(Solution Manual) Mathematical Statistics With Applications

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Math & Statistics

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lOMoAR cPSD|5574646 SOLUTION MANUAL MATHEMATICAL STATISTICS WITH APPLICATIONS 7TH EDITION, WACKERLY lOMoAR cPSD|5574646 lOMoAR cPSD|5574646 Chapter 1: What is Statistics? 1.1 a. Population : all generation X age US citizens (specifically, assign a ‘1’ to those who want to start their own business and a ‘0’ to those who do not, so that the population is the set of 1’s and 0’s). Objective : to estimate the proportion of generation X age US citizens who want to start their own business. b. Population : all healthy adults in the US. Objective : to estimate the true mean body temperature c. Population : single family dwelling units in the city. Objective : to estimate the true mean water consumption d. Populati on: all tires manufactured by the company for the specific year. Objective : to estimate the proportion of tires with unsafe tread. e. Population : all adult residents of the particular state. Objective : to estimate the proportion who favor a unicameral legislature. f. Population : times until recurrence for all people who have had a particular disease. Objective : to estimate the true average time until recurrence. g. Population : lifetime measurements for all resistors of this type. Objective : to esti mate the true mean lifetime (in hours). Histogram of wind 1.2 a. This histogram is above. b. Yes, it is quite windy there. c. 11/45, or approx. 24.4% 5 10 15 20 25 30 35 wind d. it is not especially windy in the overall sample. 1 lOMoAR cPSD|5574646 2 Chapter 1: What is Statistics? Instructor’s Solutions Manual Histogram of U235 1.3 The histogram is above. 0 2 4 6 8 10 12 U235 Histogram of stocks 1.4 a. The histogram is above. b. 18/40 = 45% c. 29/40 = 72.5% 2 4 6 8 10 12 stocks 1.5 a. The categories with the largest grouping of students are 2.45 to 2.65 and 2.65 to 2.85. (both have 7 students). b. 7/30 c. 7/30 + 3/30 + 3/30 + 3/30 = 16/30 1.6 a. The modal category is 2 (quarts of milk). About 36% (9 people) of the 25 are in this category. b. .2 + .12 + .04 = .36 c. Note that 8% purchased 0 while 4% purchased 5. Thus, 1 – .08 – .04 = .88 purchased between 1 and 4 quarts. lOMoAR cPSD|5574646 Downloaded by Gavin Singh (gavin.ricardo123@gmail.com) n Chapter 1: What is Statistics? 3 Instructor’s Solutions Manual 1.7 a. There is a possibility of bimodality in the distribution. b. There is a dip in heights at 68 inches. c. If all of the students are roughly the same age, the bimodality could be a result of the men/women distributions. Histogram of AIO 1.8 a. The histogram is above. 10 12 14 16 18 20 AIO b. The data appears to be bimodal. Llanederyn and Caldicot have lower sample values than the other two. 1.9 a. Note that 9.7 = 12 – 2.3 and 14.3 = 12 + 2.3. So, (9.7, 14.3) should contain approximately 68% of the values. b. Note that 7.4 = 12 – 2(2.3) and 16.6 = 12 + 2(2.3). So, (7.4, 16.6) should contain approximately 95% of the values. c. From parts (a) and (b) above, 95% - 68% = 27% lie in both (14.3. 16.6) and (7.4, 9.7). By symmetry, 13.5% should lie in (14.3, 16.6) so that 68% + 13.5% = 81.5% are in (9.7, 16.6) d. Since 5.1 and 18.9 represent three standard deviations away from the mean, the proportion outsid e of these limits is approximately 0. 1.10 a. 14 – 17 = -3. b. Since 68% lie within one standard deviation of the mean, 32% should lie outside. By symmetry, 16% should lie below one standard deviation from the mean. c. If normally distributed, approximately 16% of people would spend less than –3 hours on the internet. Since this doesn’t make sense, the population is not normal. 