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# Solucionario Walpole 8 ED

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\u2202eb0+b1k1+b2k2 \u2202bi )2\u2223\u2223\u2223\u2223\u2223 bi=\u3b2i, 0\u2264i\u22642 \u3c32bi = e 2(\u3b20+k1\u3b21+k2\u3b22)(\u3c320 + k 2 1\u3c3 2 1 + k 2 2\u3c3 2 2). 4.103 (a) E(Y ) = 10 \u222b 1 0 y(1\u2212 y)9 dy = \u2212 y(1\u2212 y)10|10 + \u222b 1 0 (1\u2212 y)10 dy = 1 11 . (b) E(1\u2212 Y ) = 1\u2212 E(Y ) = 10 11 . (c) V ar(Z) = V ar(1\u2212 Y ) = V ar(Y ) = E(Y 2)\u2212 [E(Y )]2 = 10 112×12 = 0.006887. Chapter 5 Some Discrete Probability Distributions 5.1 This is a uniform distribution: f(x) = 1 10 , for x = 1, 2, . . . , 10. Therefore P (X < 4) = 3\u2211 x=1 f(x) = 3 10 . 5.2 Binomial distribution with n = 12 and p = 0.5. Hence P (X = 3) = P (X \u2264 3)\u2212 P (X \u2264 2) = 0.0730\u2212 0.0193 = 0.0537. 5.3 µ = 10\u2211 x=1 x 10 = 5.5, and \u3c32 = 10\u2211 x=1 (x\u22125.5)2 10 = 8.25. 5.4 For n = 5 and p = 3/4, we have (a) P (X = 2) = ( 5 2 ) (3/4)2(1/4)3 = 0.0879, (b) P (X \u2264 3) = 3\u2211 x=0 b(x; 5, 3/4) = 1\u2212 P (X = 4)\u2212 P (X = 5) = 1\u2212 (5 4 ) (3/4)4(1/4)1 \u2212 (5 5 ) (3/4)5(1/4)0 = 0.3672. 5.5 We are considering a b(x; 20, 0.3). (a) P (X \u2265 10) = 1\u2212 P (X \u2264 9) = 1\u2212 0.9520 = 0.0480. (b) P (X \u2264 4) = 0.2375. (c) P (X = 5) = 0.1789. This probability is not very small so this is not a rare event. Therefore, P = 0.30 is reasonable. 5.6 For n = 6 and p = 1/2. (a) P (2 \u2264 X \u2264 5) = P (X \u2264 5)\u2212 P (X \u2264 1) = 0.9844\u2212 0.1094. (b) P (X < 3) = P (X \u2264 2) = 0.3438. 5.7 p = 0.7. 59 60 Chapter 5 Some Discrete Probability Distributions (a) For n = 10, P (X < 5) = P (X \u2264 4) = 0.0474. (b) For n = 20, P (X < 10) = P (X \u2264 9) = 0.0171. 5.8 For n = 8 and p = 0.6, we have (a) P (X = 3) = b(3; 8, 0.6) = P (X \u2264 3)\u2212 P (X \u2264 2) = 0.1737\u2212 0.0498 = 0.1239. (b) P (X \u2265 5) = 1\u2212 P (X \u2264 4) = 1\u2212 0.4059 = 0.5941. 5.9 For n = 15 and p = 0.25, we have (a) P (3 \u2264 X \u2264 6) = P (X \u2264 6)\u2212 P (X \u2264 2) = 0.9434\u2212 0.2361 = 0.7073. (b) P (X < 4) = P (X \u2264 3) = 0.4613. (c) P (X > 5) = 1\u2212 P (X \u2264 5) = 1\u2212 0.8516 = 0.1484. 5.10 From Table A.1 with n = 12 and p = 0.7, we have (a) P (7 \u2264 X \u2264 9) = P (X \u2264 9)\u2212 P (X \u2264 6) = 0.7472\u2212 0.1178 = 0.6294. (b) P (X \u2264 5) = 0.0386. (c) P (X \u2265 8) = 1\u2212 P (X \u2264 7) = 1\u2212 0.2763 = 0.7237. 5.11 From Table A.1 with n = 7 and p = 0.9, we have P (X = 5) = P (X \u2264 5)\u2212 P (X \u2264 4) = 0.1497\u2212 0.0257 = 0.1240. 5.12 From Table A.1 with n = 9 and p = 0.25, we have P (X < 4) = 0.8343. 5.13 From Table A.1 with n = 5 and p = 0.7, we have P (X \u2265 3) = 1\u2212 P (X \u2264 2) = 1\u2212 0.1631 = 0.8369. 