Quantitative Methods For Business, 12th Edition Solution Manual

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Solutions Manual to Accompany
Quantitative Methods for
Business
Twelfth Edition

Contents
Preface
Chapter 1: Introduction
Chapter 2: Introduction to Probability
Chapter 3: Probability Distributions
Chapter 4: Decision Analysis
Chapter 5: Utility and Game Theory
Chapter 6: Forecasting
Chapter 7: Introduction to Linear Programming
Chapter 8: Linear Programming: Sensitivity Analysis and Interpretation of Solution
Chapter 9: Linear Programming Applications in Marketing, Finance, and Operations Management
Chapter 10: Distribution and Network Models
Chapter 11: Integer Linear Programming
Chapter 12: Advanced Optimization Applications
Chapter 13: Project Scheduling: PERT/CPM
Chapter 14: Inventory Models
Chapter 15: Waiting Line Models
Chapter 16: Simulation
Chapter 17: Markov Processes
Appendix A: Building Spreadsheet Models

Preface
The purpose of Quantitative Methods for Business is to provide students with a sound
conceptual understanding of the role quantitative methods play in the decision-making
process. The text emphasizes the application of quantitative methods by using
problem situations to introduce each of the quantitative methods, concepts and
techniques. The book has been specifically designed to meet the needs of
nonmathematicians who are studying business.
The Solutions Manual provides assistance in the form of identifying learning
objectives and providing detailed solutions for all problems in the text. Specific
sections of the manual are as follows:
Learning Objectives: Each chapter begins with an itemized list of learning
objectives and key terms.
Problem Solutions: The major portion of the manual is devoted to providing
completely worked out solutions to all text problems. We have included the equations
used as well as the computational results and answers.
Note: The solutions to the case problems are included in the Solutions to Case
Problems Manual.
David R. Anderson
Dennis J. Sweeney
Thomas A. Williams
Jeffrey D. Camm
James J. Cochran
Michael J. Fry
Jeffrey W. Ohlmann

1 - 1
Chapter 1
Introduction
Learning Objectives
1. Develop a general understanding of the management science/operations research approach to decision
making.
2. Realize that quantitative applications begin with a problem situation.
3. Obtain a brief introduction to quantitative techniques and their frequency of use in practice.
4. Understand that managerial problem situations have both quantitative and qualitative considerations
that are important in the decision making process.
5. Learn about models in terms of what they are and why they are useful (the emphasis is on mathematical
models).
6. Identify the step-by-step procedure that is used in most quantitative approaches to decision making.
7. Learn about basic models of cost, revenue, and profit and be able to compute the breakeven point.
8. Obtain an introduction to the use of computer software packages such as Microsoft Excel in applying
quantitative methods to decision making.
9. Understand the following terms:
model infeasible solution
objective function management science
constraint operations research
deterministic model fixed cost
stochastic model variable cost
feasible solution breakeven point
Solutions:

Chapter 1
1 - 2
1. Management science and operations research, terms used almost interchangeably, are broad
disciplines that employ scientific methodology in managerial decision making or problem
solving. Drawing upon a variety of disciplines (behavioral, mathematical, etc.), management
science and operations research combine quantitative and qualitative considerations in order to
establish policies and decisions that are in the best interest of the organization.
2. Define the problem
Identify the alternatives
Determine the criteria
Evaluate the alternatives
Choose an alternative
For further discussion see section 1.3
3. See section 1.2.
4. A quantitative approach should be considered because the problem is large, complex, important,
new and repetitive.
5. Models usually have time, cost, and risk advantages over experimenting with actual situations.
6. Model (a) may be quicker to formulate, easier to solve, and/or more easily understood.
7. Let d = distance
m = miles per gallon
c = cost per gallon,
Therefore Total Cost = 2d c
m
 
 
 
We must be willing to treat m and c as known and not subject to variation.
8. a. Maximize 10x + 5y
s.t.
5x + 2y  40
x  0, y  0
b. Controllable inputs: x and y
Uncontrollable inputs: profit (10,5), labor hours (5,2) and labor-hour availability (40)
c.

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Introduction
1 - 3
P rofit :
Labor Hours: 5/unit forx
2/ unit fory
$10/unit forx
$ 5/ unit fory
40 labor-hour capacit y
Un con trol lable Inputs
P roduct ion Quant ities
x and y
Con troll abl e
Input
P roject ed P rofit and
check on product ion
t ime const raint
O u tpu t
Max 10x + 5y
s.t.
10 x y+ 5 40
x
y
0
0

Mathe matical
Model
d. x = 0, y = 20 Profit = $100
(Solution by trial-and-error)
e. Deterministic - all uncontrollable inputs are fixed and known.
9. If a = 3, x = 13 1/3 and profit = 133
If a = 4, x = 10 and profit = 100
If a = 5, x = 8 and profit = 80
If a = 6, x = 6 2/3 and profit = 67
Since a is unknown, the actual values of x and profit are not known with certainty.
10. a. Total Units Received = x + y
b. Total Cost = 0.20x +0.25y
c. x + y = 5000
d. x  4000 Kansas City Constraint
y  3000 Minneapolis Constraint
e. Min 0.20x + 0.25y
s.t.
x + y = 5000
x  4000
y  3000
x, y  0

