chapter 1

Chapter 5: Joint Probability Distributions and Random Samples57. E(X) = 100, Var(X) = 200, = 14. 14= P( Z ≤ 1.77) = .9616⎛ 125 −100⎞σx, so P( X ≤ 125) ≈ P ⎜Z≤ ⎟⎝ 14.14 ⎠Section 5.558.a. E( 27X 1 + 125X 2 + 512X 3 ) = 27 E(X 1 ) + 125 E(X 2 ) + 512 E(X 3 )= 27(200) + 125(250) + 512(100) = 87,850V(27X 1 + 125X 2 + 512X 3 ) = 27 2 V(X 1 ) + 125 2 V(X 2 ) + 512 2 V(X 3 )= 27 2 (10) 2 + 125 2 (12) 2 + 512 2 (8) 2 = 19,100,116b. The expected value is still correct, but the variance is not because the covariances nowalso contribute to the variance.59.a. E( X 1 + X 2 + X 3 ) = 180, V(X 1 + X 2 + X 3 ) = 45, σ 6. 708x1 + x=2 + x3⎛ 200 −180⎞P(X 1 + X 2 + X 3 ≤ 200) = P ⎜Z≤ ⎟ = P(Z ≤ 2.98) = . 9986⎝ 6.708 ⎠P(150 ≤ X 1 + X 2 + X 3 ≤ 200) = P ( −4.47≤ Z ≤ 2.98) ≈ . 9986σ 15µ X, =xσx = = 2. 236n 3⎛ 55 − 60 ⎞P ( X ≥ 55) = P⎜Z≥ ⎟ = P(Z ≥ −2.236)= .9875⎝ 2.236 ⎠P ( 58 ≤ X ≤ 62) = P − .89 ≤ Z ≤ .89 = .b. = µ = 60( ) 6266c. E( X 1 - .5X 2 -.5X 3 ) = 0;222V( X 1 - .5X 2 -.5X 3 ) = σ + .25σ+ .25σ22.5,sd = 4.7434P(-10 ≤ X 1 - .5X 2 -.5X 3 ≤ 5) =( − 2.11≤ ≤ 1.05)= Z1 23=⎛ −10− 0 5 − 0 ⎞P ⎜ ≤ Z ≤ ⎟⎝ 4.7434 4.7434 ⎠P = .8531 - .0174 = .8357194