If X is a Poisson variate such that  $2P(X=1)=5P(X=5)+2P(X=3)$ then the standard deviation of X is

X is a poisson variate 

$\Rightarrow P(X=r)=\frac{e^{-\lambda }{\lambda }^r}{r!}$

For $r=0,1,2\dots$

Given $2P(X=1)=5P(X=5)+2P(X=3)$

$\Rightarrow 2\frac{e^{-\lambda }{\lambda }^1}{1!}=5\frac{e^{-\lambda }{\lambda }^5}{5!}+2\frac{e^{-\lambda }{\lambda }^3}{3!}$

$\Rightarrow 2=\frac{\lambda }{24}+\frac{{\lambda }^2}{3}$

$$\begin{gathered} \Rightarrow {\lambda }^4+8{\lambda }^2-48\Rightarrow 0 \\ \Rightarrow \left({\lambda }^2+12\right)\left({\lambda }^2-4\right)=0 \end{gathered}$$

$\Rightarrow {\lambda }^2=4\Rightarrow \lambda =2$

Variance  of X is$\lambda$=2 $\Rightarrow$standard deviation of X is $\sqrt{2}$