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Chapter 1
1 - 4
11. a. At the $20 per-unit price, the firm can sell 800 – 10(20) = 600 units. At the $70 per-unit price, the firm can
sell 800 – 10(70) = 100 units.
b. If the firm increases the per-unit price from $26 to $27, the annual demand for the product in units
decreases from 800 – 10(26) = 540 to 800 – 10(27) = 530. If the firm increases the per-unit price
from $42 to $43, the annual demand for the product in units decreases from 800 – 10(42) = 380 to
800 – 10(43) = 370. If the firm increases the per-unit price from $68 to $69, the annual demand for
the product in units decreases from 800 – 10(68) = 120 to 800 – 10(69) = 110. These results suggest
that a one dollar increase in price results in a 10 unit decrease in demand.
c. Since total revenue (TR) is the product of demand (d) and price (p), we have that TR = dp = (800 –
10p)p = 800p – 10p2.
d. At $30, TR = 800(30) – 10(30)2 = 24,000 – 10(900) = 15,000.
At $40, TR = 800(40) – 10(40)2 = 32,000 – 10(1600) = 16,000.
At $50, TR = 800(50) – 10(50)2 = 40,000 – 10(2500) = 15,000.
When considering price alternatives of $30, $40, and $50, total revenue is maximized at the $40
price (i.e., p = 40).
e. The expected annual demand and the total revenue according to the recommended price of $40
(from part d) are:
d = 800 – 10p = 800 – 10(40) = 400 units
TR = 800p – 10p2 = 800(40) – 10(40)2 = $16,000
12. a. If x represents the number of pairs of shoes produced, a mathematical model for the total cost of producing
x pairs of shoes is TC = 2000 + 60x. The two components of total cost in this model are fixed cost ($2,000)
and variable cost (60x).
b. If P represents the total profit, the total revenue (TR) is 80x and a mathematical model for the total
profit realized from an order for x pairs of shoes is P = TR – TC = 80x – (2000+60x) = 20x – 2000.
c. The breakeven point is the number of shoes produced (x) at the point of no profit (P = 0).
Thus the breakeven point is the value of x when P = 20x – 2000 = 0. This occurs when 20x = 2000 or
x = 100, i.e., the breakeven point is 100 pairs of shoes.
13. a. If x represents the number of students who enroll in the seminar, a model for the total cost to put on the
seminar is 9600 + 60(2x) = 9600 + 120x (note that the variable cost per student is $60 per day, and the
seminar is scheduled to last for two days, so total variable cost per student will be $120).
b. A model for the total profit if x students enroll in the seminar is total revenue (TR) – total cost (TC),
which is 600x – (9600 + 120x) = 480x - 9600.
c. For Micromedia’s forecasted enrollment of 30 students, the seminar will earn 480(30) – 9600 = 4800
if its forecast is accurate.
d. The breakeven point is the number of shoes produced (x) at the point of no profit (P = 0).
Thus the breakeven point is the value of x when P = 480x - 9600 = 0. This occurs when 480x = 9600
or x = 20, i.e., the breakeven point for the seminar is 20 students.
14. a. If x represents the number of copies of the book that are sold, total revenue (TR) = 46x and total cost (TC)
= 160,000 + 6x, so Profit = TR – TC = 46x – (160,000 + 6x) = 40x – 160,000. The breakeven point is the
number of books produced (x) at the point of no profit (P = 0). Thus the breakeven point is the value of x
when P = 40x - 160,000 = 0. This occurs when 40x = 160,000 or x = 4000, i.e., the breakeven point is 4000
copies of the book.

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Introduction
1 - 5
b. At a demand of 3800 copies, the publisher can expect a profit of 40(3800) – 160,000 = 152,000 – 160,000
= -8000, i.e., a loss of $8,000.
c. Here we know demand (d = 3800) and want to determine the price p at which we will breakeven (the
point at which profit is 0). The minimum price per copy that the publisher must charge to break even
is Profit = p(3800) – (160,000 + 6(3800)) = 3800p - 182,800. This occurs whre 3800p = 182,800 or
p = 48.10526316 or a price of approximately $48.
d. If the publisher believes demand will remain at 4000 copies if the price per copy is increased to
$50.95 , then the publisher could anticipate a profit of TR – TC = 50.95(4000) – (160,000 + 6(4000))
= 203,800 - 184,000 = 19,800 or a profit of $19,800. This is a return of p/TC = 10.8% on the total
cost of $184,000, and the publisher should procede if this return is sufficient.
15. a. If x represents the number of luxury boxes that are constructed, total revenue (TR) = 300,000x and total
cost (TC) = 4,500,000 + 150,000x, so Profit = TR – TC = 300,000x – (4,500,000 + 150,000x) = 150,000x –
4,500,000. The breakeven point is the number of luxury boxes produced (x) at the point of no profit (P = 0).
Thus the breakeven point is the value of x when P = 150,000x – 4,500,000 = 0. This occurs when 150,000x
= 4,500,000 or x = 30, i.e., the breakeven point is 30 luxury boxes.
b. The anticipated profit from this decision is P = 150,000(50) – 4,500,000 = 7,500,000-4,500,000 =
$3,000,000. Since this represents a return of p/TC = 25% on the total cost of $12,000,000, city
officials should authorize construction of 50 luxury boxes.
16. a. The annual return per share of Oil Alaska is $6.00 and the annual return per share of Southwest Petroleum
is $4.00, so the objective function that maximizes the total annual return is Max 6x + 4y.
b. The price per share of Oil Alaska is $50.00 and the price per share of Southwest Petroleum is $30.00,
so
(1) the mathematical expression for the constraint that limits total investment funds to $800,000 is
50x + 30y ≤ 800000,
(2) the mathematical expression for the constraint that limits investment in Oil Alaska to $500,000
is 50x ≤ 500000, and
(3) the mathematical expression for the constraint that limits investment in Southwest Petroleum to
$450,000 is 30y ≤ 450000.
17. a. sj = sj - 1 + xj - dj
or sj - sj-1 - xj + dj = 0
b. xj  cj
c. sj  Ij