1.11 a. c = c + c + … + c = nc. i1 b. cyi i1 n = c(y1 + … + yn) = c yi i1 c. xi  yi  = x1 + y1 + x2 + y2 + … + xn + yn = (x1 + x2 + … + xn) + (y1 + y2 + … + yn) i1 n n lOMoAR cPSD|5574646 www.elsolucionario.net n n n 4 Chapter 1: What is Statistics? Instructor’s Solutions Manual Using the above, the numerator of s2 is ( y  y)2 = ( y 2  2 y y  y 2 ) =  y 2  i i1 n n n 2 i i i1 n 2 2 2 i i1 1 n 2 y yi  ny Since ny  yi , we have ( yi  y) =  yi  ny . Let y  n  yi i1 to get the result. i1 i1 i1 i1 6 6 1.12 Using the data,  y = 14 and  y 2 = 40. So, s2 = (40 - 142/6)/5 = 1.47. So, s = 1.21. i i1 i i1 45 45 1.13 a. With  y = 440.6 and  y 2 = 5067.38, we have that y = 9.79 and s = 4.14. i i1 b. i i1 k interval frequency Exp. frequency 1 5.65, 13.93 44 30.6 2 1.51, 18.07 44 42.75 3 -2.63, 22.21 44 45 25 25 1.14 a. With  y = 80.63 and  y 2 = 500.7459, we have that y = 3.23 and s = 3.17. i i1 b. i i1 k interval frequency Exp. frequency 1 0.063, 6.397 21 17 2 -3.104, 9.564 23 23.75 3 -6.271, 12.731 25 25 40 40 1.15 a. With  y = 175.48 and  y 2 = 906.4118, we have that y = 4.39 and s = 1.87. i i1 b. i i1 k interval frequency Exp. frequency 1 2.52, 6.26 35 27.2 2 0.65, 8.13 39 38 3 -1.22, 10 39 40 1.16 a. Without the extreme value, y = 4.19 and s = 1.44. b. These counts compare more favorably: k interval frequency Exp. frequency 1 2.75, 5.63 25 26.52 2 1.31, 7.07 36 37.05 3 -0.13, 8.51 39 39 lOMoAR cPSD|5574646 Chapter 1: What is Statistics? 5 Instructor’s Solutions Manual 1.17 For Ex. 1.2, range/4 = 7.35, while s = 4.14. For Ex. 1.3, range/4 = 3.04, while = s = 3.17. For Ex. 1.4, range/4 = 2.32, while s = 1.87. 1.18 The approximation is (800–200)/4 = 150. 1.19 One standard deviation below the mean is 34 – 53 = –19. The empirical rule suggests that 16% of all measurements should lie one standard deviation below the mean. Since chloroform measurements cannot be negative, this population cannot be normally distributed. 1.20 Since approximately 68% will fall between $390 ($420 – $30) to $450 ($420 + $3 0), the proportion above $450 is approximately 16%. 1.21 (Similar to exercise 1.20) Having a gain of more than 20 pounds represents all measurements greater than one standard deviation below the mean. By the empirical rule, the proportion above this value is a pproximately 84%, so the manufacturer is probably correct. n n n n 1.22 (See exercise 1.11) ( yi  y) =  yi – ny  yi  yi  0 . i1 i1 i1 i1 1.23 a. (Similar to exercise 1.20) 95 sec = 1 standard deviation above 75 sec, so this percentage is 16% by the empirical rule. b. (35 sec., 115 sec) represents an interval of 2 standard deviations about the mean, so approximately 95% c. 2 minutes = 120 sec = 2.5 standard deviations above the mean. This is unlikely. 1.24 a. (112-78)/4 = 8.5 Histogram of hr b. The histogram is above. 20 20 80 90 100 110 hr c. With  y = 1874.0 and  y 2 = 117,328.0, we have that y = 93.7 and s = 9.55. i i1 i i1 lOMoAR cPSD|5574646 6 Chapter 1: What is Statistics? Instructor’s Solutions Manual d. 1.25 a. (716-8)/4 = 177 b. The figure is omitted. 88 88 c. With  y = 18,550 and  y 2 = 6,198,356, we have that y = 210.8 and s = 162.17. i i1 d. i i1 1.26 For Ex. 1.12, 3/1.21 = 2.48. For Ex. 1.24, 34/9.55 = 3.56. For Ex. 1.25, 708/162.17 = 4.37. The ratio increases as the sample size increases. 1.27 (64, 80) is one standard deviation about the mean, so 68% of 340 or approx. 231 scores. (56, 88) is two standard deviations about the mean, so 95% of 340 or 323 scores. 1.28 (Similar to 1.23) 13 mg/L is one standard deviation below the mean, so 16%. 