Question

If probability density function of a random variable X is $f\left(x\right)=x^2\text{for}-1\le x\le 1\text{and}f\left(x\right)=0$ for any other values of x, then the percentage probability $P(-\frac{1}{3}\le x\le \frac{1}{3})\text{is}$

• A. 0.247
• B. 2.47
• C. 24.7
• D. 247

Answer 1

The correct option is B 2.47

Given,
$\mathrm{PDF}f\left(x\right)=(\begin{array}{cc}{x}^2;& -1\le x\le 1\\ 0;& \mathrm{otherwise}\end{array}$Required probability is
$P(-\frac{1}{3}\le x\le \frac{1}{3})=\left({\int }_{-\frac{1}{3}}^{\frac{1}{3}}f\right(x\left)dx\right)$

$=2{\int }_0^{\frac{1}{3}}{x}^{2}dx$

$=\frac{2}{3}[{\left(\frac{1}{3}\right)}^3-0]$

$=\frac{2}{81}$
i.e. 2.47%