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Unformatted text preview: STAT 400 Fall 2011 Homework #5 (10 points) (due Friday, October 7th, 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 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 . 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 3  ) ( x f  2x 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  2x  C C C dx dx dx dx x f . 2 5 1 3 2 2 3 b) Find the cumulative distribution function F ( x ) = P ( X x ). . ) ( 3 1 3 2 2 5 4 4 2 5 2 4 ( x x x x x x x x x dx dx x x x F dy y f y)2 5 2  2y  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 5 . 2 4 2 5 . 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 3 ) 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 momentgenerating function of X, M X ( t ). M X ( t ) = E ( e t X ) = dx x f x t e = 3 5 2 dx x t e  2x  ....
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This note was uploaded on 10/19/2011 for the course CS 231 taught by Professor  during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 

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