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2 - 1
Chapter 2
Introduction to Probability
Learning Objectives
1. Obtain an understanding of the role probability information plays in the decision making process.
2. Understand probability as a numerical measure of the likelihood of occurrence.
3. Be able to use the three methods (classical, relative frequency, and subjective) commonly used for
assigning probabilities and understand when they should be used.
4. Be able to use the addition law and be able to compute the probabilities of events using conditional
probability and the multiplication law.
5. Be able to use new information to revise initial (prior) probability estimates using Bayes' theorem.
6. Know the definition of the following terms:
experiment addition law
sample space mutually exclusive
event conditional probability
complement independent events
Venn Diagram multiplication law
union of events prior probability
intersection of events posterior probability
Bayes' theorem Simpson’s Paradox

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Chapter 2
2 - 2
Solutions:
1. a. Go to the x-ray department at 9:00 a.m. and record the number of persons waiting.
b. The experimental outcomes (sample points) are the number of people waiting: 0, 1, 2, 3, and 4.
Note: While it is theoretically possible for more than 4 people to be waiting, we use what has
actually been observed to define the experimental outcomes.
c.
Number Waiting Probability
0 .10
1 .25
2 .30
3 .20
4 .15
Total: 1.00
d. The relative frequency method was used.
2. a. Choose a person at random, have her/ him taste the 4 blends and state a preference.
b. Assign a probability of 1/4 to each blend. We use the classical method of equally likely outcomes
here.
c.
Blend Probability
1 .20
2 .30
3 .35
4 .15
Total: 1.00
The relative frequency method was used.
3. Initially a probability of .20 would be assigned if selection is equally likely. Data does not appear to
confirm the belief of equal consumer preference. For example using the relative frequency method
we would assign a probability of 5 / 100 = .05 to the design 1 outcome, .15 to design 2, .30 to
design 3, .40 to design 4, and .10 to design 5.
4. a. Of the 132,275,830 individual tax returns received by the IRS in 2006, 31,675,935were in the
1040A, Income Under $25,000 category. Using the relative frequency approach, the probability a
return from the 1040A, Income Under $25,000 category would be chosen at random is
31675935/132275830 = 0.239.
b. Of the 132,275,830 individual tax returns received by the IRS in 2006, 3,376,943 were in the
Schedule C, Reciepts Under $25,000 category; 3,867,743 were in the Schedule C, Reciepts $25,000-
$100,000 category; and were 2,288,550 in the Schedule C, Reciepts $100,000 & Over category.
Therefore, 9,533,236 Schedule Cs were filed in 2006, and the remaining 132,275,830 - 9,533,236 =
122,742,594 individual returns did not use Schedule C. By the relative frequency approach, the
probability the chosen return did not use Schedule C is 122742594/132275830 = 0.928.
c. Of the 132,275,830 individual tax returns received by the IRS in 2006, 12,893,802 were in the Non
1040A, Income $100,000 & Over category; 2,288,550 were in the Schedule C, Reciepts $100,000 &

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Introduction to Probability
2 - 3
Over category; and 265,612 were in the Schedule F, Reciepts $100,000 & Over category. By the
relative frequency approach, the probability the chosen return reported income/reciepts of $100,000
and over is (12893802 + 2288550 + 265612)/132275830 = 15447964/132275830 = 0.117.
d. 26,463,973 Non 1040A, Income $50,000-$100,000 returns were filed in 2006, so assuming
examined returns were evenly distributed across the ten categories (i.e., the IRS examined 1% of
individual returns in each category), the number of returns from the Non 1040A, Income $50,000-
$100,000 category that were examined is 0.01(26463973) = 264,639.73 (or 264,640).
e. The proportion of total 2006 returns in the Schedule C, reciepts $100,000 & Over is
2,288,550/132,275,830 = 0.0173. Therefore, if we assume the recommended additional taxes are
evenly distributed across the ten categories, the amount of recommended additional taxes for the
Schedule C, Reciepts $100,000 & Over category is 0.0173($13,045,221,000.00) = $225,699,891.81.
5. a. No, the probabilities do not sum to one. They sum to 0.85.
b. Owner must revise the probabilities so that they sum to 1.00.
6. a. P(A) = P(150 - 199) + P(200 and over)
= 26 5
100 100

= 0.31
b. P(B) = P(less than 50) + P(50 - 99) + P(100 - 149)
= 0.13 + 0.22 + 0.34
= 0.69
7. a. P(A) = .40, P(B) = .40, P(C) = .60
b. P(A  B) = P(E1, E2, E3, E4) = .80. Yes P(A  B) = P(A) + P(B).
c. Ac = {E3, E4, E5} Cc = {E1, E4} P(Ac) = .60 P(Cc) = .40
d. A  Bc = {E1, E2, E5} P(A  Bc) = .60
e. P(B  C) = P(E2, E3, E4, E5) = .80
8. a. Let P(A) be the probability a hospital had a daily inpatient volume of at least 200 and P(B) be the
probability a hospital had a nurse to patient ratio of at least 3.0. From the list of thirty hospitals, sixteen
had a daily inpatient volume of at least 200, so by the relative frequency approach the probability one of
these hospitals had a daily inpatient volume of at least 200 is P(A) = 16/30 = 0.533, Similarly, since ten
(one-third) of the hospitals had a nurse-to-patient ratio of at least 3.0, the probability of a hospital having
a nurse-to-patient ratio of at least 3.0 is P(B) = 10/30 = 0.333. Finally, since seven of the hospitals had
both a daily inpatient volume of at least 200 and a nurse-to-patient ratio of at least 3.0, the probability of a
hospital having both a daily inpatient volume of at least 200 and a nurse-to-patient ratio of at least 3.0
is P(A∩B) = 7/30 = 0.233.
b. The probability that a hospital had a daily inpatient volume of at least 200 or a nurse to patient ratio of
at least 3.0 or both is P(A U B) = P(A) + P(B) - P(A∩B) = 16/30 + 10/30 – 7/30 = (16 + 10 – 7)/30 = 19/30
= 0.633.
c. The probability that a hospital had neither a daily inpatient volume of at least 200 nor a nurse to patient
ratio of at least 3.0 is 1 – P(A U B) = 1 - 19/30 = 11/30 = 0.367.

