chapter 1

Chapter 5: Joint Probability Distributions and Random SamplesSection 5.337.P(x 1 ) .20 .50 .30P(x 2 ) x 2 | x 1 25 40 65.20 25 .04 .10 .06.50 40 .10 .25 .15.30 65 .06 .15 .09a.x 25 32.5 40 45 52.5 65p ( x).04 .20 .25 .12 .30 .09E( x) = (25)(.04) + 32.5(.20) + ... + 65(.09) = 44. 5 = µb.s 2 0 112.5 312.5 800P(s 2 ) .38 .20 .30 .12E(s 2 ) = 212.25 = σ 238.a.T 0 0 1 2 3 4P(T 0 ) .04 .20 .37 .30 .09b. µ = E ) = 2.2 = 2 ⋅ µT( T0 0c.σ22222T= E(T0) − E(T0) = 5.82 −(2.2)= .98 = 2 ⋅σ0187