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# HW3sol - STAT 401 Homework 3 Solutions(Fall 2008 3.2.1 1 2...

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STAT 401: Homework 3 Solutions (Fall 2008) 3.2.1 1. i. { 0 , 1 , 2 , 3 , 4 , 5 } ii. Binomial. 2. i. { 0 , 1 , 2 , 3 , 4 , · · · } ii. Negative Binomial. 4. i. { 0 , 1 , 2 , 3 } ii. Hypergeometric. 6. i. { 0 , 1 , 2 , · · · , 10 } ii. Binomial. 8. i. { 0 , 1 , 2 , · · · , 15 } ii. Binomial. 9. i. { 1 , 2 , 3 , 4 , · · · } ii. Negative Binomial. 3.3.4 1. i. P ( X = i ) = i 21 > 0, for all i 6 i =1 p ( x i ) = 6 i =1 i 21 = 1 21 · 6 · 7 2 = 1 Therefore, p ( x i ) = 1 21 , i = 1 , 2 , . . . , 6 is a legitimate pmf. ii. The cumulative distribution function is x 1 2 3 4 5 6 F ( x ) 1 21 3 21 6 21 10 21 15 21 1 4. i. P ( X 2) = 1 - P ( X < 2) = 1 - 0 . 7 = 0 . 3 ii. The probability mass function is x 0 1 2 3 p ( x ) 0.2 0.5 0.2 0.1 5. i. R 1 . 5 x 2 dx = x 2 4 1 . 5 = 1 4 - 1 16 = 3 16 ii. f ( x ) = x 2 , 0 x 2 7. i. R 10 8 kxdx = kx 2 2 10 8 = 1 k = 1 18 ii. F ( x ) = x 2 36 iii. P (8 . 6 X 9 . 8) = R 9 . 8 8 . 6 x 2 dx = x 2 4 9 . 8 8 . 6 = . 6133 iv. P ( X 9 . 8 | X 8 . 6) = P (8 . 6 X 9 . 8) P ( X 8 . 6) = . 6133 . 7233 = . 8479 1

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8. i. P (2 < X < 5) = P ( X < 5) - P ( X < 2) = F (5) - F (2) = 1 - exp ( - 5 λ ) - (1 - exp ( - 2 λ )) = exp ( - 2 λ ) - exp ( - 5 λ ) ii. Since Y = 60 X , F Y ( y ) = P ( Y y ) = P (60 X y ) = P ( X y 60 ) = F ( y 60 ) = 1 - exp ( - λ 60 y ) 3.4.3 2. i. E ( X ) = 3 x =0 xp ( x ) = 0 × . 2 + 1 × . 5 + 2 × . 2 + 3 × . 1 = 1 . 2 E ( X 2 ) = 3 x =0 x 2 p ( x ) = 0
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HW3sol - STAT 401 Homework 3 Solutions(Fall 2008 3.2.1 1 2...

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