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# Solucionario Walpole 8 ED

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X(0) = µ e µ(et\u22121)+t \u2223\u2223\u2223 t=0 = µ, µ \u2032 2 = M \u2032\u2032 X(0) = µe µ(et\u22121)+t(µet + 1) \u2223\u2223\u2223 t=0 = µ(µ+ 1), and \u3c32 = µ \u2032 2 \u2212 µ2 = µ(µ+ 1)\u2212 µ2 = µ. 7.20 From MX(t) = e 4(et\u22121) we obtain µ = 6, \u3c32 = 4, and \u3c3 = 2. Therefore, P (µ\u2212 2\u3c3 < X < µ+ 2\u3c3) = P (0 < X < 8) = 7\u2211 x=1 p(x; 4) = 0.9489\u2212 0.0183 = 0.9306. 90 Chapter 7 Functions of Random Variables 7.21 Using the moment-generating function of the chi-squared distribution, we obtain µ =M \u2032 X(0) = v(1\u2212 2t)\u2212v/2\u22121 \u2223\u2223 t=0 = v, µ \u2032 2 =M \u2032\u2032 X(0) = v(v + 2) (1\u2212 2t)\u2212v/2\u22122 \u2223\u2223 t=0 = v(v + 2). So, \u3c32 = µ \u2032 2 \u2212 µ2 = v(v + 2)\u2212 v2 = 2v. 7.22 MX(t) = \u222b \u221e \u2212\u221e etxf(x) dx = \u222b \u221e \u2212\u221e ( 1 + tx+ t2x2 2! + · · ·+ t rxr r! + · · · ) f(x) dx = \u222b \u221e \u2212\u221e f(x) dx+ t \u222b \u221e \u2212\u221e xf(x) dx+ t2 2 \u222b \u221e \u2212\u221e x2f(x) dx + · · ·+ t r r! \u222b \u221e \u2212\u221e xrf(x) dx+ · · · = 1 + µt+ µ\u20321 t2 2! + · · ·+ µ\u2032r tr r! + · · · . 7.23 The joint distribution of X and Y is fX,Y (x, y) = e \u2212x\u2212y for x > 0 and y > 0. The inverse functions of u = x + y and v = x/(x + y) are x = uv and y = u(1 \u2212 v) with J = \u2223\u2223\u2223\u2223 v u1\u2212 v \u2212u \u2223\u2223\u2223\u2223 = u for u > 0 and 0 < v < 1. So, the joint distribution of U and V is fU,V (u, v) = ue \u2212uv · e\u2212u(1\u2212v) = ue\u2212u, for u > 0 and 0 < v < 1. (a) fU (u) = \u222b 1 0 ue\u2212u dv = ue\u2212u for u > 0, which is a gamma distribution with parameters 2 and 1. (b) fV (v) = \u222b\u221e 0 ue\u2212u du = 1 for 0 < v < 1. This is a uniform (0,1) distribution. Chapter 8 Fundamental Sampling Distributions and Data Descriptions 8.1 (a) Responses of all people in Richmond who have telephones. (b) Outcomes for a large or infinite number of tosses of a coin. (c) Length of life of such tennis shoes when worn on the professional tour. (d) All possible time intervals for this lawyer to drive from her home to her office. 8.2 (a) Number of tickets issued by all state troopers in Montgomery County during the Memorial holiday weekend. (b) Number of tickets issued by all state troopers in South Carolina during the Memo- rial holiday weekend. 8.3 (a) x¯ = 2.4. (b) x¯ = 2. (c) m = 3. 8.4 (a) x¯ = 8.6 minutes. (b) x¯ = 9.5 minutes. (c) Mode are 5 and 10 minutes. 8.5 (a) x¯ = 3.2 seconds. (b) x¯ = 3.1 seconds. 8.6 (a) x¯ = 35.7 grams. (b) x¯ = 32.5 grams. (c) Mode=29 grams. 8.7 (a) x¯ = 53.75. 91 92 Chapter 8 Fundamental Sampling Distributions and Data Descriptions (b) Modes are 75 and 100. 