Quantitative Analysis For Management, 13th Edition Solution Manual

CHAPTER 2 Probability Concepts and Applications TEACHING SUGGESTIONS Teaching Suggestion 2.1: Concept of Probabilities Ranging From 0 to 1. People often misuse probabilities by such statements as, “I’m 110% sure we’re going to win the big game.” The two basic rules of probability should be stressed. Teaching Suggestion 2.2: Where Do Probabilities Come From? Students need to understand where probabilities come from. Sometimes they are subjective and based on personal experiences. Other times they are objectively based on logical observations such as the roll of a die. Often, probabilities are derived from historical data—if we can assume the future will be about the same as the past. Teaching Suggestion 2.3: Confusion Over Mutually Exclusive and Collectively Exhaustive Events. This concept is often foggy to even the best of students—even if they just completed a course in statistics. Use practical examples and drills to force the point home. Teaching Suggestion 2.4: Addition of Events That Are Not Mutually Exclusive. The formula for adding events that are not mutually exclusive is P(A or B) = P(A) + P(B) – P(A and B). Students must understand why we subtract P(A and B). Explain that the intersection has been counted twice. Teaching Suggestion 2.5: Expected Value of a Probability Distribution. A probability distribution is often described by its mean and variance. These important terms should be discussed with such practical examples as heights or weights of students. But students need to be reminded that even if most of the men in class (or the United States) have heights between 5 feet 6 inches and 6 feet 2 inches, there is still some small probability of outliers. Teaching Suggestion 2.6: Bell-Shaped Curve. Stress how important the normal distribution is to a large number of processes in our lives (for example, filling boxes of cereal with 32 ounces of cornflakes). Each normal distribution depends on the mean and standard deviation. Discuss Figures 2.7 and 2.8 to show how these relate to the shape and position of a normal distribution. 2-1 Copyright © 2018 Pearson Education, Inc. Teaching Suggestion 2.7: Three Symmetrical Areas Under the Normal Curve. Figure 2.13 is very important, and students should be encouraged to truly comprehend the meanings of ±1, 2, and 3 standard deviation symmetrical areas. They should especially know that managers often speak of 95% and 99% confidence intervals, which roughly refer to ±2 and 3 standard deviation graphs. Clarify that 95% confidence is actually ±1.96 standard deviations, while ±3 standard deviations is actually a 99.7% spread. Teaching Suggestion 2.8: Using the Normal Table to Answer Probability Questions. The IQ example in Figure 2.9 is a particularly good way to treat the subject since everyone can relate to it. Students are typically curious about the chances of reaching certain scores. Go through at least a half-dozen examples until it’s clear that everyone can use Table 2.10. Students get especially confused answering questions such as P(X 1/2 As | regular class) = 0.25 