1.29 If the empirical rule is assumed, approximately 95% of all bearing should lie in (2.98, 3.02) – this interval represents two standard deviations about the mean. So, approximately 5% will lie outside of this interval. 1.30 If た = 0 and j = 1.2, we expect 34% to be between 0 and 0 + 1.2 = 1.2. Also, approximately 95%/2 = 47.5% will lie between 0 an d 2.4. So, 47.5% – 34% = 13.5% should lie between 1.2 and 2.4. 1.31 Assuming normality, approximately 95% will lie between 40 and 80 (the standard deviation is 10). The percent below 40 is approximately 2.5% which is relatively unlikely. 1.32 For a sample of size n, let nガ denote the number of measurements that fall outside the interval y ± ks, so that (n – nガ)/n is the fraction that falls inside the interval. To show this fraction is greater than or equal to 1 – 1/k2, note that (n – 1)s2 = ( y iA  y)2 + ( yi ib  y)2 , (both sums must be positive) where A = {i: |yi - y | ≥ ks} and B = {i: |yi – y | < ks}. We have that ( yi iA  y)2 ≥ k 2 s2 = nガk2s2, since if i is in A, |yi – y | ≥ ks and there are nガ elements in iA A. Thus , we have that s2 ≥ k2s2nガ/(n-1), or 1 ≥ k2nガ/(n–1) ≥ k2nガ/n. Thus, 1/ k2 ≥ nガ/n or (n – nガ)/n ≥ 1 – 1/k2. i k interval frequency Exp. frequency 1 84.1, 103.2 13 13.6 2 74.6, 112.8 20 19 3 65.0, 122.4 20 20 k interval frequency Exp. frequency 1 48.6, 373 63 59.84 2 -113.5, 535.1 82 83.6 3 -275.7, 697.3 87 88 lOMoAR cPSD|5574646 Chapter 1: What is Statistics? 7 Instructor’s Solutions Manual 1.33 With k =2, at least 1 – 1/4 = 75% should lie within 2 standard deviations of the mean. The interval is (0.5, 10.5). 1.34 The point 13 is 13 – 5.5 = 7.5 units above the mean, or 7.5/2.5 = 3 standard deviations above the mean. By Tchebysheff’s theorem, at least 1 – 1/32 = 8/9 will lie within 3 standard deviations of the mean. Thus, at most 1/9 of the values will exceed 13. 1.35 a. (172 – 108)/4 =16 15 15 b. With  y = 2041 and  y 2 = 281,807 we have that y = 136.1 and s = 17.1 i i1 i i1 c. a = 136.1 – 2(17.1) = 101.9, b = 136.1 + 2(17.1) = 170.3. d. There are 14 observations contained in this interval, and 14/15 = 93.3%. 75% is a lower bound. 1.36 a. The histogram is above. 0 1 2 3 4 5 6 8 ex1.36 100 100 b. With  y = 66 and  y 2 = 234 we have that y = 0.66 and s = 1.39. i i1 i i1 c. Within two standard deviations: 95, within three standard deviations: 96. The calculations agree with Tchebysheff’s theorem. 1.37 Since the lead readings must be non negative, 0 (the smallest possible value) is only 0.33 standard deviations from the mean. This indicates that the distribution is skewed. 1.38 By Tchebysheff’s theorem, at least 3/4 = 75% lie between (0, 140), at least 8/9 lie between (0, 193), and at least 15/16 lie between (0, 246). The lower bounds are all truncated a 0 since the measurement cannot be negative. lOMoAR cPSD|5574646 Chapter 2: Probability 2.1 A = {FF}, B = {MM}, C = {MF, FM, MM}. Then, AyB = 0 , ByC = {MM}, C ∩ B = {MF, FM}, A  B ={FF,MM}, A  C = S, B  C = C. 2.2 a. AyB b. A  B c. A  B d. ( A  B )  ( A  B) 2.3 2.4 a. b. 8 lOMoAR cPSD|5574646 Chapter 2: Probability 9 Instructor's Solutions Manual 2.5 a. ( A  B)  ( A  B )  A  (B  B )  A  S  A . b. B  ( A  B )  (B  A)  (B  B )  (B  A)  A . c. ( A  B)  ( A  B )  A  (B  B )  0 . The result follows from part a. d. B  ( A  B )  A  (B  B ) = 0 . The result follows from part b. 2.6 A = {(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (1,6), (2,6), (3,6), (4,6), (5,6), (6,6)} C = {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)} AyB = {(2,2), (4,2), (6,2), (2,4), (4,4), (6,4), (2,6), (4,6), (6,6)} A  B = {(1,2), (3,2), (5,2), (1,4), (3,4), (5,4), (1,6), (3,6), (5,6)} A  