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# Mathematical Statistics with Applications, Seventh Edition

www.downloadslide.com Answers 885 5.79 0 5.81 1 5.83 1 5.85 a E(Y 1 ) = E(Y 2 ) = 1 (both marginal distributions are exponential with mean 1) b V (Y 1 ) = V (Y 2 ) = 1 c E(Y 1 − Y 2 ) = 0 d E(Y 1 Y 2 ) = 1 − α 4 ,so e Cov(Y 1 , Y 2 ) =− α ( 4 −2 √2 + α √ 2 , 2 2 + α ) 2 5.87 a E(Y 1 + Y 2 ) = ν 1 + ν 2 b V (Y 1 + Y 2 ) = 2ν 1 + 2ν 2 5.89 Cov(Y 1, Y 2 ) =− 2 9 . As the value of Y 1 increases, the value of Y 2 tends to decrease. 5.91 Cov(Y 1, Y 2 ) = 0 5.93 a 0 b Dependent c 0 d Not necessarily independent 5.95 The marginal distributions for Y 1 and Y 2 are y 1 −1 0 1 y 2 0 1 1 1 1 p 1 (y 1 ) p 2 (y 2 ) 2 1 3 3 3 3 3 Cov(Y 1, Y 2 ) = 0 5.97 a 2 b Impossible c 4 (a perfect positive linear association) d −4 (a perfect negative linear association) 5.99 0 5.101 a − α 4 5.103 E(3Y 1 + 4Y 2 − 6Y 3 ) =−22, V (3Y 1 + 4Y 2 − 6Y 3 ) = 480 5.105 1 9 5.107 E(Y 1 + Y 2 ) = 2/3 and V (Y 1 + Y 2 ) = 1 18 5.109 (11.48, 52.68) 5.113 E(G) = 42, V (G) = 25; the value \$70 70 − 42 is = 7.2 standard deviations 5 above the mean, an unlikely value. 5.115 b V (Y ) = 38.99 c The interval is 14.7 ± 2 √ 38.99 or (0, 27.188) 5.117 p 1 − p 2 , N − n n(N − 1) [p 1 + p 2 − (p 1 − p 2 ) 2 ] 5.119 a .0823 b E(Y 1 ) = n 3 , V (Y 1) = 2n 9 c Cov(Y 2 , Y 3 ) =− n 9 d E(Y 2 − Y 3 ) = 0, V (Y 2 − Y 3 ) = 2n 3 5.121 a .0972 b .2; .072 5.123 .08953 5.125 a .046 b .2262 5.127 a .2759 b .8031 5.133 a y 2 2 b 1 4 5.135 a 3 2 b 1.25 5.137 3 8 5.139 a nαβ b λαβ 5.141 E(Y 2 ) = λ 2 , V (Y 2) = 2λ2 3 5.143 m U (t) = (1 − t 2 ) −1/2 , E(U) = 0, V (U) = 1 5.145 1 3 5.147 11 36 5.149 a f (y 1 ) = 3y1 2,0≤ y 1 ≤ 1 f (y 2 ) = 3 2 (1 − y2 2 ),0≤ y 2 ≤ 1 b 23 44 c f (y 1 |y 2 ) = 2y 1 (1 − y2 2) , y 2 ≤ y 1 ≤ 1 d 5 12 5.157 ( p(y) = )( ) y + α − 1 β y ( ) 1 α , y β + 1 β + 1 y = 0, 1, 2, ... 5.161 E(Ȳ − ¯X) = µ 1 − µ 2 , V (Ȳ − ¯X) = σ1 2/n + σ 2 2/m

