Question

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

Answer 1

c.d.f. of the continuous variable is given by,

$f\left(x\right)={\int }_{-1}^x\frac{y^2}{3}dy$

$={\left[\frac{y^3}{9}\right]}_{-1}^x$

$=\frac{(x^3+1)}{9},x\in R$

Consider $P(x<1)=f\left(1\right)=\frac{(1{)}^3+1}{9}=\frac{2}{9}$

$P(x\le -2)=0$

$P(x>0)=1-P(x\le 0)$

$=1-f\left(0\right)$

$=1-\left(\frac{1}{9}\right)=\frac{8}{9}$

$P(1<x<2)=f\left(2\right)-f\left(1\right)$

$=1-\frac{2}{9}$

$=\frac{7}{9}$