8.8 x¯ = 22.2 days, x\u2dc = 14 days and m = 8 days. x\u2dc is the best measure of the center of the data. The mean should not be used on account of the extreme value 95, and the mode is not desirable because the sample size is too small. 8.9 (a) Range = 15\u2212 5 = 10. (b) s2 = n nP i=1 x2i\u2212( nP i=1 xi) 2 n(n\u22121) = (10)(838)\u2212862 (10)(9) = 10.933. Taking the square root, we have s = 3.307. 8.10 (a) Range = 4.3\u2212 2.3 = 2.0. (b) s2 = n nP i=1 x2i\u2212( nP i=1 xi) 2 n(n\u22121) = (9)(96.14)\u221228.82 (9)(8) = 0.498. 8.11 (a) s2 = 1 n\u22121 n\u2211 x=1 (xi \u2212 x¯)2 = 114 [(2\u2212 2.4)2 + (1\u2212 2.4)2 + · · ·+ (2\u2212 2.4)2] = 2.971. (b) s2 = n nP i=1 x2i\u2212( nP i=1 xi) 2 n(n\u22121) = (15)(128)\u2212362 (15)(14) = 2.971. 8.12 (a) x¯ = 11.69 milligrams. (b) s2 = n nP i=1 x2i\u2212( nP i=1 xi)2 n(n\u22121) = (8)(1168.21)\u221293.52 (8)(7) = 10.776. 8.13 s2 = n nP i=1 x2i\u2212( nP i=1 xi) 2 n(n\u22121) = (2)(148.55)\u221253.32 (20)(19) = 0.342 and hence s = 0.585. 8.14 (a) Replace Xi in S 2 by Xi + c for i = 1, 2, . . . , n. Then X¯ becomes X¯ + c and S2 = 1 n\u2212 1 n\u2211 i=1 [(Xi + c)\u2212 (X¯ + c)]2 = 1 n\u2212 1 n\u2211 i=1 (Xi \u2212 X¯)2. (b) Replace Xi by cXi in S 2 for i = 1, 2, . . . , n. Then X¯ becomes cX¯ and S2 = 1 n\u2212 1 n\u2211 i=1 (cXi \u2212 cX¯)2 = c 2 n\u2212 1 n\u2211 i=1 (Xi \u2212 X¯)2. 8.15 s2 = n nP i=1 x2i\u2212( nP i=1 xi)2 n(n\u22121) = (6)(207)\u2212332 (6)(5) = 5.1. (a) Multiplying each observation by 3 gives s2 = (9)(5.1) = 45.9. (b) Adding 5 to each observation does not change the variance. Hence s2 = 5.1. 8.16 Denote by D the difference in scores. Solutions for Exercises in Chapter 8 93 (a) D¯ = 25.15. (b) D\u2dc = 31.00. 8.17 z1 = \u22121.9, z2 = \u22120.4. Hence, P (µX¯ \u2212 1.9\u3c3X¯ < X¯ < µX¯ \u2212 0.4\u3c3X¯) = P (\u22121.9 < Z < \u22120.4) = 0.3446\u2212 0.0287 = 0.3159. 8.18 n = 54, µX¯ = 4, \u3c3 2 X¯ = \u3c32/n = (8/3)/54 = 4/81 with \u3c3X¯ = 2/9. So, z1 = (4.15\u2212 4)/(2/9) = 0.68, and z2 = (4.35\u2212 4)/(2/9) = 1.58, and P (4.15 < X¯ < 4.35) = P (0.68 < Z < 1.58) = 0.9429\u2212 0.7517 = 0.1912. 8.19 (a) For n = 64, \u3c3X¯ = 5.6/8 = 0.7, whereas for n = 196, \u3c3X¯ = 5.6/14 = 0.4. Therefore, the variance of the sample mean is reduced from 0.49 to 0.16 when the sample size is increased from 64 to 196. (b) For n = 784, \u3c3X¯ = 5.6/28 = 0.2, whereas for n = 49, \u3c3X¯ = 5.6/7 = 0.8. Therefore, the variance of the sample mean is increased from 0.04 to 0.64 when the sample size is decreased from 784 to 49. 8.20 n = 36, \u3c3X¯ = 2. Hence \u3c3 = \u221a n\u3c3X¯ = (6)(2) = 12. If \u3c3X¯ = 1.2, then 1.2 = 12/ \u221a n and n = 100. 8.21 µX¯ = µ = 240, \u3c3X¯ = 15/ \u221a 40 = 2.372. Therefore, µX¯±2\u3c3X¯ = 240± (2)(2.372) or from 235.257 to 244.743, which indicates that a value of x = 236 milliliters is reasonable and hence the machine needs not be adjusted. 