P(>1/2 As | advanced class) = 0.50 P(>1/2 As and regular class) = P(>1/2 As | regular ) × P(regular) = (0.25)(0.50) = 0.125 P(>1/2 As and advanced class) = P(>1/2 As | advanced) × P(advanced) = (0.50)(0.5) = 0.25 So P(>1/2 As) = 0.125 + 0.25 = 0.375 P ( advanced > 1/2 As ) = = P ( advanced and > 1/2 As ) P ( > 1/ 2 As ) 0.25 = 2/3 0.375 So there is a 66% chance the class tested was the advanced one. 2-5 Copyright © 2018 Pearson Education, Inc. Alternative Example 2.8: Students in a statistics class were asked how many “away” football games they expected to attend in the upcoming season. The number of students responding to each possibility is shown below: Number of games Number of students 5 40 4 30 3 20 2 10 1 0 100 A probability distribution of the results would be: Number of games Probability P(X) 5 0.4 = 40/100 4 0.3 = 30/100 3 0.2 = 20/100 2 0.1 = 10/100 1 0.0 = 0/100 1.0 = 100/100 This discrete probability distribution is computed using the relative frequency approach. Probabilities are shown in graph form below. 2-6 Copyright © 2018 Pearson Education, Inc. Alternative Example 2.9: Here is how the expected outcome can be computed for the question in Alternative Example 2.8. 5 E ( x ) = ∑ xi P ( X i ) = xi P ( X 1 ) + x2 P ( X 2 ) i =1 + x3P(X3) +x4P(X4) + x5P(X5) = 5(0.4) + 4(0.3) + 3(0.2) + 2(0.1) + 1(0) = 4.0 Alternative Example 2.10: Here is how variance is computed for the question in Alternative Example 2.8: 5 variance = ∑ ( xi − E ( x ) ) P ( xi ) 2 i =1 2 = (5 – 4) (0.4) + (4 – 4)2(0.3) + (3 – 4)2(0.2) + (2 – 4)2(0.1) + (1 – 4)2(0) = (1)2(0.4) + (0)2(0.3) + (–1)2(0.2) + (–2)2(0.1) + (-3)2(0) = 0.4 + 0.0 + 0.2 + 0.4 + 0.0 = 1.0 The standard deviation is σ = variance = 1 =1 Alternative Example 2.11: The length of the rods coming out of our new cutting machine can be said to approximate a normal distribution with a mean of 10 inches and a standard deviation of 0.2 inch. Find the probability that a rod selected randomly will have a length a. of less than 10.0 inches b. between 10.0 and 10.4 inches c. between 10.0 and 10.1 inches d. between 10.1 and 10.4 inches e. between 9.9 and 9.6 inches f. between 9.9 and 10.4 inches g. between 9.886 and 10.406 inches 2-7 Copyright © 2018 Pearson Education, Inc. First compute the standard normal distribution, the Z-value: z= x−μ σ Next, find the area under the curve for the given Z-value by using a standard normal distribution table. a. P(x < 10.0) = 0.50000 b. P(10.0 < x < 10.4) = 0.97725 – 0.50000 = 0.47725 c. P(10.0 < x < 10.1) = 0.69146 – 0.50000 = 0.19146 d. P(10.1 < x < 10.4) = 0.97725 – 0.69146 = 0.28579 e. P(9.6 < x < 9.9) = 0.97725 – 0.69146 = 0.28579 f. P(9.9 < x < 10.4) = 0.19146 + 0.47725 = 0.66871 g. P(9.886 < x 6) = P(r ≥ 7) = P(r = 7) + P(r = 8) + P(r = 9) + P(r = 10) = 0.1172 + 0.0439 + 0.0098 + 0.0010 = 0.1719 2-20 Copyright © 2018 Pearson Education, Inc. 4.04753 2-33. This is a binomial distribution with n=4, p=0.7, and q=0.3. P ( r = 3) = 4! 3 4−3 ( 0.7 ) ( 0.3) = 0.4116 3!( 4 − 3) ! P ( r = 4) = 4! 4 4−4 ( 0.7 ) ( 0.3) = 0.2401 4!( 4 − 4 ) ! 2-34. This is a binomial distribution with n =5, p=0.1, and q=0.9. P ( r = 1) = 5! 