Question

The probability density function of a random variables x is
$\begin{array}{cc}f\left(x\right)=\frac{x}{4}(4-x^2);& \mathrm{for}0\le x\le 2\\ =0;& \mathrm{otherwise}\end{array}$
The mean ${\mu }_x$ of the random variable is

• A. 16.15

Answer 1

The correct option is A 16.15

$f\left(x\right)=\{\begin{array}{cc}\frac{x}{4}(4-x^2),& 0\le x\le 2\\ 0,& \text{otherwise}\end{array}$

${\mu }_x=E\left(x\right)={\int }_{-\infty }^{\infty }xf\left(x\right)\mathrm{dx}$

$={\int }_0^{2}x\frac{x}{4}(4-x^2)\mathrm{dx}$

$={\int }_0^2\left(x^2-\frac{x^4}{4}\right)\mathrm{dx}$

$={\left(\frac{x^3}{3}-\frac{x^5}{20}\right]}_0^2=\frac{8}{3}-\frac{32}{20}=\frac{16}{15}$