Chapter 9, Problem 1CRE

To determine

To compute: $P(X\le 1)$

Answer to Problem 1CRE

  $P(X\le 1)=\frac{3}{4}$

Explanation of Solution

Given information:

X is a continuous random variable with probability density $p(x)=\frac{1}{\pi (x^2+1)}$ .

Formula used: $P(X\le a)={\displaystyle \underset{-\infty }{\overset{a}{\int }}p(x)dx}$

Calculation:

  $p(x)=\frac{1}{\pi (x^2+1)}$   (Given)

  $\begin{array}{c}P(X\le 1)={\displaystyle \underset{-\infty }{\overset{1}{\int }}p(x)dx}\\ ={\displaystyle \underset{-\infty }{\overset{1}{\int }}\frac{1}{\pi (x^2+1)}dx}\\ =\frac{1}{\pi }{\displaystyle \underset{-\infty }{\overset{1}{\int }}\frac{1}{(x^2+1)}dx}\\ =\frac{1}{\pi }{\left[{\tan }^{-1}x\right]}_{-\infty }^1\\ =\frac{1}{\pi }\left[{\tan }^{-1}1-{\tan }^{-1}\left(-\infty \right)\right]\\ =\frac{1}{\pi }\left[\frac{\pi }{4}-\left(-\frac{\pi }{2}\right)\right]\\ =\frac{1}{\pi }\left[\frac{3\pi }{4}\right]\\ =\frac{3}{4}\end{array}$

Hence, $P(X\le 1)=\frac{3}{4}$ .