1 5−1 ( 0.1) ( 0.9 ) = 0.328 1!( 5 − 1) ! P ( r = 0) = 5! 0 5− 0 ( 0.1) ( 0.9 ) = 0.590 0!( 5 − 0 ) ! 2-35. This is a binomial distribution with n=6, p=0.05, and q=0.95. P ( r = 0) = 6! 0 6−0 ( 0.05) ( 0.95) = 0.735 0!( 6 − 0 ) ! P ( r = 1) = 6! 1 6 −1 ( 0.05) ( 0.95) = 0.232 1!( 6 − 1) ! 2-36. This is a binomial distribution with n=6, p=0.15, and q=0.85. P ( r = 0) = 6! 0 6−0 ( 0.15) ( 0.85) = 0.377 0!( 6 − 0 ) ! P ( r = 1) = 6! 1 6 −1 ( 0.15) ( 0.85) = 0.399 1!( 6 − 1) ! Probability of 0 or 1 defective = P(0) + P(1) = 0.377 + 0.399 = 0.776. 2-21 Copyright © 2018 Pearson Education, Inc. 2-37. μ = 450 degrees σ = 25 degrees X = 475 degrees Z= X −μ σ = 475 − 450 =1 25 The area to the left of 475 is 0.8413 from Table 2.9, where σ = 1. The area to the right of 475 is 1 – 0.84134 = 0.15866. Thus, the probability of the oven getting hotter than 475 is 0.1587. To determine the probability of the oven temperature being between 460 and 470, we need to compute two areas. X1 = 460 X2 = 470 Z1 = 460 − 450 10 = = 0.4 25 25 area X1 = 0.65542 Z2 = 470 − 450 20 = = 0.8 25 25 area X2 = 0.78814 The area between X1 and X2 is 0.78814 – 0.65542 = 0.13272. Thus, the probability of being between 460 and 470 degrees is = 0.1327. 2-22 Copyright © 2018 Pearson Education, Inc. 2-38. μ = 4,700; σ = 500 a. The sale of 5,500 oranges (X = 5,500) is the equivalent of some Z value which may be obtained from Z= = X −μ σ 5,500 − 4, 700 500 = 1.6 The area under the curve lying to the left of 1.6σ = 0.94520. Therefore, the area to the right of 1.6σ = 1 – 0.94520, or 0.0548. Therefore, the probability of sales being greater than 5,500 oranges is 0.0548. b. Z= area 4,500 − 4, 700 200 =− = −0.4 500 500 = 0.65542 probability = 0.65542 2-23 Copyright © 2018 Pearson Education, Inc. c. Z= 4,900 − 4, 700 200 = = 0.4 500 500 area = 0.65542 = probability This answer is the same as the answer to part (b) because the normal curve is symmetrical. d. Z= 4,300 − 4, 700 400 =− = −0.8 500 500 Area to the right of 4,300 is 0.78814, from Table 2.9. The area to the left of 4,300 is 1 – 0.78814 = 0.21186 = the probability that sales will be fewer than 4,300 oranges. 2-39. μ = 87,000 σ = 4,000 X = 81,000 Z= 81, 000 − 87, 000 6 = − = −1.5 4, 000 4 Area to the right of 81,000 = 0.93319, from Table 2.9, where Z = 1.5. Thus, the area to the left of 81,000 = 1 – 0.93319 = 0.06681 = the probability that sales will be fewer than 81,000 packages. 2-24 Copyright © 2018 Pearson Education, Inc. 2-40. μ = 457,000 Ninety percent of the time, sales have been between 460,000 and 454,000 pencils. This means that 10% of the time sales have exceeded 460,000 or fallen below 454,000. Since the curve is symmetrical, we assume that 5% of the area lies to the right of 460,000 and 5% lies to the left of 454,000. Thus, 95% of the area under the curve lies to the left of 460,000. From Table 2.9, we note that the number nearest 0.9500 is 0.94950. This corresponds to a Z value of 1.64. Therefore, we may conclude that the Z value corresponding to a sale of 460,000 pencils is 1.64. Using Equation 2-13, we get Z = X −μ σ X = 460,000 μ = 457,000 σ is unknown Z = 1.64 