HW5_Solutions

# HW5_Solutions - PSTAT 120A Spring 2010 HW 5 Solutions...

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Unformatted text preview: PSTAT 120A Spring 2010 HW 5 Solutions 1. (Ross 5.3) For a function to be a valid pdf, it must be nonnegative everywhere, and integrate to 1. For the first function, check the endpoints of the domain (0 , 5 / 2) and notice that 2 x- x 3 > 0 for small x , e.g. x = . 1, and is negative for x just less than 5 / 2. Therefore, whether C is positive or negative, the function f can’t be nonnegative everywhere, so can’t be a pdf. For the second case, similarly 2 x- x 2 > 0 for small x and is negative for x just less than 5 / 2, so again can’t be a pdf. 2. (Ross 5.4) (a) P ( X > 20) = ˆ ∞ 20 f ( x ) dx = 10 ˆ ∞ 20 dx x 2 =- 10 x | ∞ x =20 = 1 2 . (b) F ( x ) = ˆ x-∞ f ( y ) dy = ( ´ x-∞ dy = 0 for x ≤ 10 ´ x 10 10 y 2 dy = 1- 10 x for x > 10 (c) Assuming that the lifetimes of the 6 are independent, then let Z be a random variable counting the number that last at least 15 hours. Then Z has a binomial distribution with parameters n = 6, and p = P ( X ≥ 15), where X is the random lifetime of any one of them.is the random lifetime of any one of them....
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HW5_Solutions - PSTAT 120A Spring 2010 HW 5 Solutions...

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