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STAT 400 hw5_ans

# STAT 400 hw5_ans - STAT 400 Spring 2012 Homework#5(10...

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STAT 400 Spring 2012 Homework #5 (10 points) (due Friday, Feb 24th, by 3:00 p.m.) No credit will be given without supporting work. 1. Suppose a random variable X has the following probability density function: otherwise 0 1 1 ) ( C x x x f a) What must the value of C be so that f ( x ) is a probability density function? For f ( x ) to be a probability density function, we must have: 1) f ( x ) 0, 2)   1 dx x f .   C C C dx x dx x f ln ln ln 1 1 1 1 . Therefore, C = e . b) Find P ( X < 2 ) . P ( X < 2 ) =   1 2 1 ln ln 2 1 2 dx x dx x f = ln 2 . c) Find P ( X < 3 ) . P ( X < 3 ) =   1 1 ln ln 1 3 e e dx x dx x f = 1 . d) Find X = E ( X ) . X = E ( X ) =   e e dx dx x x dx x f x 1 1 1 1 = e 1 .

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e) Find X 2 = Var ( X ) . E ( X 2 ) =   e e dx x dx x x dx x f x 1 1 2 2 1 = 2 1 2 e . X 2 = Var ( X ) = E ( X 2 ) [ E ( X ) ] 2 = 2 3 4 2 e e 0.242 . 2. Suppose a random variable X has the following probability density function: otherwise x 0 3 0 | ) ( x f | 2 - x C a) What must the value of C be so that f ( x ) is a probability density function?   . 5 2 2) - (x x) - (2 C | 2 - x | C C C dx dx dx dx x f . 2 5 1 3 2 2 0 3 0 b) Find the cumulative distribution function F ( x ) = P ( X x ) . . ) ( 3 1 0 0 3 2 2 0 5 4 4 2 5 2 4 ( 0 0 x x x x x x x x x dx dx x x x F dy y f y) - 2 5 2 | 2 - y | 5 2
c) Find the median of the probability distribution of X . Need m = ? such that ( Area to the left of m ) =   2 1 F(m) m x x f d . First we found that F(2)=4/5 > 0.5, so m should be smaller than 2, namely m is in [0,2). Then solve the equation 2 6 4 0 5 . 2 4 2 5 . 0 5 2 4 x x x x x The median is less than 2, so m = 2 6 4 = 0.775 . d) Find X = E ( X ) . 15 16 5 2 5 2 5 2 ) ( 3 2 2 0 3 0 ) 2 ( ) 2 ( | 2 | x x x x x x x x f x d d d d X x x x = 1.067 . e) Find the moment-generating function of X, M X ( t ) . M X ( t ) = E ( e t X ) =   - dx x f x t e = 3 0 5 2 dx x t e | 2 - x | . 3 2 2 0 ) 2 ( ) 2 ( 5 2 dx x dx x x t x t e e t t t t tx tx tx tx tx tx e t e t e t e t t t e t e t xe t e t e t e t 3 2 3 2 2 2 2 2 2 1 1 2 4 1 2 2 3 2 1 1 0 2 1 2 1 t 0. M X ( 0 ) =1.

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3. Let X be a continuous random variable with the probability density function f ( x ) = k x 2 , 0 x 1, f ( x ) = 0, otherwise.
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