1.64 = 460,000 − 457,000 σ 1.64 σ = 3000 ∴σ = 3,000 1.64 = 1829.27 2-41. The time to complete the project (X) is normally distributed with μ = 60 and σ = 4. a) P(X ≤ 62) = P(Z ≤ (62 – 60)/4) = P(Z ≤ 0.5) = 0.69146 b) P(X ≤ 66) = P(Z ≤ (66 – 60)/4) = P(Z ≤ 1.5) = 0.93319 c) P(X > 65) = 1 – P(X ≤ 65) = 1 – P(Z ≤ (65 – 60)/4) = 1 – P(Z ≤ 1.25) = 1 – 0.89435 = 0.10565 2-25 Copyright © 2018 Pearson Education, Inc. 2-42. The time to complete the project (X) is normally distributed with μ = 40 and σ = 5. A penalty must be paid if the project takes longer than the due date (or if X > due date). a) P(X > 40) = 1 – (X ≤ 40) = 1 – P(Z ≤ (40 – 40)/5) = 1 – P(Z ≤ 0) = 1 – 0.5 = 0.5 b) P(X > 43) = 1 – P(X ≤ 43) = 1 – P(Z ≤ (43 – 40)/5) = 1 – P(Z ≤ 0.6) = 1 – 0.72575 = 0.27425 c) If there is a 5% chance that the project will be late, there is a 95% chance the project will be finished by the due date. So, P(X ≤ due date) = 0.95 or P(X ≤ _____) = 0.95 The Z-value for a probability of 0.95 is approximately 1.64, so the due date (X) should have a Z-value of 1.64. Thus, 1.64 = X − 40 5 5(1.64) = X – 40 X = 48.2. The due date should be 48.2 weeks from the start of the project 2-43. λ = 5/day; e–λ = 0.0067 (from Appendix C) a. P ( 0 ) = λ x e− λ X! = (1)( 0.0067 ) = 0.0067 1 P (1) = ( 5)( 0.0067 ) = 0.0335 P ( 2) = 25 ( 0.0067 ) = 0.0837 2 P ( 3) = 125 ( 0.0067 ) = 0.1396 6 P ( 4) = 625 ( 0.0067 ) = 0.1745 24 P ( 5) = 3125 ( 0.0067 ) = 0.1745 120 1 b. These sum to 0.6125, not 1, because there are more possible arrivals. For example, 6 or 7 patients might arrive in one day. 2-26 Copyright © 2018 Pearson Education, Inc. 2-44. P(X > 3) = 1 – P(X ≤ 3) = 1 – [P(0) + P(1) + P(2) + P(3)] = 1 – [0.0067 + 0.0335 + 0.0837 + 0.1396] = 1 – 0.2635 = 0.7365 = 73.65% 2-45. μ = 3/hour a. Expected time 1 = μ = 1 hour 3 = 20 minutes b. Variance = 1 μ 2 = 1 9 2-46. Let S = steroids present N = steroids not present TP = test is positive for steroids TN = test is negative for steroids P(S) = 0.02 P(N) = 0.98 P(TP | S) = 0.95 P(TN | S) = 0.05 P(TP | N) = 0.10 P(TN | N) = 0.90 P (TP S ) P ( S ) P ( S TP ) = = P (TP S ) P ( S ) + P (TP N ) P ( N ) 0.95 ( 0.02 ) = 0.16 0.95 ( 0.02 ) + 0.10 ( 0.98 ) 2-47. Let G = market is good P = market is poor PG = test predicts good market PP = test predicts poor market P(G) = 0.70 P(P) = 0.30 P(PG | G) = 0.85 P(PP | G) = 0.15 P(PG | P) = 0.20 P(PP | P) = 0.80 P ( G PG ) = = P ( PG G ) P ( G ) P ( PG G ) P ( G ) + P ( PG P ) P ( P ) 0.85 ( 0.70 ) = 0.91 0.85 ( 0.70 ) + 0.20 ( 0.30 ) 2-27 Copyright © 2018 Pearson Education, Inc. 2-48. Let W = candidate wins the election L = candidate loses the election PW = poll predicts win PL = poll predicts loss P(W) = 0.50 P(L) = 0.50 P(PW | W) = 0.80 P(PL | W) = 0.20 P(PW | L) = 0.10 P(PL | L) = 0.90 P (W PW ) = = P ( L PL ) = = P ( PW W ) P (W ) P ( PW W ) P (W ) + P ( PW L ) P ( L ) 0.80 ( 0.50 ) 0.80 ( 0.50 ) + 0.10 ( 0.50 ) = 0.89 P ( PL L ) P ( L ) P ( PL L ) P ( L ) + P ( PL W ) P (W ) 0.90 ( 0.50 ) = 0.82 0.90 ( 0.50 ) + 0.20 ( 0.50 ) 2-49. Let S = successful restaurant U = unsuccessful restaurant PS = model predicts successful restaurant PU = model