Question

Let $f\left(x\right)=\left\{\begin{array}{c}\frac{1}{x^2};1<x<\infty \\ 0;\text{otherwise}\end{array}\right\}$, be the probability density function of random variable X, then the value of $P(-2\le x\le 2)$ is

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

Answer 1

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

Given $f\left(x\right)=\left\{\begin{array}{c}\frac{1}{x^2};1<x<\infty \\ 0;\text{otherwise}\end{array}\right\}$

$P={\int }_{-2}^{2}f\left(x\right)dx$

$\Rightarrow P={\int }_{-2}^{1}0dx+{\int }_1^2\frac{1}{x^2}dx=0+{[-\frac{1}{x}]}_1^2$

$\Rightarrow P=0+(\frac{-1}{2}+1)$

$\Rightarrow P=\frac{1}{2}$