Question

Given the probability density function (p.d.f) of a continuous random variable $X$ as,
$f\left(x\right)=\frac{x^2}{3},-1<x<2=1$, otherwise
Determine the cumulative distribution function (c.d.f.) of $X$ and hence find $P(X<1),P(X>0),P(1<X<2)$

Answer 1

Given the PDF of a continuous random variable $X.$

$\Rightarrow f\left(x\right)=\frac{x^3}{3},-1<x<2=1$

$f\left(x\right)={\int }_{-\infty }^{\alpha }f\left(x\right)dx={\int }_{\alpha }^{\beta }\frac{1}{\beta -\alpha }dx={\left[\frac{1}{\beta -\alpha }x\right]}_{\infty }^x=\frac{x-\alpha }{\beta -\alpha }$

$f\left(x\right)=\int \frac{1}{2-(-1)},-1<x<2$ $=\int \frac{1}{3},-1<x<2$

$0,otherwise$ $0,otherwise$

$\Rightarrow$ When, $P(1<x<2)={\int }_1^2\frac{1}{3}dx={\left[\frac{1}{3}x\right]}_0^1=\frac{1}{3}$

When $P(x>0)=0$

When $P(1<x<2)={\int }_1^2\frac{1}{3}dx={\left[\frac{x}{3}x\right]}_1^2=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$

Hence, solved.