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Chapter 2
2 - 4
9. Let E = event patient treated experienced eye relief.
S = event patient treated had skin rash clear up.
Given:
P (E) = 90 / 250 = 0.36
P (S) = 135 / 250 = 0.54
P (E  S) = 45 / 250 = 0.18
P (E  S ) = P (E) + P (S) - P (E  S)
= 0.36 + 0.54 - 0.18
= 0.72
10. P(Defective and Minor) = 4/25
P(Defective and Major) = 2/25
P(Defective) = (4/25) + (2/25) = 6/25
P(Major Defect | Defective) = P(Defective and Major) / P(Defective) = (2/25)/(6/25) = 2/6 = 1/3.
11. a. Yes; the person cannot be in an automobile and a bus at the same time.
b. P(Bc) = 1 - P(B) = 1 - 0.35 = 0.65
12. a. P(A B) 0.40
P(A B) 0.6667
P(B) 0.60

  
b. P(A B) 0.40
P(B A) 0.80
P(A) 0.50

  
c. No because P(A | B)  P(A)
13. a.
Reason for Applying
Quality Cost/Convenience Other Total
Full Time 0.218 0.204 0.039 0.461
Part Time 0.208 0.307 0.024 0.539
Total 0.426 0.511 0.063 1.00
b. It is most likely a student will cite cost or convenience as the first reason: probability = 0.511. School
quality is the first reason cited by the second largest number of students: probability = 0.426.
c. P (Qualityfull time) = 0.218/0.461 = 0.473
d. P (Qualitypart time) = 0.208/0.539 = 0.386
e. P (B) = 0.426 and P (BA) = 0.473
Since P (B)  P (BA), the events are dependent.

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Introduction to Probability
2 - 5
14.
$0-$499 $500-$999 >=$1000
<2 yrs 120 240 90 450
>= 2 yrs 75 275 200 550
195 515 290 1000
$0-$499 $500-$999 >=$1000
<2 yrs 0.12 0.24 0.09 0.45
>= 2 yrs 0.075 0.275 0.2 0.55
0.195 0.515 0.29 1.00
a. P(< 2 yrs) = .45
b. P(>= $1000) = .29
c. P(2 accounts have > = $1000) = (.29)(.29) = .0841
d. P($500-$999 | >= 2 yrs) = P($500-$999 and >= 2 yrs) / P(>=2yrs) = .275/.55 = .5
e. P(< 2 yrs and >=$1000) = .09
f. P(>=2 yrs | $500-$999) = .275/.515 = .533981
15. a. A joint probability table for these data looks like this:
Automobile Insurance Coverage
Yes No Total
Age 18 to 34 .375 .085 .46
35 and over .475 .065 .54
Total .850 .150 1.00
For parts (b) through (g):
Let A= 18 to 34 age group
B= 35 and over age group
Y = Has automobile insurance coverage
N = Does not have automobile insurance coverage
b. We have P(A) = .46 and P(B) = .54, so of the population age 18 and over, 46% are ages 18 to 34 and
54% are ages 35 and over.
c. The probability a randomly selected individual does not have automobile insurance coverage is P(N)
= .15.
d. If the individual is between the ages of 18 and 34, the probability the individual does not have
automobile insurance coverage is
   
 
P .085
P = = = .1848.
P .46
N A
N A A

e. If the individual is age 35 or over, the probability the individual does not have automobile insurance
coverage is

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Chapter 2
2 - 6
   
 
P .065
P B = = = .1204.
P .54
N B
N B

f. If the individual does not have automobile insurance, the probability that the individual is in the 18–
34 age group is
   
 
P .085
P = = = .5667.
P .15
A N
A N N

g. The probability information tells us that in the US, younger drivers are less likely to have automobile
insurance coverage.
16. a. P(A  B) = P(A)P(B) = (0.55)(0.35) = 0.19
b. P(A  B) = P(A) + P(B) - P(A  B) = 0.90 - 0.19 = 0.71
c. 1 - 0.71 = 0.29
17. a. P(attend multiple games) = 196 / 989 ≈ 19.8%.
b. P(male | attend multiple games) = 177 / 196 ≈ 90.3%.
c. P(male and attend multiple games) = P(male | attend multiple games) × P(attend multiple games) =
(177 / 196) × (196 / 989) = 177 / 989 ≈ 17.9%.
d. P(attend multiple games | male) = P(attend multiple games and male) / P(male) = (177 / 989) / (759 /
989) = 177 / 759 ≈ 23.3%.
e. P(male or attend multiple games) = P(male) + P(attend multiple games) – P(male and attend multiple
games) = (759 / 989) + (196 / 989) – (177 / 989) = 778 / 989 ≈ 78.7%.
18. a. P(B) = 0.25
P(SB) = 0.40
P(S  B) = 0.25(0.40) = 0.10
b. P(S B) 0.10
P(B S) 0.25
P(S) 0.40

  
c. B and S are independent. The program appears to have no effect.
19. Let: A = lost time accident in current year
B = lost time accident previous year
 Given: P(B) = 0.06, P(A) = 0.05, P(AB) = 0.15
a. P(A  B) = P(AB)P(B) = 0.15(0.06) = 0.009
b. P(A  B) = P(A) + P(B) - P(A B)
= 0.06 + 0.05 - 0.009 = 0.101 or 10.1%
20. a. P(B  A1) = P(A1)P(BA1) = (0.20)(0.50) = 0.10
P(B  A2) = P(A2)P(BA2) = (0.50)(0.40) = 0.20

