 3 standard deviations of the mean and at least 3/4 of the observations will fall within 2 standard deviations of the mean. For this problem,

  2  52.5  2(28.83)  52.5  57.66  (5.16, 110.16) . Thus, at least 3/4 of the data will fall between 5.16 and 110.16. For our problem, all of the observations will fall within 2 standard deviations of the mean. Thus, x is just as likely to fall within any interval of equal length. d.

If a player spins the wheel twice, the total number of outcomes will be 20(20) = 400. The sample space is: 5, 5 10, 5 15, 5 20, 5 25, 5... 100, 5 5,10 10,10 15,10 20,10 25,10... 100,10 5,15 10,15 15,15 20,15 25,15... 100,15 . . . . . . . . . . . . . . . . . . 5,100 10,100 15,100 20,100 25,100... 100,100 Each of these outcomes are equally likely, so each has a probability of 1/400 = .0025. Now, let x equal the sum of the two numbers in each sample. There is one sample with a sum of 10, two samples with a sum of 15, three samples with a sum of 20, etc. If the sum of the two numbers exceeds 100, then x is zero. The probability distribution of x is: x 0 10 15 20 25 30 35 40 45 50

e. f.

p(x) .5250 .0025 .0050 .0075 .0100 .0125 .0150 .0175 .0200 .0225

x 55 60 65 70 75 80 85 90 95 100

p(x) .0250 .0275 .0300 .0325 .0350 .0375 .0400 .0425 .0450 .0475

We assumed that the wheel is fair, or that all outcomes are equally likely.

  E( x)   xp( x)  0(.5250)  10(.0025)  15(.0050)  20(.0075)    100(.0475)  33.25

 2  E ( x   ) 2   ( x -  ) 2 p ( x)  (0  33.25) 2 (.5250)  (10  33.25) 2 (.0025)

 (15  33.25) 2 (.0050)  (20  33.25) 2 (.0075)    (100  33.25) 2 (.0475)  1, 471.3125

  1,471.3125  38.3577 g.

P ( x  0)  .525

Copyright © 2014 Pearson Education, Inc.

Random Variables and Probability Distributions

h.

Given that the player obtains a 20 on the first spin, the possible values for x (sum of the two spins) are 0 (player spins 85, 90, 95, or 100 on the second spin), 25, 30, ..., 100. To get an x of 25, the player would spin a 5 on the second spin. Similarly, the player would have to spin a 10 on the second spin order to get an x of 30, etc. Since all of the outcomes are equally likely on the second spin, the distribution of x is: x 0 25 30 35 40 45 50 55 60

p(x) .20 .05 .05 .05 .05 .05 .05 .05 .05

x 65 70 75 80 85 90 95 100

p(x) .05 .05 .05 .05 .05 .05 .05 .05

i.

The probability that the players total score will exceed one dollar is the probability that x is zero. P( x  0)  .20

j.

Given that the player obtains a 65 on the first spin, the possible values for x (sum of the two spins) are 0 (player spins 40, 45, 50, up to 100 on second spin), 70, 75, 80,..., 100. In order to get an x of 70, the player would spin a 5 on the second spin. Similarly, the player would have to spin a 10 on the second spin in order to get an x of 75, etc. Since all of the outcomes are equally likely on the second spin, the distribution of x is: x 0 70 75 80 85 90 95 100

p(x) .65 .05 .05 .05 .05 .05 .05 .05

The probability that the players total score will exceed one dollar is the probability that x is zero. P( x  0)  .65 . 4.38

163

a.

Each point in the system can have one of 2 status levels, “free” or “obstacle”. Define the following events: AF: {Point A is free} BF: {Point B is free} CF: {Point C is free}

AO: {Point A is obstacle} BO: {Point B is obstacle} CO: {Point C is obstacle}

Thus, the sample points for the space are: AFBFCF, AFBFCO, AFBOCF, AFBOCO, AOBFCF, AOBFCO, AOBOCF, AOBOCO Copyright © 2014 Pearson Education, Inc.

164

Chapter 4

b.