B = everything but {(1,2), (1,4), (1,6), (3,2), (3,4), (3,6), (5,2), (5,4), (5,6)} A  C  A 2.7 A = {two males} = {M 1, M2), (M1,M3), (M2,M3)} B = {at least one female} = {(M 1,W1), (M2,W1), (M3,W1), (M1,W2), (M2,W2), (M3,W2), {W 1,W2)} B = {no females} = A A  B  S A  B  0 A  B  A 2.8 a. 36 + 6 = 42 b. 33 c. 18 2.9 S = {A+, B+, AB+, O+, A-, B-, AB-, O-} 2.10 a. S = {A, B, AB, O} b. P({A}) = 0.41, P({B}) = 0.10, P({AB}) = 0.04, P({O}) = 0.45. c. P({A} or {B}) = P({A}) + P({B}) = 0.51, since the events are mutually exclusive. 2.11 a. Since P(S) = P(E1) + … + P(E5) = 1, 1 = .15 + .15 + .40 + 3P(E5). So, P(E5) = .10 and P(E4) = .20. b. Obviously, P(E3) + P(E4) + P(E5) = .6. Thus, they are all equal to .2 2.12 a. Let L = {left tern}, R = {right turn}, C = {continues straight}. b. P(vehicle turns) = P(L) + P(R) = 1/3 + 1/3 = 2/3. 2.13 a. Denote the events as very likely (VL), somewhat likely (SL), unlikely (U), other (O). b. Not equally likely: P(VL) = .24, P(SL) = .24, P(U) = .40, P(O) = .12. c. P(at least SL) = P(SL) + P(VL) = .48. 2.14 a. P(needs glasses) = .44 + .14 = .48 b. P(needs glasses but doesn't use them) = .14 c. P(uses glasses) = .44 + .02 = .46 2.15 a. Since the events are M.E., P(S) = P(E1) + … + P(E4) = 1. So, P(E2) = 1 – .01 – .09 – .81 = .09. b. P(at least one hit) = P(E1) + P(E2) + P(E3) = .19. lOMoAR cPSD|5574646 10 Chapter 2: Probability Instructor's Solutions Manual 2.16 a. 1/3 b. 1/3 + 1/15 = 6/15 c. 1/3 + 1/16 = 19/48 d. 49/240 2.17 Let B = bushing defect, SH = shaft defect. a. P(B) = .06 + .02 = .08 b. P(B or SH) = .06 + .08 + .02 = .16 c. P(exactly one defect) = .06 + .08 = .14 d. P(neither defect) = 1 – P(B or SH) = 1 – .16 = .84 2.18 a. S = {HH, TH, HT, TT} b. if the coin is fair, all events have probability .25. c. A = {HT, TH}, B = {HT, TH, HH} d. P(A) = .5, P(B) = .75, P( A  B ) = P(A) = .5, P( A  B ) = P(B) = .75, P( A  B ) = 1. 2.19 a. (V1, V1), (V1, V2), (V1, V3), (V2, V1), (V2, V2), (V2, V3), (V3, V1), (V3, V2), (V3, V3) b. if equally likely, all have probability of 1/9. c. A = {same vendor gets both} = {(V 1, V1), (V2, V2), (V3, V3)} B = {at least one V2} = {(V 1, V2), (V 2, V1), (V 2, V2), (V 2, V3), (V 3, V2)} So, P(A) = 1/3, P(B) = 5/9, P( A  B ) = 7/9, P( A  B ) = 1/9. 2.20 a. P(G) = P(D1) = P(D2) = 1/3. b. i. The probability of selecting the good prize is 1/3. ii. She will get the other dud. iii. She will get the good prize. iv. Her probability of winning is now 2/3. v. The best strategy is to switch. 2.21 P(A) = P(( A  B)  ( A  B ) ) = P ( A  B) + P ( A  B ) since these are M.E. by Ex. 2.5. 2.22 P(A) = P( B  ( A  B ) ) = P(B) + P ( A  B ) since these are M.E. by Ex. 2.5. 2.23 All elements in B are in A, so that when B occurs, A must also occur. However, it is possible for A to occur and B not to occur. 2.24 From the relation in Ex. 2.22, P ( A  B ) ≥ 0, so P(B) ≤ P(A). 2.25 Unless exactly 1/2 of all cars in the lot are Volkswagens, the claim is not true. 2.26 a. Let N1, N2 denote the empty cans and W1, W 2 denote the cans filled with water. Thus, S = {N1N2, N1W2, N2W2, N1W1, N2W1, W1W2} b. If this a merely a guess, the events are equally likely. So, P(W 1W2) = 1/6. 