Question-84

Let \(X\) be a continuous random variable with the following PDF:

\[ f_{X}( x) =\begin{cases} \cfrac{x}{2} , & 0\leqslant x\leqslant 2\\ 0, & \text{otherwise} \end{cases} \]

Find the variance of \(X\). Enter your answer correct to two decimal places.

NoteAnswer

\(0.22\)

NoteSolution

We know that \(\sigma ^{2} =E\left[ X^{2}\right] -E[ X]^{2}\). So we first compute \(E[ X]\) and \(E\left[ X^{2}\right]\):

\[ \begin{aligned} E[ X] & =\int\limits _{0}^{2}\frac{x^{2}}{2} dx\\ & =\frac{4}{3} \end{aligned} \] And

\[ \begin{aligned} E\left[ X^{2}\right] & =\int\limits _{0}^{2}\frac{x^{3}}{2} dx\\ & =2 \end{aligned} \]

Therefore \(\sigma ^{2} =2-\left(\frac{4}{3}\right)^{2} =\frac{2}{9}\).