Since it is stated that the probability of any point in the system having a “free” status is .5, the probability of any point having an “obstacle” status is also .5. Thus, the probability of each of the sample points above is P( Ai Bi Ci )  .5(.5)(.5)  .125 . The values of Y, the number of free links in the system, for each sample point are listed below. A link is free if both the points are free. Thus, a link from A to B is free if A is free and B is free. A link from B to C is free if B is free and C is free. Sample point

Y

Probability

AFBFCF AFBFCO AFBOCF AFBOCO AOBFCF AOBFCO AOBOCF AOBOCO

2 1 0 0 1 0 0 0

.125 .125 .125 .125 .125 .125 .125 .125

The probability distribution for Y is: Y 0 1 2

4.39

Probability .625 .250 .125

Let x = bookie's earnings per dollar wagered. Then x can take on values $1 (you lose) and $-5 (you win). The only way you win is if you pick 3 winners in 3 games. If the probability of picking 1 winner in 1 game is .5, then P( www)  p( w) p( w) p( w)  .5(.5)(.5)  .125 (assuming games are independent). Thus, the probability distribution for x is: x

p(x)

$1 .875 $-5 .125

E( x)   xp( x)1(.875)  5(.125)  .875  .625  $.25 4.40

a.

6! 6! 6  5  4  3  2 1    15 2!(6  2)! 2!4! (2  1)(4  3  2  1)

b.

 5 5! 5! 5  4  3  2 1    10   2   1)(3  2  1) 2!(5 2)! 2!3! (2  

c.

7 7! 7! 7  6  5  4  3  2 1   1    0  0!(7  0)! 0!7! (1)(7  6  5  4  3  2  1)

(Note: 0! = 1)

Copyright © 2014 Pearson Education, Inc.

Random Variables and Probability Distributions

4.41

d.

6 6! 6! 6  5  4  3  2 1   1    6  6!(6  6)! 6!0! (6  5  4  3  2  1)(1)

e.

 4 4! 4! 4  3  2 1   4   3 3!(4 3)! 3!1! (3   2  1)(1)  

a.

x is discrete. It can take on only six values.

b.

This is a binomial distribution.

c.

165

 5 5! 5  4  3  2 1 (.7) 0 (.3)5  (1)(.00243)  .00243 p (0)    (.7) 0 (.3)5  0  0!5! 1  5  4  3  2 1 0 5 5! (.7)1 (.3) 4  .02835 p (1)    (.7)1 (.3)5 1  1 1!4!  

 5 5! (.7) 2 (.3)3  .1323 p (2)    (.7) 2 (.3)5  2  2 2!3!  

5 5! (.7)3 (.3) 2  .3087 p (3)    (.7)3 (.3)5  3  3 3!2!  

 5 5! (.7) 4 (.3)1  .36015 p (4)    (.7) 4 (.3)5 4  4 4!1!  

5 5! (.7)5 (.3) 0  .16807 p (5)    (.7)5 (.3)5  5  5!0! 5

Histogram of x .4

p(x)

.3

.2

.1

0 0

1

2

  2

4.42

3

4

  3.5

5

  2

  npq  5(.7)(.3)  1.0247

d.

  np  5(.7)  3.5

e.

  2  3.5  2(1.0247)  3.5  2.0494  (1.4506, 5.5494)

a.

 3 3! 3  2 1 p (0)    (.3) 0 (.7)3 0  (.3)0 (.7)3  (1)(.7)3  .343 0!3! 1  3  2 1 0  3 3! (.3)1 (.7) 2  .441 p (1)    (.3)1 (.7)31  1 1!2!  

 3 3! (.3) 2 (.7)1  .189 p (2)    (.3) 2 (.7)3 2  2 2!1!  

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166

Chapter 4  3 3! (.3)3 (.7) 0  .027 p (3)    (.3)3 (.7)3  3  3!0!  3

b.

4.43

4.44

The probability distribution in tabular form is: x

p(x)

0 1 2 3

.343 .441 .189 .027

a.

