chapter 1

Chapter 5: Joint Probability Distributions and Random SamplesSection 5.222.∑∑x ya. E( X + Y ) = ( x + y)p(x,y)= (0 + 0)(.02)+ ( 0 + 5)(.06) + ... + (10 + 15)(.01) = 14.10b. E[max (X,Y)] = ∑∑ max( + y)⋅x yx p(x,y)= ( 0)(.02) + (5)(.06) + ... + (15)(.01) = 9.60∑ ∑23. E(X 1 – X 2 ) = ( − x )43x1= 0 x2= 0x ⋅ p(x , x ) =(0 – 0)(.08) + (0 – 1)(.07) + … + (4 – 3)(.06) = .15(which also equals E(X 1 ) – E(X 2 ) = 1.70 – 1.55)121224. Let h(X,Y) = # of individuals who handle the message.yh(x,y) 1 2 3 4 5 61 - 2 3 4 3 22 2 - 2 3 4 3x 3 3 2 - 2 3 44 4 3 2 - 2 35 3 4 3 2 - 26 2 3 4 3 2 -Since p(x,y) = 30∑∑x y1 for each possible (x,y), E[h(X,Y)] = 1 84h x,y)⋅ = = 2. 80(30 3025. E(XY) = E(X) ⋅ E(Y) = L ⋅ L = L 226. Revenue = 3X + 10Y, so E (revenue) = E (3X + 10Y)52= ∑∑x= 0 y=0(3x+ 10y)⋅ p(x,y)= 0⋅p(0,0)+ ... + 35⋅p(5,2)= 15.4184