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 ( ±
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This note was uploaded on 10/08/2009 for the course STAT 401 taught by Professor Akritas during the Spring '00 term at Pennsylvania State University, University Park.

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

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