Question

The probability distribution function of continuous random variable $X$ is given by $f\left(x\right)=\frac{x}{4},0<x<2$
$=0$, otherwise
Find $P(x\le 1)$.

Answer 1

$f\left(x\right)=\frac{x}{4},0<x<2$

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

$P(x>1)={\int }_1^{2}f\left(x\right)dx$

$=\frac{1}{4}{\int }_1^{2}xdx$

$=\frac{1}{4}{\left[\frac{x^2}{2}\right]}_1^2$

$=\frac{1}{4}[2-\frac{1}{2}]$

$=\frac{1}{4}\left[\frac{3}{2}\right]$

Or $\frac{3}{8}$

$\therefore P(x\le 1)=1-\frac{3}{8}=\frac{5}{8}$