#### 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)

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#### 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)=(1^{3}+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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