hw8_418_10 - 1. 1 a) 1 c(1 x2 )dx = 1, then c = 3 . 4 x x...

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1. a) 1 - 1 c (1 - x 2 ) dx = 1, then c = 3 4 . b) F ( x ) = x -∞ f ( t ) dt = x - 1 3 4 (1 - t 2 ) dt = - 1 / 4 x 3 + 3 / 4 x + 1 / 2 , F ( x ) = 0 if x ≤ - 1; - 1 / 4 x 3 + 3 / 4 x + 1 / 2 if - 1 < x < 1; 1 if x 1 . 3. For f to be a valid pdf, it must satisfy 2 conditions:1) f ( x ) 0 , for any x; 2) -∞ = 1. a) f (1) = c,f (2) = - 4 c . So, there is no c that makes f (1) and f (2) be positive at the same time. So, f cannot be a pdf. b) f (1 / 2) = 3 c 4 ,f (7 / 3) = - 7 c 9 , but there is no c that makes f (1 / 2) and f (7 / 3) be positive at the same time. So, f cannot be a pdf. 4. a) p ( x > 20) = 20 10 x 2 dx = 1 / 2 b) F ( x ) = x 10 10 t 2 dt = 1 - 10 /x , hence F ( x ) = { 0 if x 10; 1 - 10 /x if x > 1 . c) p ( x 15) = 1 - p ( x 15) = 1 - (1 - 10 / 15) = 2 / 3 . Let X denotes the lifetime of a certain type of electronic device, then X B (6 , 2 / 3), then 1 - p ( X 2) = 1 - (0 . 00137 + 0 . 1646 + 0 . 0823) = 0 . 899. 5.
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hw8_418_10 - 1. 1 a) 1 c(1 x2 )dx = 1, then c = 3 . 4 x x...

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