chapter 1

Chapter 5: Joint Probability Distributions and Random Samples25−xc. P ( X + Y ≤ 25) = ∫ ∫ kxydydx+∫ ∫20020−x25 25−x200kxydydx3 230,625= ⋅ = .35581,250 24202d. E X + Y ) = E(X ) + E(Y)= 2{ ∫ x⋅k( 250x−10x)0+ ⋅ ( − ) }302 1 3k 450x30xx dx = k2( dx∫ x +2 (351,666.67) = 25. 96920∞∞e. E ( XY)= ∫−∞∫ xy⋅f ( x,y)dxdy =−∞∫0∫30 30+2 2∫ ∫ − xkkx y dydx = ⋅20032030−x20−xkx2y2dydx33,250,000= 136.41033Cov(X,Y) = 136.4103 – (12.9845) 2 = -32.19, and E(X 2 ) = E(Y 2 ) = 204.6154, so2 22− 32.19σ x= σ y= 204.6154 − (12.9845) = 36.0182 and ρ = = −.89436.0182f. Var (X + Y) = Var(X) + Var(Y) + 2Cov(X,Y) = 7.66, sony78. F Y (y) = P( max(X 1 , …, X n ) ≤ y) = P( X 1 ≤ y, …, X n ≤ y) = [P(X 1 ≤ y)] n ⎛ −100⎞100 ≤ y ≤ 200.ny −100for 100 ≤ y ≤ 200.n100nnE100100n 100nn 2n+ 1= 100 += 100 + 100 = ⋅100100∫ u dun 0n + 1 n + 1n−Thus f Y (y) = ( ) 1n−1100( y −100) dy = ( u + 100)= ⎜⎝200n−1( Y ) = ∫ y ⋅100∫ u dunn 0100⎟⎠for79. E ( X + Y + Z ) = 500 + 900 + 2000 = 34002 2 250 100 180Var ( X + Y + Z ) = + + = 123.014 , and the std dev = 11.09.365 365 365P ( X + Y + Z ≤ 3500) = P(Z ≤ 9.0) ≈ 1200