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# hw4sol - HW4 Solution Section 2.5 8(a 5 5 1= y =0 5 x=0...

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HW4 Solution Section 2.5 8. (a) 1 = Z 5 y =0 Z 5 x =0 A (20 - x - 2 y ) dx dy = A Z 5 y =0 20 x - x 2 2 - 2 xy 5 x =0 dy = A Z 5 y =0 (87 . 5 - 10 y ) dy = 312 . 5 A Since 312 . 5 A = 1, that is A = 1 / 312 . 5 = 0 . 0032. (b) P (1 X 2 , 2 Y 3) = 0 . 0432. (c) f X ( x ) = 0 . 016(15 - x ) , 0 x 5 f Y ( y ) = 0 . 008(35 - 4 y ) , 0 y 5 (d) Since f ( x, y ) 6 = f X ( x ) f Y ( y ), X, Y are NOT indepenent. (e) EX = 7 3 , EX 2 = 15 2 , V arX = 37 18 . (f) EY = 13 6 , EY 2 = 20 3 , V arY = 71 36 . (g) f Y | X =3 ( y ) = 17 - 2 y 60 , 0 y 5 . (h) EXY = 5 , Cov ( X, Y ) = - 1 18 . (i) Corr ( X, Y ) = - 0 . 0276 9. (Check solution of Textbook.) Section 2.6 1. (Check solution of Textbook.) 3. (Check solution of Textbook.) 6. μ = 1 . 12 , σ = 0 . 03 (a) n = 25, E ¯ X = μ = 1 . 12, V ar ( ¯ X ) = σ 2 n = . 03 2 25 = 3 . 6 × 10 - 5 (b) To have standard deviation no more than 0.005, that is V ar ( ¯ X ) 0 . 005 2 σ 2 n 0 . 005 2 n 0 . 03 2 . 005 2 = 36 9. (a) A = 3 4 . 1

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(b) V = 4 3 πR 3 , find the cdf of V first F V ( v ) = P ( V v ) = P ( 4 3 πR 3 v ) = P R 3 v 4 π 1 / 3 ! = F R 3 v 4 π 1 / 3 ! Where F R ( r ) is the cdf for R, F R ( r ) = Z r 0 3 4 (1 - ( y - 1) 2 ) dy = - 1 4 r 3 + 3 4 r 2 Therefore, F V ( v ) = - 3 v 16 π + 3 4 3 v 4 π 2 / 3 The pdf of V is f V ( v ) = d dv F V ( v ) = -
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