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Introduction to Probability
2 - 7
P(B  A3) = P(A3)P(BA3) = (0.30)(0.30) = 0.09
b. 2
0.20
P(A B) 0.51
0.10 0.20 0.09
 
 
c.
Events P(Ai) P(BAi) P(Ai  B) P(Ai B)
A1 0.20 0.50 0.10 0.26
A2 0.50 0.40 0.20 0.51
A3 0.30 0.30 0.09 0.23
1.00 0.39 1.00
21. S1 = successful, S2 = not successful and B = request received for additional information.
a. P(S1) = 0.50
b. P(BS1) = 0.75
c. 1
(0.50)(0.75) 0.375
P(S B) 0.65
(0.50)(0.75) (0.50)(0.40) 0.575
  

22. a. Let F = female. Using past history as a guide, P(F) = .40
b. Let D = Dillard's
.40(3/ 4) .30
P(F D) .67
.40(3/ 4) .60(1/ 4) .30 .15
  
 
The revised (posterior) probability that the visitor is female is .67.
We should display the offer that appeals to female visitors.
23. a. P(Oil) = 0.50 + 0.20 = 0.70
b. Let S = Soil test results
Events P(Ai) P(SAi) P(Ai  S) P(Ai S)
High Quality (A1) 0.50 0.20 0.10 0.31
Medium Quality (A2) 0.20 0.80 0.16 0.50
No Oil (A3) 0.30 0.20 0.06 0.19
1.00 P(S) = 0.32 1.00
P(Oil) = 0.81 which is good; however, probabilities now favor medium quality rather than high
quality oil.
24. Let S= speeding is reported
SC= speeding is not reported
F = Accident results in fatality for vehicle occupant
We have P(S) = .129, so P(SC) = .871. Also P(F|S) = .196 and P(F|SC) = .05. Using the tabular form

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Chapter 2
2 - 8
of Bayes’ Theorem provides:
Prior Conditional Joint Posterior
Events Probabilities Probabilities Probabilities Probabilities
S .129 .196 .0384 .939
SC .871 .050 .0025 .061
1.000 P(F) = .0409 1.000
P(S | F) = .2195, i.e., if an accident involved a fatality. the probability speeding was reported is
0.939.
25.
Events P(Ai) P(DAi) P(AiD) P(AiD)
Supplier A 0.60 0.0025 0.0015 0.23
Supplier B 0.30 0.0100 0.0030 0.46
Supplier C 0.10 0.0200 0.0020 0.31
1.00 P(D) = 0.0065 1.00
a. P(D) = 0.0065
b. B is the most likely supplier if a defect is found.
26. a.
Events P(Di) P(S1|Di) P(Di S1) P(Di |S1)
D1 .60 .15 .090 .2195
D2 .40 .80 .320 .7805
1.00 P(S1) = .410 1.0000
P(D1 | S1) = .2195
P(D2 | S1) = .7805
b.
Events P(Di) P(S2 |Di) P(Di S2) P(Di |S2)
D1 .60 .10 .060 .500
D2 .40 .15 .060 .500
1.00 P(S2) = .120 1.000
P(D1 | S2) = .50
P(D2 | S2) = .50
c.
Events P(Di) P(S3 |Di) P(Di S3) P(Di |S3)
D1 .60 .15 .090 .8824
D2 .40 .03 .012 .1176
1.00 P(S3) = .102 1.0000
P(D1 | S3) = .8824

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Introduction to Probability
2 - 9
P(D2 | S3) = .1176
d. Use the posterior probabilities from part (a) as the prior probabilities here.
Events P(Di) P(S2 | Di) P(Di  S2) P(Di | S2)
D1 .2195 .10 .0220 .1582
D2 .7805 .15 .1171 .8418
1.0000 .1391 1.0000
P(D1 | S1 and S2) = .1582
P(D2 | S1 and S2) = .8418
27. a.
Gender Too Fast Acceptable
Male Golfers 35 65
Female Golfers 40 60
The proportion of male golfers who say the greens are too fast is 35/(35 + 65) = 0.35, while the
proportion of female golfers who say the greens are too fast is 40/(40 + 60) = 0.40. There is a higher
percentage of female golfers who say the greens are too fast.
b. There are 50 male golfers with low handicaps, and 10 of these golfers say the greens are too fast, so
for male golfers the proportion with low handicaps who say the greens are too fast is 10/50 = 0.20.
On the other hand, there are 10 female golfers with low handicaps, and 1 of these golfers says the
greens are too fast, so for female golfers the proportion with low handicaps who say the greens are
too fast is 1/10 = 0.10.
c. There are 50 male golfers with higher handicaps, and 25 of these golfers say the greens are too fast,
so for male golfers the proportion with higher handicaps who say the greens are too fast is 25/50 =
0.50. On the other hand, there are 90 female golfers with higher handicaps, and 39 of these golfers
says the greens are too fast, so for female golfers the proportion with higher handicaps who say the
greens are too fast is 39/90 = 0.43.
d. When the data are aggregated across handicap categories, the proportion of female golfers who say
the greens are too fast exceeds the proportion of male golfers who say the greens are too fast.
However, when we introduce a third variable, handicap, we see different results. When sorted by
handicap categories, we see that the proportion of male golfers who find the greens too fast is higher
than female golfers for both low and high handicap categories. This is an example of Simpson’s
paradox.
28. a.
Male Applicants Female Applicants
Accept 70 40
Deny 90 80
After combining these two crosstabulations into a single crosstabulation with Accept and Deny as the row
labels and Male and Female as the column labels, we see that the rate of acceptance for males across the
university is 70/(70+90) = .4375 or approximately 44%, while the rate of acceptance for females across
the university is 40/(40+80) = .33 or 33%.
b. If we focus solely on the overall data, we would conclude the university’s admission process is biased in
favor of male applicant. However, this occurs because most females apply to the College of Business

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Chapter 2
2 - 10
(which has a far lower rate of acceptance that the College of Engineering). When we look at each
college’s acceptance rate by gender, we see the acceptance rate of males and females are equal in the
College of Engineering (75%) and the acceptance rate of males and females are equal in the College of
Business (33%). The data do not support the accusation that the university favors male applicants in its
admissions process.

