Question

The value of $\frac{1}{\sqrt{2\pi }}{\int }_0^{\infty }exp\left(\frac{-x^2}{8}\right)dx$ is.

• A. 1

Answer 1

The correct option is A 1
Let f(x) is the p.d.f. of normal distribution.
${\int }_{-\infty }^{\infty }f\left(x\right)dx=1$

${\int }_{-\infty }^{\infty }\frac{1}{\sigma \sqrt{2\pi }}exp\{-\frac{1}{2}{\left(\frac{x-\mu }{\sigma }\right)}^2\}dx=1$

$\text{Put}\mu =0$ & $\sigma =2$

${\int }_{-\infty }^{\infty }\frac{1}{2\sqrt{2\pi }}exp\{-\frac{x^2}{8}\}dx=1$

$2{\int }_0^{\infty }\frac{1}{2\sqrt{2\pi }}exp\{-\frac{x^2}{8}\}dx=1$

${\int }_0^{\infty }\frac{1}{\sqrt{2\pi }}exp(-\frac{x^2}{8})dx=1$