Homework_2_Solutions

# Homework_2_Solutions - 78 CHAPTER 4 Sampling Distributions...

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78 CHAPTER 4 Sampling Distributions and, the DF is F(x) = 0, x 0 ±(α + β) ±(α)±(β) ± x 0 t α 1 ( 1 t) β 1 dt ,0 x 1 1, x 1 In our case, f(x) = ±( 5 ) ±( 2 )±( 3 ) x 2 1 ( 1 x) 3 1 = 4 ! 1 ! 2 ! x( 1 x) 2 = 12 x( 1 x) 2 ,i f0 x 1 and = x ² 0 12 t( 1 2 dt = 12 ³ x 2 2 2 x 3 3 + x 4 4 ´ x 1 Then, the joint pdf f Y 1 , Y n (x , y) , using the 4.3.3, is given by f Y 1 , Y n (x , y) = n(n 1 ) ³ 12 µ y 2 2 2 y 3 3 + y 4 4 10 µ x 2 2 2 x 3 3 + x 4 4 ¶´ n 2 12 x( 1 x) 2 12 y( 1 y) 2 = 12 n n(n 1 ) · 1 2 (y 2 x 2 ) 2 3 (y 3 x 3 ) + 1 4 (y 4 x 4 ) ¸ n 2 xy( 1 x) 2 ( 1 y) 2 , if 0 x y 1, and f Y 1 , Y n (x , y) = 0, otherwise 4.3.12. p i = P(X = i) = pq i 1 , i = 1, 2, 3, ... <p , q< 1, q = 1 p P(Y k ) = k y) k y 1 ) , y = 1, 2, 3, where, k y) = P (At least k of X ’s are less than or equal to y ) = n ¹ i = k ( n i ) [ y ] i [ P(X > y ] n i = n ¹ i = k ( n i ) [ 1 q y ] i [ q y ] n i Then, k ) = n ¹ i = k ( n i ) [ 1 q y ] i q y(n i) n ¹ i = k ( n i ) [ 1 q y 1 ] i q (y 1 )(n i) = n ¹ i = k ( n i ) q (y 1 )(n i) { q (n i) [ 1 q y ] i −[ 1 q y 1 ] i } , y = 1, 2, 3, EXERCISES 4.4 4.4.1. X 1 , X 2 , , X n , where n = 150, μ = 8, σ 2 = 4, then μ x = 8 and σ x 2 = 4 By Theorem 4.4.1: lim n →∞ P µ Z X μ x σ x = 1 2 k ± z

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Homework_2_Solutions - 78 CHAPTER 4 Sampling Distributions...

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