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3 - 1
Chapter 3
Probability Distributions
Learning Objectives
1. Understand the concepts of a random variable and a probability distribution.
2. Be able to distinguish between discrete and continuous random variables.
3. Be able to compute and interpret the expected value, variance, and standard deviation for a discrete
random variable.
4. Be able to compute probabilities using a binomial probability distribution.
5. Be able to compute probabilities using a Poisson probability distribution.
6. Understand the difference between how probabilities are computed for discrete and continuous
random variables.
7. Know how to compute probability values for a continuous uniform probability distribution and be
able to compute the expected value and variance for such a distribution.
8. Be able to compute probabilities using a normal probability distribution. Understand the role of the
standard normal distribution in this process.
9. Be able to compute probabilities using an exponential probability distribution.

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3 - 2
Solutions:
1. a. values: 0,1,2,...,20
discrete
b. values: 0,1,2,...
discrete
c. values: 0,1,2,...,50
discrete
d. values: 0  x  8
continuous
e. values: x  0
continuous
2. a. f (200) = 1 - f (-100) - f (0) - f (50) - f (100) - f (150)
= 1 - 0.95 = 0.05
This is the probability MRA will have a $200,000 profit.
b. P(Profit) = f (50) + f (100) + f (150) + f (200)
= 0.30 + 0.25 + 0.10 + 0.05 = 0.70
c. P(at least 100) = f (100) + f (150) + f (200)
= 0.25 + 0.10 + 0.05 = 0.40
3. a.
x f (x)
1 3/20 = 0.15
2 5/20 = 0.25
3 8/20 = 0.40
4 4/20 = 0.20
1.00
b.

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Probability Distributions
3 - 3
c. f (x)  0 for x = 1,2,3,4.
 f (x) = 1
4. a.
x f (x) x f (x)
3 .25 .75
6 .50 3.00
9 .25 2.25
1.00 6.00
E (x) =  = 6.00
b.
x x -
 (x -
)2 f (x) (x -
)2 f (x)
3 -3 9 .25 2.25
6 0 0 .50 0.00
9 3 9 .25 2.25
4.50
Var (x) =
 2 = 4.50
c. 4.50 2.12
  
5. Let X = amount of money in a randomly selected bar of soap.
The following table shows the calculations for parts (a) and (b).
Distribution of Paper
Currency Prizes
Bill
Denomination (x)
Number of
Bills f(x) xf(x) (x-
)2 (x-
)2f(x)
$1 520 0.52 0.52 24.7009 12.844468
$5 260 0.26 1.3 0.9409 0.244634
$10 120 0.12 1.2 16.2409 1.948908
$20 70 0.07 1.4 196.8409 13.778863
$50 29 0.029 1.45 1938.6409 56.2205861
$100 1 0.001 0.1 8841.6409 8.8416409
Total 1000 1.00 5.97 93.8791
a. E[X] = 5.97
b. St. Dev. (X) =
 = 9.69
c. Let Y = number of bars that contain a $50 or $100 bill. Observe that Y is a binomial random variable,

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Chapter 3
3 - 4
where n = numbers of bars purchased and p = probability of a bar containing a $50 or $100 bill =
(29/1000) + (1/1000) = .03. E[Y] = n × p = n × 0.03 = 3, thus n = 100. The customer needs to buy 100
bars of soap to have, on average, three bars with a $50 or $100 bill.
d. Let W = number of bars that contain a $20 or $50 or $100 bill. Observe that W is a binomial random
variable, where n = 8 and p = (70 + 29 + 1 / 1000) = 0.10. P(W ≥ 1) = 1 – P(W = 0) = 1 - .4305 = .5695.
6. The following table shows the calculations for parts (a) and (b).
x f(x) xf(x) x-
 (x-
)2 (x-
)2f(x)
1 .97176 .97176 -.03000 .00090 .00087
2 .026675 .05333 .97000 .94090 .02509
3 .00140 .004218 1.97000 3.88090 .00544
4 .00014 .00059 2.97000 8.82090 .00131
5 .00002 .00011 3.97000 15.76089 .00034
1.0000 1.03000 .03305
If we let x = 5 represent quintuplets or more, the probability distribution of the number children born per
pregnancy in 1996 is provided in the first two columns of the table above.
a. The expected value of the number children born per pregnancy in 1996 is E[x] = 1.030.
b. The variance of the number children born per pregnancy in 1996 is V[x] =
2 = 0.0331.
The following table shows the calculations for parts (c) and (d).
y f(y) yf(y) y-
 (y-
)2 (y-
)2f(y)
1 0.965964 0.9650964 -0.0366118 0.0013404 0.001293639
2 0.0333143 0.0666286 0.9633882 0.9281168 0.030919551
3 0.0014868 0.0044604 1.9633882 3.8548932 0.005731423
4 0.0000863 0.0003451 2.9633882 8.7816695 0.000757611
5 0.0000163 0.0000814 3.9633882 15.7084459 0.000255769
1.0000000 1.0366118 0.038957993
If we let y = 5 represent quintuplets or more, the probability distribution of the number children born per
pregnancy in 2006 is provided in the first two columns of the table above.
c. The expected value of the number children born per pregnancy in 2006 is E[y] = 1.030.
d. The variance of the number children born per pregnancy in 2006 (after rounding) is V[y] =
2 = 0.0390.
e. The number of children born per pregnancy is greater in 2006 than in 1996, and the variation in the
number of children born per pregnancy is also greater in 2006 than in 1996. However, these data
provide no information on which we could base a determination of causes of this upward trend.
7. a. E(x) = xf (x)
= 300(0.20) + 400(0.30) + 500(0.35) + 600(0.15)
= 445

