chapter 1

Chapter 5: Joint Probability Distributions and Random Samples64. Let X 1 , …, X 5 denote morning times and X 6 , …, X 10 denote evening times.a. E(X 1 + …+ X 10 ) = E(X 1 ) + … + E(X 10 ) = 5 E(X 1 ) + 5 E(X 6 )= 5(4) + 5(5) = 45b. Var(X 1 + …+ X 10 ) = Var(X 1 ) + … + Var(X 10 ) = 5 Var(X 1 ) + 5Var(X 6 )⎡64100⎤820= 5 ⎢ + = = 68.3312 12 ⎥⎣ ⎦ 12c. E(X 1 – X 6 ) = E(X 1 ) - E(X 6 ) = 4 – 5 = -164 100 164Var(X 1 – X 6 ) = Var(X 1 ) + Var(X 6 ) = + = = 13. 6712 12 12d. E[(X 1 + … + X 5 ) – (X 6 + … + X 10 )] = 5(4) – 5(5) = -5Var[(X 1 + … + X 5 ) – (X 6 + … + X 10 )]= Var(X 1 + … + X 5 ) + Var(X 6 + … + X 10 )] = 68.3365. µ = 5.00, σ = .22 2σ σE V ( X −Y) = + = . 0032 , σ = . 0566X −Y25 25a. ( X − Y ) = 0;⇒( − . 1 ≤ X − Y ≤ .1) ≈ P( −1.77≤ Z ≤1.77) = . 9232P (by the CLT)2 2σ σb. V ( X − Y ) = + = . 0022222 , σ = . 0471X −Y36 36⇒ P − . 1 ≤ X − Y ≤ .1 ≈ P − 2.12 ≤ Z ≤ 2.12 = .( ) ( ) 966066.a. With M = 5X 1 + 10X 2 , E(M) = 5(2) + 10(4) = 50,Var(M) = 5 2 (.5) 2 + 10 2 (1) 2 = 106.25, σ M = 10.308.⎛ 75 − 50⎜⎝ 10.308b. P( 75 < M ) = P < Z = P(2.43< Z ) = . 0075c. M = A 1 X 1 + A 2 X 2 with the A I ’s and X I ’s all independent, soE(M) = E(A 1 X 1 ) + E(A 2 X 2 ) = E(A 1 )E(X 1 ) + E(A 2 )E(X 2 ) = 50d. Var(M) = E(M 2 ) – [E(M)] 2 . Recall that for any r.v. Y,⎞⎟⎠E(Y 2 ) = Var(Y) + [E(Y)] 2 . Thus, E(M 2 2 22 2) = E ( A1 X1+ 2A1X1A2X2+ A2X2)2 22 2= E ( A ) E( X ) + E( A ) E( X ) E( A ) E( X ) + E( A ) E( )1 121 1 2 22X2(by independence)= (.25 + 25)(.25 + 4) + 2(5)(2)(10)(4) + (.25 + 100)(1 + 16) = 2611.5625, so Var(M) =2611.5625 – (50) 2 = 111.5625196