P( x  1) 

5! 5  4  3  2 1 (.2)1(.8) 4  (.2)1(.8) 4  5(.2)1 (.8)4  .4096 1!4! (1)(4  3  2  1)

b.

P( x  2) 

4! 4  3  2 1 (.6) 2(.4) 2  (.6) 2(.4) 2  6(.6) 2 (.4) 2  .3456 2!2! (2  1)(2  1)

c.

P( x  0) 

3! 3  2 1 (.7) 0(.3) 3  (.7) 0(.3) 3  1(.7)0 (.3)3  .027 0!3! (1)(3  2  1)

d.

P( x  3) 

5! 5  4  3  2 1 (.1) 3(.9) 2  (.1) 3(.9) 2  10(.1)3 (.9) 2  .0081 3!2! (3  2  1)(2  1)

e.

P( x  2) 

4! 4  3  2 1 (.4) 2(.6) 2  (.4) 2(.6) 2  6(.4)2 (.6) 2  .3456 2!2! (2  1)(2  1)

f.

P( x  1) 

a.

P  x  2  P( x  2)  P( x  1)  .167  .046  .121 (from Table I, Appendix D with n = 10 and p = .4)

b.

P ( x  5)  .034

c.

P( x  1)  1- P( x  1)  1  .919  .081

d.

P( x  10)  P( x  9)  0

e.

P( x  10)  1  P( x  9)  1  .002  .998

f.

P( x  2)  P( x  2)  P( x  1)  .206  .069  .137

3! 3  2 1 (.9)1(.1) 2  (.9)1(.1) 2  3(.9)1 (.1)2  .027 1!2! (1)(2  1)

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Random Variables and Probability Distributions

4.45

a.

167

  np  25(.5)  12.5

 2  np(1  p)  25(.5)(.5)  6.25 and    2  6.25  2.5 b.

  np  80(.2)  16

 2  np(1  p)  80(.2)(.8)  12.8 and    2  12.8  3.578 c.

  np  100(.6)  60

 2  np(1  p)  100(.6)(.4)  24 and    2  24  4.899 d.

  np  70(.9)  63

 2  np(1  p)  70(.9)(.1)  6.3 and    2  6.3  2.510 e.

  np  60(.8)  48

 2  np(1  p)  60(.8)(.2)  9.6 and    2  9.6  3.098 f.

  np  1,000(.04)  40

 2  np(1  p)  1,000(.04)(.96)  38.4 and    2  38.4  6.197 4.46

x is a binomial random variable with n = 4. a.

If the probability distribution of x is symmetric, p(0) = p(4) and p(1) = p(3). n We know p ( x )    p x q n  x x = 0, 1, ... , n,  x When n = 4,

 4 4 4! 0 4 4! 4 0 4 4 p (0)  p (4)    p 0 q 4    p 4 q 0  p q  p q q  p  pq 0!4! 4!0!  0 4

Since p  q  1 , p = .5 Therefore, the probability distribution of x is symmetric when p = .5. b.

If the probability distribution of x is skewed to the right, then the mean is greater than the median. Therefore, there are more small values in the distribution (0, 1) than large values (3, 4). For this to happen, p must be smaller than .5. If we pick p  .2 , the probability distribution of x will be skewed to the right.

c.

If the probability distribution of x is skewed to the left, then the mean is smaller than the median. Therefore, there are more large values in the distribution (3, 4) than small values (0, 1). For this to happen, p must be larger than .5. If we pick p  .8 , the probability distribution of x will be skewed to the left. Copyright © 2014 Pearson Education, Inc.

Chapter 4

d.

In part a, x is a binomial random variable with n = 4 and p = .5.  4 p ( x )    .5 x.54  x  x

x = 0, 1, 2, 3, 4

4 4! 4 .5  1(.5) 4  .0625 p (0)    .5 0.5 4  0 0!4!  

 4 4! 4 .5  4(.5) 4  .25 p (1)    .51.5 3  1 1!3!  