predicts unsuccessful restaurant P(S) = 0.70 P(U) = 0.30 P(PS | S) = 0.90 P(PU | S) = 0.10 P(PS | U) = 0.20 P(PU | U) = 0.80 P ( S PS ) = = P ( PS S ) P ( S ) P ( PS S ) P ( S ) + P ( PS U ) P (U ) 0.90 ( 0.70 ) = 0.91 0.90 ( 0.70 ) + 0.20 ( 0.30 ) 2-28 Copyright © 2018 Pearson Education, Inc. 2-50. Let D = Default on loan; D’ = No default; R = Loan rejected; R′ = Loan approved Given: P(D) = 0.2 P(D’) = 0.8 P(R | D) = 0.9 P(R’ | D’) = 0.7 (a) P(R | D’) = 1 – 0.7 = 0.3 (b) P ( D′ R ) = = P ( R D ′ ) P ( D′ ) P ( R D′ ) P ( D ′ ) + P ( R D ) P ( D ) 0.3 ( 0.8 ) = 0.57 0.3 ( 0.8) + 0.9 ( 0.2 ) 2-51. (a) F0.05, 5, 10 = 3.33 (b) F0.05, 8.7 = 3.73 (c) F0.05, 3, 5 = 5.41 (d) F0.05, 10. 4 = 5.96 2-52. (a) F0.01, 15, 6 = 7.56 (b) F0.01, 12, 8 = 5.67 (c) F0.01, 3, 5 = 12.06 (d) F0.01, 9, 7 = 6.72 2-53. (a) From the appendix, P(F3,4 > 6.59) = 0.05, so P(F > 6.8) must be less than 0.05. (b) From the appendix, P(F7,3 > 8.89) = 0.05, so P(F > 3.6) must be greater than 0.05. (c) From the appendix, P(F20,20 > 2.12) = 0.05, so P(F > 2.6) must be less than 0.05. (d) From the appendix, P(F7,5 > 4.88) = 0.05, so P(F > 5.1) must be less than 0.05. (e) From the appendix, P(F7,5 > 4.88) = 0.05, so P(F 15.52) = 0.01, so P(F > 14) must be greater than 0.01. (b) From the appendix, P(F6,3 > 27.91) = 0.01, so P(F > 30) must be less than 0.01. (c) From the appendix, P(F10,12 > 4.30) = 0.01, so P(F > 4.2) must be greater than 0.01. (d) From the appendix, P(F2,3 > 30.82) = 0.01, so P(F > 35) must be less than 0.01. (e) From the appendix, P(F2,3 > 30.82) = 0.01, so P(F < 35) must be greater than 0.01. 2-55. Average time = 4 minutes = 1/µ. So µ = ¼ = 0.25 (a) P(X < 3) = 1 – e-0.25(3) = 1 – 0.4724 = 0.5276 (b) P(X < 4) = 1 – e-0.25(4) = 1 – 0.3679 = 0.6321 (c) P(X 5) = e-0.25(5) = 0.2865 2-56. Average number per minute = 5. So λ = 5 (a) P(X is exactly 5) = P(5) = (55e-5)/5! = 0.1755 (b) P(X is exactly 4) = P(4) = (54e-5)/4! = 0.1755 (c) P(X is exactly 3) = P(3) = (53e-5)/3! = 0.1404 (d) P(X is exactly 6) = P(6) = (56e-5)/6! = 0.1462 (e) P(X< 2) = P(X is 0 or 1) = P(0) + P(1) = 0.0067 + 0.0337 = 0.0404 2-57. The average time to service a customer is 1/3 hour or 20 minutes. The average number that would be served per minute (µ) is 1/20 = 0.05 per minute. P(time < ½ hour) = P(time < 30 minutes) = P(X < 30) = 1 – e-0.05(30) = 0.7769 P(time < 1/3 hour) = P(time < 20 minutes) = P(X < 20) = 1 – e-0.05(20) = 0.6321 P(time < 2/3 hour) = P(time < 40 minutes) = P(X < 40) = 1 – e-0.05(40) = 0.8647 These are exactly the same probabilities shown in the example. 2-30 Copyright © 2018 Pearson Education, Inc. 2-58. X = 280 μ = 250 σ = 25 ∴z = X −μ σ 280 − 250 25 30 = 25 = = 1.20 standard deviations From Table 2.9, the area under the curve corresponding to a Z of 1.20 = 0.88493. Therefore, the probability that the sales will be less than 280 boats is 0.8849. 