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# Solucionario Walpole 8 ED

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prospects for bankruptcy. Also, the probability that the first success appears in the 11th drill is pq10 = 0.014 which is even smaller. 5.95 It is a negative binomial distribution. ( x\u22121 k\u22121 ) pkqx\u2212k = ( 6\u22121 2\u22121 ) (0.25)2(0.75)4 = 0.0989. 5.96 It is a negative binomial distribution. ( x\u22121 k\u22121 ) pkqx\u2212k = ( 4\u22121 2\u22121 ) (0.5)2(0.5)2 = 0.1875. 5.97 n = 1000 and p = 0.01, with µ = (1000)(0.01) = 10. P (X < 7) = P (X \u2264 6) = 0.1301. 5.98 n = 500; (a) If p = 0.01, P (X \u2265 15) = 1\u2212 P (X \u2264 14) = 1\u2212 14\u2211 x=0 ( 500 x ) (0.01)x(0.99)500\u2212x = 0.00021. This is a very rare probability and thus the original claim that p = 0.01 is ques- tionable. (b) P (X = 3) = ( 500 3 ) (0.01)3(0.99)497 = 0.1402. (c) For (a), if p = 0.01, µ = (500)(0.01) = 5. So P (X \u2265 15) = 1\u2212 P (X \u2264 14) = 1\u2212 0.9998 = 0.0002. For (b), P (X = 3) = 0.2650\u2212 0.1247 = 0.1403. 5.99 N = 50 and n = 10. 70 Chapter 5 Some Discrete Probability Distributions (a) k = 2; P (X \u2265 1) = 1\u2212 P (X = 0) = 1\u2212 ( 2 0)( 48 10) (5010) = 1\u2212 0.6367 = 0.3633. (b) Even though the lot contains 2 defectives, the probability of reject the lot is not very high. Perhaps more items should be sampled. (c) µ = (10)(2/50) = 0.4. 5.100 Suppose n items need to be sampled. P (X \u2265 1) = 1\u2212 ( 2 0)( 48 n ) (50n ) = 1\u2212 (50\u2212n)(49\u2212n) (50)(49) \u2265 0.9. The solution is n = 34. 5.101 Define X = number of screens will detect. Then X \u223c b(x; 3, 0.8). (a) P (X = 0) = (1\u2212 0.8)3 = 0.008. (b) P (X = 1) = (3)(0.2)2(0.8) = 0.096. (c) P (X \u2265 2) = P (X = 2) + P (X = 3) = (3)(0.8)2(0.2) + (0.8)3 = 0.896. 5.102 (a) P (X = 0) = (1\u2212 0.8)n \u2264 0.0001 implies that n \u2265 6. (b) (1\u2212 p)3 \u2264 0.0001 implies p \u2265 0.9536. 5.103 n = 10 and p = 2 50 = 0.04. P (X \u2265 1) = 1\u2212 P (X = 0) \u2248 1\u2212 ( 10 0 ) (0.04)0(1\u2212 0.04)10 = 1\u2212 0.6648 = 0.3351. The approximation is not that good due to n N = 0.2 is too large. 5.104 (a) P = (22)( 3 0) (52) = 0.1. (b) P = (21)( 1 1) (52) = 0.2. 5.105 n = 200 with p = 0.00001. (a) P (X \u2265 5) = 1\u2212 P (X \u2264 4) = 1\u2212 4\u2211 x=0 ( 200 x ) (0.00001)x(1\u2212 0.00001)200\u2212x \u2248 0. This is a rare event. Therefore, the claim does not seem right. (b) µ = np = (200)(0.00001) = 0.02. Using Poisson approximation, P (X \u2265 5) = 1\u2212 P (X \u2264 4) \u2248 1\u2212 4\u2211 x=0 e\u22120.02 (0.02)x x! = 0. Chapter 6 Some Continuous Probability Distributions 6.1 (a) Area=0.9236. (b) Area=1\u2212 0.1867 = 0.8133. (c) Area=0.2578\u2212 0.0154 = 0.2424. (d) Area=0.0823. (e) Area=1\u2212 0.9750 = 0.0250. (f) Area=0.9591\u2212 0.3156 = 0.6435. 