5.14 (a) n = 4, P (X = 4) = 1\u2212 0.3439 = 0.6561. (b) Assuming the series went to the seventh game, the probability that the Bulls won 3 of the first 6 games and then the seventh game is given by[( 6 3 ) (0.9)3(0.1)3 ] (0.9) = 0.0131. (c) The probability that the Bulls win is always 0.9. 5.15 p = 0.4 and n = 5. (a) P (X = 0) = 0.0778. (b) P (X < 2) = P (X \u2264 1) = 0.3370. (c) P (X > 3) = 1\u2212 P (X \u2264 3) = 1\u2212 0.9130 = 0.0870. Solutions for Exercises in Chapter 5 61 5.16 Probability of 2 or more of 4 engines operating when p = 0.6 is P (X \u2265 2) = 1\u2212 P (X \u2264 1) = 0.8208, and the probability of 1 or more of 2 engines operating when p = 0.6 is P (X \u2265 1) = 1\u2212 P (X = 0) = 0.8400. The 2-engine plane has a slightly higher probability for a successful flight when p = 0.6. 5.17 Since µ = np = (5)(0.7) = 3.5 and \u3c32 = npq = (5)(0.7)(0.3) = 1.05 with \u3c3 = 1.025. Then µ± 2\u3c3 = 3.5± (2)(1.025) = 3.5± 2.050 or from 1.45 to 5.55. Therefore, at least 3/4 of the time when 5 people are selected at random, anywhere from 2 to 5 are of the opinion that tranquilizers do not cure but only cover up the real problem. 5.18 (a) µ = np = (15)(0.25) = 3.75. (b) With k = 2 and \u3c3 = \u221a npq = \u221a (15)(0.25)(0.75) = 1.677, µ ± 2\u3c3 = 3.75± 3.354 or from 0.396 to 7.104. 5.19 Let X1 = number of times encountered green light with P (Green) = 0.35, X2 = number of times encountered yellow light with P (Yellow) = 0.05, and X3 = number of times encountered red light with P (Red) = 0.60. Then f(x1, x2, x3) = ( n x1, x2, x3 ) (0.35)x1(0.05)x2(0.60)x3. 5.20 (a) ( 10 2,5,3 ) (0.225)2(0.544)5(0.231)3 = 0.0749. (b) ( 10 10 ) (0.544)10(0.456)0 = 0.0023. (c) ( 10 0 ) (0.225)0(0.775)10 = 0.0782. 5.21 Using the multinomial distribution with required probability is( 7 0, 0, 1, 4, 2 ) (0.02)(0.82)4(0.1)2 = 0.0095. 5.22 Using the multinomial distribution, we have ( 8 5,2,1 ) (1/2)5(1/4)2(1/4) = 21/256. 5.23 Using the multinomial distribution, we have( 9 3, 3, 1, 2 ) (0.4)3(0.2)3(0.3)(0.2)2 = 0.0077. 5.24 p = 0.40 and n = 6, so P (X = 4) = P (X \u2264 4)\u2212P (X \u2264 3) = 0.9590\u22120.8208 = 0.1382. 5.25 n = 20 and the probability of a defective is p = 0.10. So, P (X \u2264 3) = 0.8670. 62 Chapter 5 Some Discrete Probability Distributions 5.26 n = 8 and p = 0.60; (a) P (X = 6) = ( 8 6 ) (0.6)6(0.4)2 = 0.2090. (b) P (X = 6) = P (X \u2264 6)\u2212 P (X \u2264 5) = 0.8936\u2212 0.6846 = 0.2090. 5.27 n = 20 and p = 0.90; (a) P (X = 18) = P (X \u2264 18)\u2212 P (X \u2264 17) = 0.6083\u2212 0.3231 = 0.2852. (b) P (X \u2265 15) = 1\u2212 P (X \u2264 14) = 1\u2212 0.0113 = 0.9887. (c) P (X \u2264 18) = 0.6083. 