8.22 (a) µX¯ = µ = 174.5, \u3c3X¯ = \u3c3/ \u221a n = 6.9/5 = 1.38. (b) z1 = (172.45\u2212 174.5)/1.38 = \u22121.49, z2 = (175.85\u2212 174.5)/1.38 = 0.98. So, P (172.45 < X¯ < 175.85) = P (\u22121.49 < Z < 0.98) = 0.8365\u2212 0.0681 = 0.7684. Therefore, the number of sample means between 172.5 and 175.8 inclusive is (200)(0.7684) = 154. (c) z = (171.95\u2212 174.5)/1.38 = \u22121.85. So, P (X¯ < 171.95) = P (Z < \u22121.85) = 0.0322. Therefore, about (200)(0.0322) = 6 sample means fall below 172.0 centimeters. 8.23 (a) µ = \u2211 xf(x) = (4)(0.2) + (5)(0.4) + (6)(0.3) + (7)(0.1) = 5.3, and \u3c32 = \u2211 (x \u2212 µ)2f(x) = (4 \u2212 5.3)2(0.2) + (5 \u2212 5.3)2(0.4) + (6 \u2212 5.3)2(0.3) + (7 \u2212 5.3)2(0.1) = 0.81. 94 Chapter 8 Fundamental Sampling Distributions and Data Descriptions (b) With n = 36, µX¯ = µ = 5.3 and \u3c3X¯ = \u3c3 2/n = 0.81/36 = 0.0225. (c) n = 36, µX¯ = 5.3, \u3c3X¯ = 0.9/6 = 0.15, and z = (5.5\u2212 5.3)/0.15 = 1.33. So, P (X¯ < 5.5) = P (Z < 1.33) = 0.9082. 8.24 n = 36, µX¯ = 40, \u3c3X¯ = 2/6 = 1/3 and z = (40.5\u2212 40)/(1/3) = 1.5. So, P ( 36\u2211 i=1 Xi > 1458 ) = P (X¯ > 40.5) = P (Z > 1.5) = 1\u2212 0.9332 = 0.0668. 8.25 (a) P (6.4 < X¯ < 7.2) = P (\u22121.8 < Z < 0.6) = 0.6898. (b) z = 1.04, x¯ = z(\u3c3/ \u221a n) + µ = (1.04)(1/3) + 7 = 7.35. 8.26 n = 64, µX¯ = 3.2, \u3c3X¯ = \u3c3/ \u221a n = 1.6/8 = 0.2. (a) z = (2.7\u2212 3.2)/0.2 = \u22122.5, P (X¯ < 2.7) = P (Z < \u22122.5) = 0.0062. (b) z = (3.5\u2212 3.2)/0.2 = 1.5, P (X¯ > 3.5) = P (Z > 1.5) = 1\u2212 0.9332 = 0.0668. (c) z1 = (3.2\u2212 3.2)/0.2 = 0, z2 = (3.4\u2212 3.2)/0.2 = 1.0, P (3.2 < X¯ < 3.4) = P (0 < Z < 1.0) = 0.9413\u2212 0.5000 = 0.3413. 8.27 n = 50, x¯ = 0.23 and \u3c3 = 0.1. Now, z = (0.23\u2212 0.2)/(0.1/\u221a50) = 2.12; so P (X¯ \u2265 0.23) = P (Z \u2265 2.12) = 0.0170. Hence the probability of having such observations, given the mean µ = 0.20, is small. Therefore, the mean amount to be 0.20 is not likely to be true. 8.28 µ1\u2212µ2 = 80\u221275 = 5, \u3c3X¯1\u2212X¯2 = \u221a 25/25 + 9/36 = 1.118, z1 = (3.35\u22125)/1.118 = \u22121.48 and z2 = (5.85\u2212 5)/1.118 = 0.76. So, P (3.35 < X¯1 \u2212 X¯2 < 5.85) = P (\u22121.48 < Z < 0.76) = 0.7764\u2212 0.0694 = 0.7070. 8.29 µX¯1\u2212X¯2 = 72 \u2212 28 = 44, \u3c3X¯1\u2212X¯2 = \u221a 100/64 + 25/100 = 1.346 and z = (44.2 \u2212 44)/1.346 = 0.15. So, P (X¯1 \u2212 X¯2 < 44.2) = P (Z < 0.15) = 0.5596. 8.30 µ1 \u2212 µ2 = 0, \u3c3X¯1\u2212X¯2 = 50 \u221a 1/32 + 1/50 = 11.319. (a) z1 = \u221220/11.319 = \u22121.77, z2 = 20/11.319 = 1.77, so P (|X¯1 \u2212 X¯2| > 20) = 2P (Z < \u22121.77) = (2)(0.0384) = 0.0768. (b) z1 = 5/11.319 = 0.44 and z2 = 10/11.319 = 0.88. So, P (\u221210 < X¯1 \u2212 X¯2 < \u22125) + P (5 < X¯1 \u2212 X¯2 < 10) = 2P (5 < X¯1 \u2212 X¯2 < 10) = 2P (0.44 < Z < 0.88) = 2(0.8106\u2212 0.6700) = 0.2812. 8.31 The normal quantile-quantile