6.2 (a) The area to the left of z is 1\u2212 0.3622 = 0.6378 which is closer to the tabled value 0.6368 than to 0.6406. Therefore, we choose z = 0.35. (b) From Table A.3, z = \u22121.21. (c) The total area to the left of z is 0.5000+0.4838=0.9838. Therefore, from Table A.3, z = 2.14. (d) The distribution contains an area of 0.025 to the left of \u2212z and therefore a total area of 0.025+0.95=0.975 to the left of z. From Table A.3, z = 1.96. 6.3 (a) From Table A.3, k = \u22121.72. (b) Since P (Z > k) = 0.2946, then P (Z < k) = 0.7054/ From Table A.3, we find k = 0.54. (c) The area to the left of z = \u22120.93 is found from Table A.3 to be 0.1762. Therefore, the total area to the left of k is 0.1762+0.7235=0.8997, and hence k = 1.28. 6.4 (a) z = (17\u2212 30)/6 = \u22122.17. Area=1\u2212 0.0150 = 0.9850. (b) z = (22\u2212 30)/6 = \u22121.33. Area=0.0918. (c) z1 = (32\u22123)/6 = 0.33, z2 = (41\u221230)/6 = 1.83. Area = 0.9664\u22120.6293 = 0.3371. (d) z = 0.84. Therefore, x = 30 + (6)(0.84) = 35.04. 71 72 Chapter 6 Some Continuous Probability Distributions (e) z1 = \u22121.15, z2 = 1.15. Therefore, x1 = 30 + (6)(\u22121.15) = 23.1 and x2 = 30 + (6)(1.15) = 36.9. 6.5 (a) z = (15\u2212 18)/2.5 = \u22121.2; P (X < 15) = P (Z < \u22121.2) = 0.1151. (b) z = \u22120.76, k = (2.5)(\u22120.76) + 18 = 16.1. (c) z = 0.91, k = (2.5)(0.91) + 18 = 20.275. (d) z1 = (17\u2212 18)/2.5 = \u22120.4, z2 = (21\u2212 18)/2.5 = 1.2; P (17 < X < 21) = P (\u22120.4 < Z < 1.2) = 0.8849\u2212 0.3446 = 0.5403. 6.6 z1 = [(µ\u2212 3\u3c3)\u2212 µ]/\u3c3 = \u22123, z2 = [(µ+ 3\u3c3)\u2212 µ]/\u3c3 = 3; P (µ\u2212 3\u3c3 < Z < µ+ 3\u3c3) = P (\u22123 < Z < 3) = 0.9987\u2212 0.0013 = 0.9974. 6.7 (a) z = (32\u2212 40)/6.3 = \u22121.27; P (X > 32) = P (Z > \u22121.27) = 1\u2212 0.1020 = 0.8980. (b) z = (28\u2212 40)/6.3 = \u22121.90, P (X < 28) = P (Z < \u22121.90) = 0.0287. (c) z1 = (37\u2212 40)/6.3 = \u22120.48, z2 = (49\u2212 40)/6.3 = 1.43; So, P (37 < X < 49) = P (\u22120.48 < Z < 1.43) = 0.9236\u2212 0.3156 = 0.6080. 6.8 (a) z = (31.7\u2212 30)/2 = 0.85; P (X > 31.7) = P (Z > 0.85) = 0.1977. Therefore, 19.77% of the loaves are longer than 31.7 centimeters. (b) z1 = (29.3\u2212 30)/2 = \u22120.35, z2 = (33.5\u2212 30)/2 = 1.75; P (29.3 < X < 33.5) = P (\u22120.35 < Z < 1.75) = 0.9599\u2212 0.3632 = 0.5967. Therefore, 59.67% of the loaves are between 29.3 and 33.5 centimeters in length. (c) z = (25.5\u2212 30)/2 = \u22122.25; P (X < 25.5) = P (Z < \u22122.25) = 0.0122. Therefore, 1.22% of the loaves are shorter than 25.5 centimeters in length. 6.9 (a) z = (224 \u2212 200)/15 = 1.6. Fraction of the cups containing more than 224 mil- limeters is P (Z > 1.6) = 0.0548. (b) z1 = (191\u2212 200)/15 = \u22120.6, Z2 = (209\u2212 200)/15 = 0.6; P (191 < X < 209) = P (\u22120.6 < Z < 0.6) = 0.7257\u2212 0.2743 = 0.4514. (c) z = (230 \u2212 200)/15 = 2.0; P (X > 230) = P (Z > 2.0) = 0.0228. Therefore, (1000)(0.0228) = 22.8 or approximately 23 cups will overflow. (d) z = \u22120.67, x = (15)(\u22120.67) + 200 = 189.95 millimeters. 