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Probability Distributions
3 - 5
The monthly order quantity should be 445 units.
b. Cost: 445 @ $50 = $22,250
Revenue: 300 @ $70 = 21,000
$ 1,250 Loss
c.
x f (x) (x - μ) (x - μ)2 (x - μ)2f (x)
300 0.20 -145 21025 4205.00
400 0.30 - 45 2025 607.50
500 0.35 + 55 3025 1058.75
600 0.15 +155 24025 3603.75
 2 = 9475.00
 = 9475 97.34
8. a.
Medium: E(x) = xf (x)
= 50(0.20) + 150(0.50) + 200(3.0) = 145
Large: E(x) = xf (x)
= 0(0.20) + 100(0.50) + 300(0.30) = 140
Medium preferred.
b. Medium
x f (x) (x - μ) (x - μ)2 (x - μ)2f (x)
50 0.20 -95 9025 1805.0
150 0.50 5 25 12.5
200 0.30 55 3025 907.5
 2 = 2725.0
Large
x f (x) (x - μ) (x - μ)2 (x - μ)2f (x)
0 0.20 -140 19600 3920
100 0.50 - 40 1600 800
300 0.30 160 25600 7680
 2 = 12,400
Medium preferred due to less variance.
9. a. 1 12 2!
(1) (0.4) (0.6) (0.4)(0.6) 0.48
1 1!1!
f  
   
 
b. 0 22 2!
(0) (0.4) (0.6) (1)(0.36) 0.36
0 0!2!
f  
   
 
c. 2 02 2!
(2) (0.4) (0.6) (0.16)(1) 0.16
2 2!0!
f  
   
 
d. P (x  1) = f (1) + f (2) = .48 + .16 = .64
e. E (x) = n p = 2 (.4) = .8

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Chapter 3
3 - 6
Var (x) = n p (1 - p) = 2 (.4) (.6) = .48
0.48 0.6928
  
10. a. f (0) = .3487
b. f (2) = .1937
c. P(x  2) = f (0) + f (1) + f (2) = .3487 + .3874 + .1937 = .9298
d. P(x  1) = 1 - f (0) = 1 - .3487 = .6513
e. E (x) = n p = 10 (.1) = 1
f. Var (x) = n p (1 - p) = 10 (.1) (.9) = .9
0.9 0.9487
  
11. a. In a sample of six British citizens, the probability that two have ever boycotted goods for ethical reasons
is
      2789.23.1.23
2
6
2 262 





 
f
b. In a sample of six British citizens, the probability that at least two reports have boycotted goods for ethical
reasons is
           
           
       
.4180.0001.0030.0249.1111.2789.
23.1.23
6
6
23.1.23
5
6
23.1.23
4
6
23.1.23
3
6
23.1.23
2
6
654322
666565
464363262




































fffffleastatP
Or we can find the compliment of this event and subtract from 1.0, i.e.
        
       
 .roundingofbecauseoccursdifferenceslighta4181.3735.2084.0.1
23.1.23
1
6
23.1.23
0
6
1
10120.12
161060
















ffthanlessPleastatP
c. In a sample of ten British citizens, the probability that none have boycotted goods for ethical reasons is
      .0733.23.1.23
0
10
0 0100 





 
f
12. a. Probability of a defective part being produced must be .03 for each trial; trials must be independent.
b. 2 outcomes result in exactly one defect.

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Probability Distributions
3 - 7
c. P (no defects) = (.97) (.97) = .9409
P (1 defect) = 2 (.03) (.97) = .0582
P (2 defects) = (.03) (.03) = .0009
13. a. 0.90
b. P(at least 1) = f (1) + f (2)
f (1) = 1 12! (0.9) (0.1)
1!1!
= 2 (0.9) (0.1) = 0.18
f (2) = 2 02! (0.9) (0.1)
2!0!
= 1 (0.81) (1) = 0.81
Therefore P(at least 1) = 0.18 + 0.81 = 0.99
Alternatively
P(at least 1) = 1 - f (0)
= 0 22! (0.9) (0.1) 0.01
0!2! 
Therefore P(at least 1) = 1 - 0.01 = 0.99
c. P(at least 1) = 1 - f (0)
f (0) = 0 33! (0.9) (0.1) 0.001
0!3! 
Therefore P(at least 1)= 1 - 0.001 = 0.999
d. Yes; P(at least 1) becomes very close to 1 with multiple systems and the inability to detect an attack
would become catastrophic.
14. a.
2
2
( ) !
x e
f x x


b.
 = 6 for 3 time periods
c.
6
6
( ) !
x e
f x x



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Chapter 3
3 - 8
d.
2 2
2 4(0.1353)
(2) 0.2706
2! 2
e
f

  
e.
6 6
6
(6) 0.1606
6!
e
f

 
f.
5 4
4
(5) 0.1563
5!
e
f

 
15. a.
 = 48 (5 / 60) = 4
3 4
4 (64)(0.0183)
(3) 0.1952
3! 6
e
f

  
b.
 = 48 (15 / 60) = 12
10 12
12
(10) 0.1048
10!
e
f

 
c.
 = 48 (5 / 60) = 4 I expect 4 callers to be waiting after 5 minutes.
0 4
4
(0) 0.0183
0!
e
f

