Question

The variance of the random variable X with probability density function $f\left(x\right)=\frac{1}{2}|x|e^{-|x|}$ is .

• A. 6

Answer 1

The correct option is A 6

$E\left(x\right)={\int }_{-\infty }^{\infty }xf\left(x\right)dx={\int }_{-\infty }^{\infty }\frac{1}{2}x|x|e^{-|x|}dx$

= 0 $(\because$ Integrand is an odd function)

$E\left(X^2\right)={\int }_{-\infty }^{\infty }\frac{1}{2}{x}^2.|x|.e^{-|x|}.dx={\int }_0^{\infty }{x}^3{x}^{-x}dx=6$

$(\because$ Integrand is an even function

Variance of

$X=E\left(X^2\right)-{\left\{E\right(X\left)\right\}}^2=6-(0{)}^2=6$