5.28 n = 20; (a) p = 0.20, P (X \u2265 x) \u2264 0.5 and P (X < x) > 0.5 yields x = 4. (b) p = 0.80, P (Y \u2265 y) \u2265 0.8 and P (Y < y) < 0.2 yields y = 14. 5.29 Using the hypergeometric distribution, we get (a) (122 )( 40 5 ) (527 ) = 0.3246. (b) 1\u2212 ( 48 7 ) (527 ) = 0.4496. 5.30 P (X \u2265 1) = 1\u2212 P (X = 0) = 1\u2212 h(0; 15, 3, 6) = 1\u2212 ( 6 0)( 9 3) (153 ) = 53 65 . 5.31 Using the hypergeometric distribution, we get h(2; 9, 6, 4) = (42)( 5 4) (96) = 5 14 . 5.32 (a) Probability that all 4 fire = h(4; 10, 4, 7) = 1 6 . (b) Probability that at most 2 will not fire = 2\u2211 x=0 h(x; 10, 4, 3) = 29 30 . 5.33 h(x; 6, 3, 4) = (4x)( 2 3\u2212x) (63) , for x = 1, 2, 3. P (2 \u2264 X \u2264 3) = h(2; 6, 3, 4) + h(3; 6, 3, 4) = 4 5 . 5.34 h(2; 9, 5, 4) = (42)( 5 3) (95) = 10 21 . 5.35 P (X \u2264 2) = 2\u2211 x=0 h(x; 50, 5, 10) = 0.9517. 5.36 (a) P (X = 0) = h(0; 25, 3, 3) = 77 115 . (b) P (X = 1) = h(1; 25, 3, 1) = 3 25 . 5.37 (a) P (X = 0) = b(0; 3, 3/25) = 0.6815. Solutions for Exercises in Chapter 5 63 (b) P (1 \u2264 X \u2264 3) = 3\u2211 x=1 b(x; 3, 1/25) = 0.1153. 5.38 Since µ = (4)(3/10) = 1.2 and \u3c32 = (4)(3/10)(7/10)(6/9) = 504/900 with \u3c3 = 0.7483, at least 3/4 of the time the number of defectives will fall in the interval µ± 2\u3c3 = 1.2± (2)(0.7483), or from \u2212 0.297 to 2.697, and at least 8/9 of the time the number of defectives will fall in the interval µ± 3\u3c3 = 1.2± (3)(0.7483) or from \u2212 1.045 to 3.445. 5.39 Since µ = (13)(13/52) = 3.25 and \u3c32 = (13)(1/4)(3/4)(39/51) = 1.864 with \u3c3 = 1.365, at least 75% of the time the number of hearts lay between µ± 2\u3c3 = 3.25± (2)(1.365) or from 0.52 to 5.98. 5.40 The binomial approximation of the hypergeometric with p = 1 \u2212 4000/10000 = 0.6 gives a probability of 7\u2211 x=0 b(x; 15, 0.6) = 0.2131. 5.41 Using the binomial approximation of the hypergeometric with p = 0.5, the probability is 1\u2212 2\u2211 x=0 b(x; 10, 0.5) = 0.9453. 5.42 Using the binomial approximation of the hypergeometric distribution with p = 30/150 = 0.2, the probability is 1\u2212 2\u2211 x=0 b(x; 10, 0.2) = 0.3222. 5.43 Using the binomial approximation of the hypergeometric distribution with 0.7, the probability is 1\u2212 13\u2211 x=10 b(x; 18, 0.7) = 0.6077. 5.44 Using the extension of the hypergeometric distribution the probability is( 13 5 )( 13 2 )( 13 3 )( 13 3 )( 52 13 ) = 0.0129. 5.45 (a) The extension of the hypergeometric distribution gives a probability( 2 1 )( 3 1 )( 5 1 )( 2 1 )( 12 4 ) = 4 33 . (b) Using the extension of the hypergeometric distribution, we have( 2 1 )( 3 1 )( 2 2 )( 12 4 ) + (22)(31)(21)(12 4 ) + (21)(32)(21)(12 4 ) = 8 165 . 64 Chapter 5 Some Discrete Probability Distributions 5.46 Using the extension of the hypergeometric distribution