 
The probability none will be waiting after 5 minutes is .0183.
d.
 = 48 (3 / 60) = 2.4
0 2.4
2.4
(0) 0.0907
0!
e
f

 
The probability of no interruptions in 3 minutes is .0907.
16. a.
0 7
77
(0) .0009
0!
e
f e


  
b. probability = 1 - [f(0) + f(1)]
1 7
77
(1) 7 .0064
1!
e
f e


  
probability = 1 - [.0009 + .0064] = .9927
c.
 = 3.5
0 3.5
3.53.5
(0) .0302
0!
e
f e


  
probability = 1 - f(0) = 1 - .0302 = .9698
d. probability = 1 - [f(0) + f(1) + f(2) + f(3) + f(4)]

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Probability Distributions
3 - 9
= 1 - [.0009 + .0064 + .0223 + .0521 + .0912]
= .8271
Note: Appendix B was used to compute the Poisson probabilities f(0), f(1), f(2), f(3) and f(4) in
part (d).
17. a.
0 10
1010
(0) 0.000045
0!
e
f e


  
b. f (0) + f (1) + f (2) + f(3)
f (0)= 0.000045 (part a)
f (1)=
1 10
10
1!
e
= 0.00045
Similarly, f (2) = 0.00225, f (3) = 0.0075
Therefore f (0) + f (1) + f (2) + f (3) = 0.010245
c. 15 second period, therefore we have 2.5 arrivals/15 second period
0 2.5
2.5
(0) 0.0821
0!
e
f

 
d. 1 - f (0) = 1 - 0.0821 = 0.9179
18. a.
1
2
3
.50 1.0 1.5 2.0
f (x)
x
b. P(x = 1.25) = 0. the probability of any single point is zero since the area under the curve above any
single point is zero.
c. P(1.0  x  1.25) = 2(0.25) = 0.50
d. P(1.20 < x < 1.5) = 2(0.30)= 0.60

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Chapter 3
3 - 10
19. a.
b. P(x  130) = (1/20) (130 - 120) = 0.50
c. P(x > 135) = (1/20) (140 - 135) = 0.25
d. 120 140
( ) 130 minutes
2
E x 
 
20. a.
.5
1 .0
1 .5
1 2 3
f (x)
x
0
b. P(0.25 < x < 0.75) = 1(0.50) = 0.50
c. P(x  0.30) = 1(0.30) = 0.30
d. P(x > 0.60) = 1(0.40) = 0.40
21. a. Area for z = 0 0.5000
Area for z = 0.83 0.7967
P(0  z  0.83) = 0.7967 – 0.5000 = 0.2967
b. Area for z = -1.57 0.0582
Area for z = 0 0.5000
P(-1.57  z  0) = 0.5000 – 0.0582 = 0.4418
c. Area for z = 0.44 0.6700
P(z>0.44) = 1.0000 – 0.6700 = 0.2300
d. Area for z = -0.23 0.4090

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Probability Distributions
3 - 11
P(z  -0.23) = 1.0000 – 0.4090 = 0.5910
e. Area for z = 1.20 0.8849
P(z  1.20) = 0.8849
f. Area for z = -0.71 0.2389
P(z  -0.71) = 0.2389
22. a. Area = 0.9750 z = 1.96
b. Area for z = 0 0.5000
Area for z must be 0.5000 + 0.4750 = 0.9750 z = 1.96
c. Area = 0.7291 z = 0.61
d. Area = 1.0000 – 0.1314 = 0.8686 z = 1.12
e. Area 0.6700 z = 0.44
f. Area = 1.0000 – 0.3300 = 0.6700 z = 0.44
23. a. Area = 0.2119 z = -0.80
b. Area outside the interval = 1.0000 – 0.9030 = 0.0970 must be split between the two tails.
Cumulative probability at z must be 0.5(0.0970) + 0.9030 = 0.9515 z = 1.66
c. Area outside the interval = 1.0000 – 0.2052 = 0.7948 must be split between the two tails.
Cumulative probability at z must be 0.5(0.7948) + 0.2052 = 0.6026 z = 0.26
d. Area = 0.9948 z = 2.56
e. Area = 1.0000 – 0.6915 = 0.3085 z = -0.50
24. a. At x = 180
180 200 0.50
40
z 
  
Cumulative probability = 0.3085
At x = 220
220 200 0.50
40
z 
  
Cumulative probability = 0.6915
P(180  x  220) = 0.6915 – 0.3085 = 0.3830

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Chapter 3
3 - 12
b. At x = 250
250 200 1.25
40
z 
 
Cumulative probability = 0.8944
P(x  250) = 1.0000 – 0.8944 = 0.1056
c. At x = 100
100 200 2.50
40
z 
  
Cumulative probability = 0.0062
P(x  100) = 0.0062
d. At x = 250
250 200 1.25
40
z 
 
Cumulative probability = 0.8944
At x = 225
225 200 0.625
40
z 
 
Cumulative probability = 0.7341
P(225  x  250) = 0.8944 – 0.7341 = 0.1603
25. a. At x = 40
40 30 10 1.22
8.20 8.20
z 
  
Cumulative probability = 0.8888
P(x  40) = 1.0000 - 0.8888 = 0.1112
b. At x = 20
20 30 10 1.22
8.20 8.20
z  
   
Cumulative probability = 0.1122
P(x < 20) = 0.1122
c. The largest 10% of the stock prices would be in the upper tail of the normal distribution with a
cumulative probability 0.90. From the table of the cumulative standard normal distribution, the area or
probability closest to 0.90 is 0.8997. The corresponding z = 1.28. Thus we have.

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Quantitative Methods For Business, 12th Edition Solution Manual

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