2-31 Copyright © 2018 Pearson Education, Inc. 2-59. The probability of sales being over 265 boats: X = 265 μ = 250 σ = 25 265 − 250 25 15 = 25 Z= = 0.60 From Table 2.9, we find that the area under the curve to the left of Z = 0.60 is 0.72575. Since we want to find the probability of selling more than 265 boats, we need the area to the right of Z = 0.60. This area is 1 – 0.72575, or 0.27425. Therefore, the probability of selling more than 265 boats = 0.2743. For a sale of fewer than 250 boats: X = 250 μ = 250 σ = 25 However, a sale of 250 boats corresponds to μ = 250. At this point, Z = 0. The area under the curve that concerns us is that half of the area lying to the left of μ = 250. This area = 0.5000. Thus, the probability of selling fewer than 250 boats = 0.5. 2-32 Copyright © 2018 Pearson Education, Inc. 2-60. μ = 0.55 inch (average shaft size) X = 0.65 inch σ = 0.10 inch Converting to a Z value yields Z= X −μ σ 0.65 − 0.55 0.10 0.10 = 0.10 = =1 We thus need to look up the area under the curve that lies to the left of 1σ. From Table 2.9, this is seen to be = 0.84134. As seen earlier, the area to the left of μ is = 0.5000. We are concerned with the area between μ and μ + 1σ. This is given by the difference between 0.84134 and 0.5000, and it is 0.34134. Thus, the probability of a shaft size between 0.55 inch and 0.65 inch = 0.3413. 2-61. Greater than 0.65 inch: area to the left of 1σ = 0.84134 area to the right of 1σ = 1 – 0.84134 = 0.15866 Thus, the probability of a shaft size being greater than 0.65 inch is 0.1587. 2-33 Copyright © 2018 Pearson Education, Inc. The shaft size between 0.53 and 0.59 inch: X2 = 0.53 inch X1 = 0.59 inch μ = 0.55 inch Converting to scores: Z1 = X1 − μ σ 0.59-0.55 = 0.10 0.04 = 0.10 =0.4 Z2 = X2 − μ σ 0.53 − 0.55 = 0.10 −0.02 = 0.10 = −0.2 Since Table 2.9 handles only positive Z values, we need to calculate the probability of the shaft size being greater than 0.55 + 0.02 = 0.57 inch. This is determined by finding the area to the left of 0.57, that is, to the left of 0.2σ. From Table 2.9, this is 0.57926. The area to the right of 0.2σ is 1 – 0.57926 = 0.42074. The area to the left of 0.53 is also 0.42074 (the curve is symmetrical). The area to the left of 0.4σ is 0.65542. The area between X1 and X2 is 0.65542 – 0.42074 = 0.23468. The probability that the shaft will be between 0.53 inch and 0.59 inch is 0.2347. Under 0.45 inch: X = 0.45 μ = 0.55 σ = 0.10 X −μ Z= σ 0.45 − 0.55 = 0.10 −0.10 = 0.10 = –1 2-34 Copyright © 2018 Pearson Education, Inc. Thus, we need to find the area to the left of –1σ. Again, since Table 2.9 handles only positive values of Z, we need to determine the area to the right of 1σ. This is obtained by 1 – 0.84134 = 0.15866 (0.84134 is the area to the left of 1σ). Therefore, the area to the left of –1σ = 0.15866 (the curve is symmetrical). Thus, the probability that the shaft size will be under 0.45 inch is 0.1587. ⎛n⎞ 2-62. ⎜ ⎟ p x q n− x ⎝ x⎠ x=3 n=4 q = 15/20 = .75 p = 5/20 = .25 3 ⎛ 4⎞ ⎜ ⎟ (.25) (.75) = ⎝ 3⎠ 4! 3 (.25) (.75) = 3!( 4 − 3) ! (4)(.0156)(.75) = .0468 [probability that Marie will win 3 games] ⎛ 4⎞ 4 0 ⎜ ⎟ (.25) (.75 ) = .003906 [Probability that Marie will win all four games against Jan] ⎝ 4⎠ Probability that Marie will be number one is .04698 + .003906 = .050886. 