Question

The probability density function (PDF) of a random variable, X is given by,
$f_X\left(x\right)=K{e}^{-\frac{(x+1{)}^2}{18}}$
The value of K will be______.

• A. 0.133

Answer 1

The correct option is A 0.133
The given random variable has PDF which is similar to that of a Gaussian random variable.
The PDF of a Gaussian random variable can be given by,

$f_X\left(x\right)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{(x+\overline{X}{)}^2}{2{\sigma }^2}}$

$\text{so,}2{\sigma }^2=18$

$K=\frac{1}{\sqrt{2\pi {\sigma }^2}}=\frac{1}{\sqrt{\pi \left(18\right)}}=0.133$