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Quantitative Methods For Business, 12th Edition Solution Manual

Quantitative Methods For Business, 12th Edition Solution Manual helps you tackle difficult exercises with expert guidance.

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    Solutions Manual to Accompany Quantitative Methods for Business Twelfth Edition

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    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

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    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

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    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:

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    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 = 2 d c m       We must be willing to treat m and c as known and not subject to variation. 8. a. Maximize 10 x + 5 y s.t. 5 x + 2 y  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 for x 2/ unit for y $10/unit for x $ 5/ unit for y 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 10 x + 5 y 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 un controllable 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 = 0.20 x +0.25 y c. x + y = 5000 d. x  4000 Kansas City Constraint y  3000 Minneapolis Constraint e. Min 0.20 x + 0.25 y 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 80 0 – 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 – 10 p ) p = 800 p – 10 p 2 . 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 – 10 p = 800 – 10(40) = 400 units TR = 800 p – 10 p 2 = 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 + 60 x . The two components of total cost in this model are fixed cost ($2,000) and variable cost (60 x ). b. If P represents the total profit, the total revenue (TR) is 80 x and a mathematical model for the total profit realized from an order for x pairs of shoes is P = TR – TC = 80 x – (2000+60 x ) = 20 x – 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 = 20 x – 2000 = 0. This occurs when 20 x = 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(2 x ) = 9600 + 120 x (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 600 x – (9600 + 120 x ) = 480 x - 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 = 480 x - 9600 = 0. This occurs when 480 x = 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) = 46 x and total cost (TC) = 160,000 + 6 x , so Profit = TR – TC = 46 x – (160,000 + 6 x ) = 40 x – 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 = 40 x - 160,000 = 0. This occurs when 40 x = 160,00 0 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)) = 3800 p - 182,800. This occurs whre 3800 p = 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 , 000 x and total cost (TC) = 4 , 500 , 000 + 150 , 000 x , so Profit = TR – TC = 300 , 000 x – (4 , 500 , 000 + 150 , 000 x ) = 150 , 000 x – 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 , 000 x – 4 , 500 , 000 = 0. This occurs when 150 , 000 x = 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 6 x + 4 y . 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 50 x + 30 y ≤ 800000, (2) the mathematical expression for the constraint that limits investment in Oil Alaska to $500,000 is 50 x ≤ 500000, and (3) the mathematical expression for the constraint that limits investment in Southwest Petroleum to $450,000 is 30 y ≤ 450000. 17. a. s j = s j - 1 + x j - d j or s j - s j - 1 - x j + d j = 0 b. x j  c j c. s j  I j

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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, 31,675,935 were 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(E 1 , E 2 , E 3 , E 4 ) = .80. Yes P(A  B) = P(A) + P(B). c. A c = {E 3 , E 4 , E 5 } C c = {E 1 , E 4 } P(A c ) = .60 P(C c ) = .40 d. A  B c = {E 1 , E 2 , E 5 } P(A  B c ) = .60 e. P(B  C) = P(E 2 , E 3 , E 4 , E 5 ) = .80 8. a. Let P( A ) be the pr obability 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(B c ) = 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 1 7 . 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 accide nt 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  A 1 ) = P(A 1 )P(B  A 1 ) = (0.20)(0.50) = 0.10 P(B  A 2 ) = P(A 2 )P(B  A 2 ) = (0.50)(0.40) = 0.20

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    Introduction to Probability 2 - 7 P(B  A 3 ) = P(A 3 )P(B  A 3 ) = (0.30)(0.30) = 0.09 b. 2 0.20 P(A B) 0.51 0.10 0.20 0.09     c. Events P(A i ) P(B  A i ) P(A i  B) P(A i  B) A 1 0.20 0.50 0.10 0.26 A 2 0.50 0.40 0.20 0.51 A 3 0.30 0.30 0.09 0.23 1.00 0.39 1.00 21. S 1 = successful, S 2 = not successful and B = request received for additional information. a. P(S 1 ) = 0.50 b. P(B  S 1 ) = 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(A i ) P(S  A i ) P(A i  S) P(A i  S) High Quality (A 1 ) 0.50 0.20 0.10 0.31 Medium Quality (A 2 ) 0.20 0.80 0.16 0.50 No Oil (A 3 ) 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 S C = speeding is not reported F = Accident results in fatality for vehicle occupant We have P(S) = .129, so P(S C ) = .871. Also P(F|S) = .196 and P(F|S C ) = .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 S C .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(A i ) P(D  A i ) P(A i  D) P(A i  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(D i ) P(S 1 |D i ) P(D i  S 1 ) P(D i |S 1 ) D 1 .60 .15 .090 .2195 D 2 .40 .80 .320 .7805 1.00 P(S 1 ) = .410 1.0000 P(D 1 | S 1 ) = .2195 P(D 2 | S 1 ) = .7805 b. Events P(D i ) P(S 2 |D i ) P(D i  S 2 ) P(D i |S 2 ) D 1 .60 .10 .060 .500 D 2 .40 .15 .060 .500 1.00 P(S 2 ) = .120 1.000 P(D 1 | S 2 ) = .50 P(D 2 | S 2 ) = .50 c. Events P(D i ) P(S 3 |D i ) P(D i  S 3 ) P(D i |S 3 ) D 1 .60 .15 .090 .8824 D 2 .40 .03 .012 .1176 1.00 P(S 3 ) = .102 1.0000 P(D 1 | S 3 ) = .8824

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    Introduction to Probability 2 - 9 P(D 2 | S 3 ) = .1176 d. Use the posterior probabilities from part (a) as the prior probabilities here. Events P(D i ) P(S 2 | D i ) P(D i  S 2 ) P(D i | S 2 ) D 1 .2195 .10 .0220 .1582 D 2 .7805 .15 .1171 .8418 1.0000 .1391 1.0000 P(D 1 | S 1 and S 2 ) = .1582 P(D 2 | S 1 and S 2 ) = .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 V ar ( 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 -  ) 2 f(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 -  ) 2 f ( 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, t he 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. T he 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 -  ) 2 f ( 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, t he 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. T he variance of the number children born per pregnancy in 2006 (aft er 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 informati on 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 ) 2 f ( 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 ) 2 f ( 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 ) 2 f ( 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 1 2 2! (1) (0.4) (0.6) (0.4)(0.6) 0.48 1 1!1! f          b. 0 2 2 2! (0) (0.4) (0.6) (1)(0.36) 0.36 0 0!2! f          c. 2 0 2 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 2 6 2             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 6 5 4 3 2 2 6 6 6 5 6 5 4 6 4 3 6 3 2 6 2                                                                   f f f f f least at P Or we can find the compliment of this event and subtract from 1.0, i.e.                     . rounding of because occurs difference slight a 4181 . 3735 . 2084 . 0 . 1 23 . 1 .23 1 6 23 . 1 .23 0 6 1 1 0 1 2 0 . 1 2 1 6 1 0 6 0                                 f f than less P least at P 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 0 10 0             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 1 2! (0.9) (0.1) 1!1! = 2 (0.9) (0.1) = 0.18 f (2) = 2 0 2! (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 2 2! (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 3 3! (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 inabil ity 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 no ne 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 7 7 (0) .0009 0! e f e      b. probability = 1 - [ f (0) + f (1)] 1 7 7 7 (1) 7 .0064 1! e f e      probability = 1 - [.0009 + .0064] = .9927 c.  = 3.5 0 3.5 3.5 3.5 (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 10 10 (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 0.50 00 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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