Question

$f\left(x\right)=\frac{A}{\pi }.\frac{1}{16+x^2},-\infty <x<\infty$ is a p.d.f of a continuous random variable $X$, then the value of $A$ is:

• A. $16$
• B. $8$
• C. $4$
• D. $1$

Answer 1

The correct option is C $4$

For a probability distribution function $f\left(x\right),a<x<b$ of a continuous random variable $X$, it is required that ${\int }_a^{b}f\left(x\right)dx=1$

$\therefore {\int }_{-\infty }^{\infty }\frac{A}{\pi }\frac{1}{16+x^2}dx=1$

$\Rightarrow \frac{A}{\pi }{\int }_{-\infty }^{\infty }\frac{1}{4^2+x^2}dx=1$

$\Rightarrow \frac{A}{\pi }\frac{1}{4}{\tan }^{-1}\left(\frac{x}{4}\right){|}_{-\infty }^{\infty }=1$

$\Rightarrow \frac{A}{4\pi }[\frac{\pi }{2}-(-\frac{\pi }{2})]=1$

$\Rightarrow \frac{A}{4}=1$

$\Rightarrow A=4$