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HW6sol - STAT401 Homework 6 Solutions(Fall 2008 5.2.1 1...

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STAT401: Homework 6 Solutions (Fall 2008) 5.2.1 1. E ( X i ) = 3 , for all i = 1 , 2 , 3 , 4 , 5 E ( Y i ) = 6 , for all i = 1 , 2 , 3 E ( X 1 + X 2 + X 3 + X 4 + X 5 + Y 1 + Y 2 + Y 3 ) = 5 × 3 + 3 × 6 = 33 3. The p.m.f of entree is given by Entree = x 7.5 10 P ( Entree = x ) .65 .35 The p.m.f of tip is given by Tip = y 1 1.5 2 P ( Tip = y ) .49 .405 .105 E ( Entree ) = 7 . 5 · . 65 + 10 · . 35 = 8 . 375 E ( Tip ) = 1 · . 49 + 1 . 5 · . 405 + 2 · . 105 = 1 . 3075 E ( T ) = E ( Entree ) + E ( Tip ) = 8 . 375 + 1 . 3075 = 9 . 6825 5.3.1 2. E ( X i ) = 3 , V ar ( X i ) = 1 . 12 2 for all i = 1 , 2 , 3 , 4 , 5 E ( Y i ) = 6 , V ar ( Y i ) = 4 2 for all i = 1 , 2 , 3 If X 1 , . . . , X 5 , Y 1 , Y 2 , Y 3 are independent, V ar ( X 1 + X 2 + X 3 + X 4 + X 5 + Y 1 + Y 2 + Y 3 ) = 5 × 1 . 12 + 3 × 4 = 17 . 6 4. Y = 9 . 3 + 1 . 5 X + E ( Y ) = E (9 . 3 + 1 . 5 X + ) = 9 . 3 + 1 . 5 × 24 = 45 . 3 V ar ( Y ) = V ar (9 . 3 + 1 . 5 X + ) = 1 . 5 2 × V ar ( X ) + V ar ( ) = 20 . 25 + 16 = 36 . 25 5. E(X)=E(Entree)=8.375. Var(X)=Var(Entree)= E ( Entree 2 ) - E ( Entree ) 2 = (7 . 5) 2 × 0 . 65 + (10) 2 × 0 . 35 - (8 . 375) 2 = 1 . 421875. E ( Y ) = E ( Tip ) = 1 . 3075. V ar ( Y ) = V ar ( Tip ) = E ( Tip 2 ) - E ( Tip ) 2 = 1 2 × 0 . 49 + (1 . 5) 2 × 0 . 405 + 2 2 × 0 . 105 - (1 . 3075) 2 = 0 . 1116937. Cov ( X, Y ) = EXY - EXEY = 7 . 50 × . 455 + 7 . 50 × 1 . 50 × . 195 + 10 × . 035 + 15 × . 210 + 20 × . 105 - 1 . 3075 × 8 . 375 = 0 . 2559375. Var(X+Y)=Var(X)+Var(Y)+2*Cov(X,Y)=1.421875+0.1116937+2 × 0 . 2559375 = 2 . 045444 .
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