Question

Let $X$ be a random variable with the following PDF:

$f_X\left(x\right)=\{\begin{array}{c}C{x}^2;|x|\le 1\\ 0;\text{otherwise}\end{array}$

Find $P(X\ge \frac{1}{2})$

• A. $\frac{5}{16}$
• B. $\frac{11}{16}$
• C. $\frac{7}{16}$
• D. None of these

Answer 1

The correct option is C $\frac{7}{16}$

Given, $X$ is a continuous random variable with PDF $f_X\left(x\right)=\{\begin{array}{c}C{x}^2,|x|\le 1\\ 0,\text{otherwise}\end{array}$

We have to find the positive constant $c$.

We know that, a PDF is valid if it satisfies the following

$\left(1\right)$ $f\left(x\right)$ is positive everywhere.

$\left(2\right)$ The area under the curve $f\left(x\right)$ in the range $(a,b)$is $1$, that is: ${\int }_a^{b}f\left(x\right)dx=1$.

Thus we have ${\int }_{-1}^1{f}_X\left(x\right)dx=1$

$\Rightarrow {\int }_{-1}^{1}C{x}^{2}dx=1$

$\Rightarrow C{\int }_{-1}^1{x}^{2}dx=1$

$\Rightarrow C{\left[\frac{x^3}{3}\right]}_{-1}^1=1$

$\Rightarrow C[\frac{1}{3}-\left(\frac{-1}{3}\right)]=1$

$\Rightarrow C\left(\frac{2}{3}\right)=1$

$\Rightarrow C=\frac{3}{2}$

Therefore, $f_X\left(x\right)=\{\begin{array}{c}\frac{3}{2}{x}^2,|x|\le 1\\ 0,\text{otherwise}\end{array}$

We have to find $P(X\ge \frac{1}{2})$

$P(X\ge \frac{1}{2})={\int }_{1/2}^1{f}_X\left(x\right)dx$

$={\int }_{1/2}^1\frac{3}{2}{x}^{2}dx$

$=\frac{3}{2}{\int }_{1/2}^1{x}^{2}dx$

$=\frac{3}{2}{\left[\frac{x^3}{3}\right]}_{1/2}^1$

$=\frac{3}{2}[\frac{1}{3}-\frac{1}{24}]$

$=\frac{3}{2}\left[\frac{7}{24}\right]$

$=\frac{1}{2}\left[\frac{7}{8}\right]$

$=\frac{7}{16}$

Thus $P(X\ge \frac{1}{2})=\frac{7}{16}$