2-35 Copyright © 2018 Pearson Education, Inc. 2-63. Probability one will be fined = P(2) + P(3) + P(4) + P(5) = 1 – P(0) – P(1) ⎛ 5⎞ ⎛ 5⎞ 1 0 5 4 = 1 − ⎜ ⎟ (.5) (.5) − ⎜ ⎟ (.5) (.5) ⎝ 0⎠ ⎝1⎠ = 1 – .03125 – .15625 = .08125 Probability of no foul outs = P(0) = 0.03125 Probability of foul out in all 5 games = P(5) = 0.03125 2-64. X P(X) 0 ⎛ 5⎞ 0 5 ⎜ ⎟ (.2 ) (.8) = .327 ⎝ 0⎠ .327 1 ⎛ 5⎞ 1 5−1 ⎜ ⎟ (.2 ) (.8) = .410 ⎝1⎠ .410 2 ⎛ 5⎞ 2 5− 2 ⎜ ⎟ (.2 ) (.8) = .205 ⎝ 2⎠ .205 3 ⎛ 5⎞ 3 5−3 ⎜ ⎟ (.2 ) (.8) = .051 ⎝ 3⎠ .051 4 ⎛5⎞ 4 5− 4 ⎜ ⎟ (.2 ) (.8) = .0064 ⎝ 4⎠ .0064 5 .00032 ⎛ 5⎞ 5 5 −5 .2 .8 .00032 = ( ) ( ) ⎜ ⎟ ⎝ 5⎠ 1.0 XP(X) 0.0 .41 .41 .153 .024 .0015 .9985 X – E(X) (X – E(X))2 (X – E(X))2P(X) –.9985 .0015 1.0015 2.0015 3.0015 4.0015 .997 0 1.003 4.006 9.009 16.012 .326 0 .2056 .2043 .0541 .0048 .7948 E(X) = .9985 ≅ 1.0 σ2 = Σ(X – E(X))2P(X) = .7948 ≅ 0.80 2-36 Copyright © 2018 Pearson Education, Inc. Using the formulas for the binomial: E(X) = np = (5)(.2) = 1.0 2 σ = np(1 – p) = 5(.2)(.8) = 0.80 The equation produced equivalent results. 2-65. a. n = 10; p = .25; q = .75; ⎛10 ⎞ x 10− x ⎜ ⎟p q ⎝x⎠ P(X) X ⎛10 ⎞ 0 10 −0 ⎜ ⎟ (.25) (.75) = ⎝0⎠ .0563 0 ⎛10 ⎞ 1 10−1 ⎜ ⎟ (.25) (.75) = ⎝1⎠ .1877 1 ⎛10 ⎞ 2 10− 2 ⎜ ⎟ (.25) (.75) = ⎝2⎠ .2816 2 ⎛10 ⎞ 3 10 −3 ⎜ ⎟ (.25 ) (.75 ) = ⎝3⎠ .2503 3 ⎛10 ⎞ 4 10− 4 ⎜ ⎟ (.25) (.75) = ⎝4⎠ .1460 4 ⎛10 ⎞ 5 10 −5 ⎜ ⎟ (.25 ) (.75) = ⎝5⎠ .0584 5 ⎛10 ⎞ 6 10 −6 ⎜ ⎟ (.25) (.75) = ⎝6⎠ .0162 6 ⎛10 ⎞ 7 10 −7 ⎜ ⎟ (.25) (.75) = ⎝7⎠ .0031 7 ⎛10 ⎞ 8 10 −8 ⎜ ⎟ (.25) (.75) = ⎝8⎠ .0004 8 ⎛10 ⎞ 9 10−9 ⎜ ⎟ (.25) (.75) = ⎝9⎠ .00003 9 ⎛10 ⎞ 10 10−10 = ⎜ ⎟ (.25) (.75) ⎝10 ⎠ .0000 10 2-37 Copyright © 2018 Pearson Education, Inc. b. E(X) = np = (10).25 = 2.5 σ2 = npq = (10)(.25)(.75) = 1.875 c. Expected weekly profit: $125. × 2.5 $312.50 SOLUTION TO WTVX CASE 1. The chances of getting 15 days of rain during the next 30 days can be computed by using the binomial theorem. The problem is well suited for solution by the theorem because there are two and only two possible outcomes (rain or shine) with given probabilities (70% and 30%, respectively). The formula used is: Probability of r successes = n! p r ( q n−r ) r !( n − r ) ! where n = the number of trials (in this case, the number of days = 30), r = the number of successes (number of rainy days = 15), p = probability of success (probability of rain = 70%), and q = probability of failure (probability of shine = 30%). n! 30! 15 15 p r ( q n−r ) = (.70 ) (.30 ) = .0106 r !( n − r ) ! 15!(15!) The probability of getting exactly 15 days of rain in the next 30 days is 0.0106 or a 1.06% chance. 2. Joe’s assumptions concerning the weather for the next 30 days state that what happens on one day is not in any way dependent on what happened the day before; what this says, for example, is that if a cold front passed through yesterday, it will not affect what happens today. But there are perhaps certain conditional probabilities associated with the weather (for example, given that it rained yesterday, the probability of rain today is 80% as opposed to 70%). Not being familiar with the field of meteorology, we cannot say precisely what these are. However, our contention is that these probabilities do exist and that Joe’s assumptions are fallacious. 2-38 Copyright © 2018 Pearson Education, Inc.