chapter 1

Chapter 5: Joint Probability Distributions and Random Samplesd. E( X 1 + X 2 + X 3 ) = 150, V(X 1 + X 2 + X 3 ) = 36, σ 6x1 + x=2 + x 3⎛ 160 −150⎞P(X 1 + X 2 + X 3 ≤ 200) = P ⎜Z≤ ⎟ = P(Z ≤ 1.67) = . 9525⎝ 6 ⎠We want P( X 1 + X 2 ≥ 2X 3 ), or written another way, P( X 1 + X 2 - 2X 3 ≥ 0)E( X 1 + X 2 - 2X 3 ) = 40 + 50 – 2(60) = -30,2 2V(X 1 + X 2 - 2X 3 ) = σ + σ + σ 78,36, sd = 8.832, so4 2 1 2 3=⎛ 0 − ( −30)⎞P( X 1 + X 2 - 2X 3 ≥ 0) = P ⎜Z≥ ⎟ = P(Z ≥ 3.40) = . 0003⎝ 8.832 ⎠60. Y is normally distributed with = ( µ + µ ) − ( µ + µ + µ ) = 111µ Y, and1 23 4 5−2 32 1 2 1 2 1 2 1 2 1 2σY= σ1+ σ2+ σ3+ σ4+ σ5= 3.167, σY= 1.7795 .4 4 9 9 9⎛ 0 − ( −1)⎞Thus, P ( 0 ≤ Y ) = P⎜≤ Z ⎟ = P(.56≤ Z ) = . 2877 and⎝ 1.7795 ⎠⎛ 2 ⎞P ( −1 ≤ Y ≤ 1) = P⎜0≤ Z ≤ ⎟ = P(0≤ Z ≤ 1.12) = . 3686⎝ 1.7795 ⎠61.a. The marginal pmf’s of X and Y are given in the solution to Exercise 7, from which E(X)= 2.8, E(Y) = .7, V(X) = 1.66, V(Y) = .61. Thus E(X+Y) = E(X) + E(Y) = 3.5, V(X+Y)= V(X) + V(Y) = 2.27, and the standard deviation of X + Y is 1.51b. E(3X+10Y) = 3E(X) + 10E(Y) = 15.4, V(3X+10Y) = 9V(X) + 100V(Y) = 75.94, and thestandard deviation of revenue is 8.7162. E( X 1 + X 2 + X 3 ) = E( X 1 ) + E(X 2 ) + E(X 3 ) = 15 + 30 + 20 = 65 min.,V(X 1 + X 2 + X 3 ) = 1 2 + 2 2 + 1.5 2 = 7.25, σ = 7.25 2. 6926x1 + x=2 + x3⎛ 60 − 65 ⎞Thus, P(X 1 + X 2 + X 3 ≤ 60) = P ⎜Z≤ ⎟ = P(Z ≤ −1.86)= . 0314⎝ 2.6926 ⎠63.∑∑a. E(X 1 ) = 1.70, E(X 2 ) = 1.55, E(X 1 X 2 ) = x p(x , x ) 3. 33xx1 2E(X 1 X 2 ) - E(X 1 ) E(X 2 ) = 3.33 – 2.635 = .695x , so Cov(X 1 ,X 2 ) =1 2 1 2=b. V(X 1 + X 2 ) = V(X 1 ) + V(X 2 ) + 2 Cov(X 1 ,X 2 )= 1.59 + 1.0875 + 2(.695) = 4.0675195