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# slides14 - Lecture Stat 302 Introduction to Probability -...

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Lecture Stat 302 Introduction to Probability - Slides 14 AD March 2010 AD () March 2010 1 / 18

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Continuous Random Variable : We say that X is a (real-valued) continuous r.v. if there exists a nonnegative function f : R ! [ 0 , ) such that for any set A of real numbers P ( X 2 A ) = Z A f ( x ) dx . f ( x ) is called the probability density function (pdf) of the r.v. X and the associated (cumulative) distribution function is F ( x ) = Pr ( X x ) = Z x ± f ( y ) dy so we have f ( x ) = dF ( x ) dx . AD () March 2010 2 / 18
Example: Insurance Policy A group insurance policy covers the medical claims of the employees of a small company. The value , V , of the claims made in one year is described by V = 100 , 000 X where X is a random variable with pdf f ( x ) = c ( 1 x ) 4 if 0 < x < 1 0 otherwise . What is the conditional probability that V exceeds 40,000 given that V exceeds 10,000? AD () March 2010 3 / 18

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Example: Insurance Policy We are interested in P ( V > 40 , 000 j V > 10 , 000 ) = P ( V > 40 , 000 \ V > 10 , 000 ) P ( V > 10 , 000 ) = P ( V > 40 , 000 ) P ( V > 10 , 000 ) where P ( V > v ) = P ( 100 , 000 X > v ) = P X > v 100 , 000 ± First we need to determine c using R 1 0 f ( x ) dx = 1 ; that is Z 1 0 f ( x ) dx = c " ( 1 u ) 5 5 # 1 0 = c 5 ) c = 5 . AD () March 2010 4 / 18
Example: Insurance Policy We need to compute the cdf F X ( x ) of X which is given by F X ( x ) = 8 > < > : 0 if x < 0 c h ( 1 u ) 5 5 i x 0 = 1 ( 1 x ) 5 if 0 ± x ± 1 1 if x > 1 So we are interested in P ( V > 40 , 000 j V > 10 , 000 ) = 1 F X ( 0 . 4 ) 1 F X ( 0 . 1 ) = 0 . 078 0 . 590 = 0 . 132 . AD () March 2010 5 / 18

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Example: Nuclear power plant Assume a nuclear power plant has three independent safety systems. These safety systems have lifetimes
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## This note was uploaded on 10/21/2010 for the course STAT Stat302 taught by Professor 222 during the Spring '10 term at UBC.

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slides14 - Lecture Stat 302 Introduction to Probability -...

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