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Question
Given the p. d. f. (probability density function) of a continuous random variable x as :
`f(x)=x^2/3, -1`
= 0 , otherwise
Determine the c. d. f. (cumulative distribution function) of x and hence find P(x < 1), P(x ≤ -2), P(x > 0), P(1 < x < 2)
Solution
c.d.f. of the continous random variable is given by
`F(x)=int_-1^xy^2/3dx`
`=[y^3/9]_-1^x`
`=(x^3+1)/9, x in R`
Consider P(X<1)=F(1)=(13+1)/9=2/9
`P(x<=-2)=0`
`P(X>0)=1-P(X<=0)`
`=1-F(0)`
`=1-(0/9+1/9)`
`=8/9`
`P(1<x<2)=F(2) - F(1)`
`=1-(1/9+1/9)`
=`7/9`
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