hw4sol - STAT 401: Homework 4 Solutions (Spring, 2009)...

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Unformatted text preview: STAT 401: Homework 4 Solutions (Spring, 2009) 3.5.5 1. i. P ( X = x ) = 5 x . 3 x . 7 5- x ii. E ( X ) = 5 . 3 = 1 . 5 iii. V ar ( X ) = 5 . 3 . 7 = 1 . 05 iv. Let Y be the cost of failure. A. P ( Y > 20) = P (9 X > 20) = P ( X 3) = P ( X = 3) + P ( X = 4) + P ( X = 5) = 5 3 . 3 3 . 7 2 + 5 4 . 3 4 . 7 1 + 5 5 . 3 5 = 0 . 1323+0 . 02835+0 . 00243 = 0 . 16308 B. E ( Y ) = E (9 X ) = 9 E ( X ) = 9 1 . 5 = 13 . 5 C. V ar ( Y ) = V ar (9 X ) = 9 2 V ar ( X ) = 85 . 05 2. X Bin (15 , . 1) i. E(X)=15*(0.1)=1.5 ii. P ( X = 3) = 15 3 . 1 3 . 9 1 2=0.1285 iii. P (2 X 5) = P ( X 5)- P ( X 1)=0.998-0.549=0.449 iv. Y represents the number of non-defective components among 15 randomly selected ones. v. Y Bin (15 , . 9) vi. P ( Y = 12)(= P ( X = 3)) = 15 12 . 9 1 20 . 1 3 =0.1285 3. i. Binomial ii. E ( X ) = np = 10 . 9 = 9 iii. P ( X 7) = 1- P ( X 6) = 1- . 013 = 0 . 987 iv. Let Y = 100 + 10 X ....
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hw4sol - STAT 401: Homework 4 Solutions (Spring, 2009)...

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