www.downloadslide.com 886 Answers 5.163 b F(y 1 , y 2 ) = y 1 y 2 [1 − α(1 − y 1 )(1 − y 2 )] c f (y 1 , y 2 ) = 1 − α[(1 − 2y 1 )(1 − 2y 2 )], 0 ≤ y 1 ≤ 1, 0 ≤ y 2 ≤ 1 d Choose two different values for α with −1 ≤ α ≤ 1. 5.165 a (p 1 e t 1 + p 2 e t 2 + p 3 e t 3) n b m(t, 0,0) c Cov(X 1, X 2 ) =−np 1 p 2 Chapter 6 6.1 a 1 − u 6.29 a f , −1 ≤ u ≤ 1 W (w) = 2 1 b u + 1 , −1 ≤ u ≤ 1 Ɣ ( ) w 1/2 e −w/kT w > 0 3 2 (kT) 3/2 2 1 b E(W ) = 3 c √ − 1, 0 ≤ u ≤ 1 2 kT u 2 d E(U 1 ) =−1/3, E(U 2 ) = 6.31 f U (u) = (1 + u) , u ≥ 0 3 1/3, E(U 3 ) = 1/6 6.33 f U (u) = 4(80 − 31u + 3u 2 ), e E(2Y − 1) =−1/3, E(1 − 2Y ) = 4.5 ≤ u ≤ 5 1/3, E(Y 2 ) = 1/6 6.35 f U (u) =−ln(u), 0≤ u ≤ 1 6.3 b { f U (u) = 6.37 a m Y1 (t) = 1 − p + pe t (u + 4)/100, −4 ≤ u ≤ 6 b m W (t) = E(e tW ) = [1 − p + pe t ] n 1/10, 6 < u ≤ 11 6.39 f U (u) = 4ue −2u , u ≥ 0 c 5.5833 6.5 f U (u) = 1 ( ) u − 3 −1/2 6.43 a Ȳ has a normal distribution , with mean µ and variance σ 2 /n 16 2 b P(|Ȳ − µ| ≤1) = .7888 5 ≤ u ≥ 53 1 c The probabilities are .8664, .9544, 6.7 a f U (u) = √ √ u −1/2 e −u/2 , .9756. So, as the sample size π 2 u ≥ 0 increases, so does the probability b U has a gamma distribution with that P(|Ȳ − µ| ≤1) α = 1/2 and β = 2 (recall that 6.45 Ɣ(1/2) = √ c = \$190.27 π). 6.47 P(U > 16.0128) = .025 6.9 a f U (u) = 2u,0≤ u ≤ 1 6.51 The distribution of Y 1 + (n 2 − Y 2 ) is b E(U) = 2/3 binomial with n 1 + n 2 trials and success c E(Y 1 + Y 2 ) = 2/3 probability p = .2 6.11 a f U (u) = 4ue −2u , u ≥ 0, a gamma 6.53 a Binomial (nm, p) where density with α = 2 n i = m and β = 1/2 b Binomial (n 1 = n 2 +···n n , p) b E(U) = 1, V (U) = 1/2 c Hypergeometric (r = n, 6.13 f U (u) = F U ′ (u) = u N = n 1 + n 2 +···n n ) β 2 e−u/β , u > 0 6.55 P(Y ≥ 20) = .077 6.15 [−ln(1 − U)] 1/2 6.65 a f (u 1 , u 2 ) = 6.17 a f (y) = αyα−1 1 ,0≤ y ≤ θ θ α 2π e−[u2 1 +(u 2−u 1 ) 2 ]/2 = b Y = θU 1/α c y = 4 √ 1 u. The values are 2.0785, 2π e−(2u2 1 −2u 1u 2 +u 2 2)/2 3.229, 1.5036, 1.5610, 2.403. b E(U 1 ) = E(Z 1 ) = 0, 6.25 f U (u) = 4ue −2u for u ≥ 0 E(U 2 ) = E(Z 1 + Z 2 ) = 0, 6.27 a f Y (y) = 2 β we−w2 /β V (U 1 ) = V (Z 1 ) = 1, , w ≥ 0, which V (U 2 ) = V (Z 1 + Z 2 ) = is Weibull density ( with ) m = 2. V (Z 1 ) + V (Z 2 ) = 2, k b E(Y k/2 ) = Ɣ 2 + 1 Cov(U β k/2 1 , U 2 ) = E(Z1 2) = 1

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