Question

A normal random variable X has the following probability density function

$f_x\left(x\right)=\frac{1}{\sqrt{8\pi }}{e}^{-\left\{\frac{(x-1{)}^2}{8}\right\}},-\infty <x<\infty$

Then ${\int }_1^{\infty }{f}_x\left(x\right)dx?$

• A. 0
• B. $\frac{1}{2}$
• C. $1-\frac{1}{e}$
• D. 1

Answer 1

The correct option is B $\frac{1}{2}$

$f\left(x\right)=\frac{1}{\sqrt{8\pi }}{e}^{-\left\{\frac{(x-1{)}^2}{8}\right\}},-\infty <x<\infty$

$\Rightarrow \mu =1$ & $\sigma =2$ (On comparision with normal distribution)


$\because$ N. curve is symmetrical about mean

So ${\int }_1^{\infty }f\left(x\right)dx=\frac{1}{2}$