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

Question No. 148 CSE | GATE 2017

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 $

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