2.27 a. S = {CC, CR, CL, RC, RR, RL, LC, LR, LL} b. 5/9 c. 5/9 lOMoAR cPSD|5574646 Chapter 2: Probability 11 Instructor's Solutions Manual 2.28 a. Denote the four candidates as A 1, A2, A3, and M. Since order is not important, the outcomes are {A 1A2, A1A3, A1M, A 2A3, A2M, A 3M}. b. assuming equally likely outcomes, all have probability 1/6. c. P(minority hired) = P(A1M) + P(A2M) + P(A3M) = .5 2.29 a. The experiment consists of randomly selecting two jurors from a group of two women and four men. b. Denoting the women as w1, w2 and the men as m1, m2, m3, m4, the sample space is w1,m1 w2,m1 m1,m2 m2,m3 m3,m4 w1,m2 w2,m2 m1,m3 m2,m4 w1,m3 w2,m3 m1,m4 w1,m4 w2,m4 w1,w2 c. P(w1,w2) = 1/15 2.30 a. Let w1 denote the first wine, w2 the second, and w3 the third. Each sample point is an ordered triple indicating the ranking. b. triples: (w1,w2,w3), (w1,w3,w2), (w2,w1,w3), (w2,w3,w1), (w3,w1,w2), (w3,w2,w1) c. For each wine, there are 4 ordered triples where it is not last. So, the probability is 2/3. 2.31 a. There are four “good” systems and two “defactive” systems. If two out of the six systems are chosen randomly, there are 15 possible unique pairs. Denoting the systems as g1, g2, g3, g4, d1, d2, the sample space is given by S = {g1g2, g1g3, g1g4, g1d1, g1d2, g2g3, g2g4, g2d1, g2d2, g3g4, g3d1, g3d2, g4g1, g4d1, d1d2}. Thus: P(at least one defective) = 9/15 P(both defective) = P(d1d2) = 1/15 b. If four are defective, P(at least one defective) = 14/15. P(both defective) = 6/15. 2.32 a. Let “1” represent a customer seeking style 1, and “2” represent a customer seeking style 2. The sample space consists of the following 16 four-tuples: 1111, 1112, 1121, 1211, 2111, 1122, 1212, 2112, 1221, 2121, 2211, 2221, 2212, 2122, 1222, 2222 b. If the styles are equally in demand, the ordering should be equally likely. So, the probability is 1/16. c. P(A) = P(1111) + P(2222) = 2/16. 2.33 a. Define the events: G = family income is greater than $43,318, N otherwise. The points are E1: GGGG E2: GGGN E3: GGNG E4: GNGG E5: NGGG E6: GGNN E7: GNGN E8: NGGN E9: GNNG E10: NGNG E11: NNGG E12: GNNN E13: NGNN E14: NNGN E15: NNNG E16: NNNN b. A = {E1, E2, …, E11} B = {E6, E7, …, E11} C = {E2, E3, E4, E5} c. If P(E) = P(N) = .5, each element in the sample space has probability 1/16. Thus, P(A) = 11/16, P(B) = 6/16, and P(C) = 4/16. lOMoAR cPSD|5574646 3 12 Chapter 2: Probability Instructor's Solutions Manual 2.34 a. Three patients enter the hospital and randomly choose stations 1, 2, or 3 for service. Then, the sample space S contains the following 27 three -tuples: 111, 112, 113, 121, 122, 123, 131, 132, 133, 211, 212, 213, 221, 222, 223, 231, 232, 233, 311, 312, 313, 321, 322, 323, 331, 332, 333 b. A = {123, 132, 213, 231, 312, 321} c. If the stations are selected at random, each sample point is equally likely. P(A) = 6/27. 2.35 The total number of flights is 6(7) = 42. 2.36 There are 3! = 6 orderings. 2.37 a. There are 6! = 720 possible itineraries. b. In the 720 orderings, exactly 360 have Denver before San Francisco and 360 have San Francisco before Denver. So, the probability is .5. 2.38 By the mn rule, 4(3)(4)(5) = 240. 2.39 a. By the mn rule, there are 6(6) = 36 possible roles. b. Define the event A = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}. Then, P(A) = 6/36. 2.40 a. By the mn rule, the dealer must stock 5(4)(2) = 40 autos. b. To have each of these in every one of the eight colors, he must stock 8*40 = 320 autos. 2.41 If the first digit cannot be zero, there are 9 possible values. For the remaining six, there are 10 possible values. Thus, the total number is 9(10)(10)(10)(10)(10)(10) = 9*106. 2.42 There are three different positions to fill using ten engineers. Then, there are = 720 different ways to fill the positions. P10 = 10!/3! 961 2.43    = 504 ways.  3 51  85 2.44 a. The number of ways the taxi needing repair can be sent to airport C is   = 56. So, the probability is 56/504 = 1/9. 6 455 b. 3   = 45, so the probability that every airport receives one of the taxis requiring  2 4 repair is 45/504.  17  2.45   = 408,408.  2 7 10      