6.10 (a) z = (10.075\u2212 10.000)/0.03 = 2.5; P (X > 10.075) = P (Z > 2.5) = 0.0062. Therefore, 0.62% of the rings have inside diameters exceeding 10.075 cm. (b) z1 = (9.97\u2212 10)/0.03 = \u22121.0, z2 = (10.03\u2212 10)/0.03 = 1.0; P (9.97 < X < 10.03) = P (\u22121.0 < Z < 1.0) = 0.8413\u2212 0.1587 = 0.6826. (c) z = \u22121.04, x = 10 + (0.03)(\u22121.04) = 9.969 cm. 6.11 (a) z = (30\u2212 24)/3.8 = 1.58; P (X > 30) = P (Z > 1.58) = 0.0571. (b) z = (15 \u2212 24)/3.8 = \u22122.37; P (X > 15) = P (Z > \u22122.37) = 0.9911. He is late 99.11% of the time. Solutions for Exercises in Chapter 6 73 (c) z = (25\u2212 24)/3.8 = 0.26; P (X > 25) = P (Z > 0.26) = 0.3974. (d) z = 1.04, x = (3.8)(1.04) + 24 = 27.952 minutes. (e) Using the binomial distribution with p = 0.0571, we get b(2; 3, 0.0571) = ( 3 2 ) (0.0571)2(0.9429) = 0.0092. 6.12 µ = 99.61 and \u3c3 = 0.08. (a) P (99.5 < X < 99.7) = P (\u22121.375 < Z < 1.125) = 0.8697\u2212 0.08455 = 0.7852. (b) P (Z > 1.645) = 0.05; x = (1.645)(0.08) + 99.61 = 99.74. 6.13 z = \u22121.88, x = (2)(\u22121.88) + 10 = 6.24 years. 6.14 (a) z = (159.75 \u2212 174.5)/6.9 = \u22122.14; P (X < 159.75) = P (Z < \u22122.14) = 0.0162. Therefore, (1000)(0.0162) = 16 students. (b) z1 = (171.25\u2212 174.5)/6.9 = \u22120.47, z2 = (182.25\u2212 174.5)/6.9 = 1.12. P (171.25 < X < 182.25) = P (\u22120.47 < Z < 1.12) = 0.8686\u2212 0.3192 = 0.5494. Therefore, (1000)(0.5494) = 549 students. (c) z1 = (174.75\u2212 174.5)/6.9 = 0.04, z2 = (175.25\u2212 174.5)/6.9 = 0.11. P (174.75 < X < 175.25) = P (0.04 < Z < 0.11) = 0.5438\u2212 0.5160 = 0.0278. Therefore, (1000)(0.0278)=28 students. (d) z = (187.75\u2212 174.5)/6.9 = 1.92; P (X > 187.75) = P (Z > 1.92) = 0.0274. Therefore, (1000)(0.0274) = 27 students. 6.15 µ = $15.90 and \u3c3 = $1.50. (a) 51%, since P (13.75 < X < 16.22) = P ( 13.745\u221215.9 1.5 < Z < 16.225\u221215.9 1.5 ) = P (\u22121.437 < Z < 0.217) = 0.5871\u2212 0.0749 = 0.5122. (b) $18.36, since P (Z > 1.645) = 0.05; x = (1.645)(1.50) + 15.90 + 0.005 = 18.37. 6.16 (a) z = (9.55 \u2212 8)/0.9 = 1.72. Fraction of poodles weighing over 9.5 kilograms = P (X > 9.55) = P (Z > 1.72) = 0.0427. (b) z = (8.65\u2212 8)/0.9 = 0.72. Fraction of poodles weighing at most 8.6 kilograms = P (X < 8.65) = P (Z < 0.72) = 0.7642. (c) z1 = (7.25\u2212 8)/0.9 = \u22120.83 and z2 = (9.15\u2212 8)/0.9 = 1.28. Fraction of poodles weighing between 7.3 and 9.1 kilograms inclusive = P (7.25 < X < 9.15) = P (\u22120.83 < Z < 1.28) = 0.8997\u2212 0.2033 = 0.6964. 6.17 (a) z = (10, 175\u2212 10, 000)/100 = 1.75. Proportion of components exceeding 10.150 kilograms in tensile strength= P (X > 10, 175) = P (Z > 1.75) = 0.0401. (b) z1 = (9, 775 \u2212 10, 000)/100 = \u22122.25