# GATE Papers >> CSE >> 2017 >> Question No 148

Question No. 148

If a random variable X has a Poisson distribution with mean 5, then the expectation E[(X+2)2] equals ______.

##### Answer : 54.0 to 54.0

Solution of Question No 148 of GATE 2017 CSE Paper

Given, Poisson distribution $\mathrm\lambda=5$

We know that in Poisson distribution

$\mathrm E\left(\mathrm X\right)=\mathrm V\left(\mathrm X\right)=\mathrm\lambda$

so here $\mathrm E\left(\mathrm X\right)=\mathrm V\left(\mathrm X\right)=5$

now, we need $\mathrm E\left[\left(\mathrm X+2\right)^2\right]$

$\begin{array}{l}=\mathrm E\left(\mathrm X^2+4\mathrm X+4\right)\\=\mathrm E\left(\mathrm X^2\right)+4\mathrm E\left(\mathrm X\right)+4\end{array}$

To find $\mathrm E\left(\mathrm X^2\right)$ we write, $\mathrm V\left(\mathrm X\right)=\mathrm E\left(\mathrm X^2\right)-\left(\mathrm E\left(\mathrm X\right)\right)^2$

$5=\mathrm E\left(\mathrm X^2\right)-5^2$

So,                            $\mathrm E\left(\mathrm X^2\right)=5^2+5=30$

required value $=30+4\times5+4=54$