 4 4! 4 .5  6(.5) 4  .375 p (2)    .5 2.5 2  2!2!  2

p(3)  p(1)  .25 (since the distribution is symmetric) p(4)  p(0)  .0625 The probability distribution of x in tabular form is: x

0

1

2

3

4

p(x)

.0625

.25

.375

.25

.0625

  np  4(.5)  2 Using MINITAB, the graph of the probability distribution of x when n  4 and p  .5 is as follows. Histogram of x .4

.3

p(x)

168

.2

.1

0 0

1

2 x

3

4

 2 In part b, x is a binomial random variable with n  4 and p  .2 .  4 p ( x )    .2 x.8 4  x  x

x = 0, 1, 2, 3, 4

 4 p (0)    .20.84  1(1).84  .4096  0

4 p (1)    .21.83  4(.2).83  .4096  1

Copyright © 2014 Pearson Education, Inc.

Random Variables and Probability Distributions  4 p (3)    .23.81  4(.2)3 (.8)  .0256  3

 4 p (2)    .2 2.82  6(.2) 2 .82  .1536  2  4 p(4)    .24.80  1(.2) 4 (1)  .0016  4

The probability distribution of x in tabular form is: x

0

1

2

3

4

p(x)

.4096

.4096

.1536

.0256

.0016

  np  4(.2)  .8 . Using MINITAB, the graph of the probability distribution of x when n  4 and p  .2 is as follows: Histogram of x .4

p(x)

.3

.2

.1

0 0

1

2 x

3

4

  .8 In part c, x is a binomial random variable with n  4 and p  .8 .  4 p ( x )    .8 x.2 4- x  x

x = 0, 1, 2, 3, 4

 4 p (0)    .80.2 4  1(1).2 4  .0016  0

4 p (1)    .81.23  4(.8).23  .0256  1

 4 p (2)    .82.2 2  6(.8) 2 .2 2  .1536  2

 4 p (3)    .83.21  4(.8)3 .2  .4096  3

 4 p(4)    .84.20  1(.8) 4 (1)  .4096  4

Copyright © 2014 Pearson Education, Inc.

169

170

Chapter 4

The probability distribution of x in tabular form is: x

0

1

2

3

4

p(x)

.0016

.0256

.1536

.4096

.4096

Note: The distribution of x when n  4 and p  .8 is the reverse of the distribution of x when n  4 and p  .8 .

  np  4(.8)  3.2 Using MINITAB, the graph of the probability distribution of x when n  4 and p  .8 is as follows: Histogram of x .4

p(x)

.3

.2

.1

0 0

1

2 x

3

4

  3.2

4.47

e.

In general, when p  .5 , a binomial distribution will be symmetric regardless of the value of n. When p is less than .5, the binomial distribution will be skewed to the right; and when p is greater than .5, it will be skewed to the left. (Refer to parts a, b, and c.)

a.

Let S = adult who does not work while on summer vacation.

b.

To see if x is approximately a binomial random variable we check the characteristics: 1.

n identical trials. Although the trials are not exactly identical, they are close. Taking a sample of reasonable size n from a very large population will result in trials being essentially identical.

2.

Two possible outcomes. The adults can either not work on their summer vacation or they can work on their summer vacation. S = adult does not work on summer vacation and F = adult does work on summer vacation.

3.

P(S) remains the same from trial to trial. If we sample without replacement, then P(S) will change slightly from trial to trial. However, the differences are extremely small and will essentially be 0.

4.

Trials are independent. Again, although the trials are not exactly independent, they are very close.

5.

The random variable x = number of adults who work on their summer vacation in n = 10 trials.

Thus, x is very close to being a binomial. We will assume that it is a binomial random variable. Copyright © 2014 Pearson Education, Inc.

Random Variables and Probability Distributions

c.

For this problem, p  .35 .

d.

Using MINITAB n  10 and p  .35 , the probability is:

171

Probability Density Function Binomial with n = 10 and p = 0.35 x 3

P( X = x ) 0.252220

Thus, P( x  3)  .2522 . e.

Using MINITAB n  10 and p  .35 , the probability is: Cumulative Distribution Function Binomial with n = 10 and p = 0.35 x 2

P( X