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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 cant be nonnegative everywhere, so cant 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 cant be a pdf. 2. (Ross 5.4) (a) P ( X > 20) = 20 f ( x ) dx = 10 20 dx x 2 =- 10 x | x =20 = 1 2 . (b) F ( x ) = 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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- Spring '10