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

Given X follows Poisson distribution.

And given that $2\mathrm{P}(\mathrm{X}=1)=5 \mathrm{P}(\mathrm{x}=5)+2\mathrm{P}(\mathrm{X}=3)$

$$\begin{gathered} \Rightarrow 2\times \frac{{\mathrm{e}}^{-\mathrm{\lambda }}·{\mathrm{\lambda }}^1}{1!}=5\times \frac{{\mathrm{e}}^{-\mathrm{\lambda }}·{\mathrm{\lambda }}^5}{5!}+2\times \frac{{\mathrm{e}}^{-\mathrm{\lambda }}·{\mathrm{\lambda }}^3}{3!} \\ \Rightarrow 2\mathrm{\lambda }=\frac{5\times {\mathrm{\lambda }}^5}{120}+\frac{2\times {\mathrm{\lambda }}^3}{6} \\ \Rightarrow 2\mathrm{\lambda }=\frac{{\mathrm{\lambda }}^5}{24}+\frac{{\mathrm{\lambda }}^3}{3} \\ \Rightarrow \frac{{\mathrm{\lambda }}^4}{24}+\frac{{\mathrm{\lambda }}^2}{3}-2=0 \\ \Rightarrow {\mathrm{\lambda }}^4+8{\mathrm{\lambda }}^2-48=0 \\ \Rightarrow {\mathrm{\lambda }}^4+12{\mathrm{\lambda }}^2-4{\mathrm{\lambda }}^2-48=0 \\ \Rightarrow ({\mathrm{\lambda }}^2-4) ({\mathrm{\lambda }}^2+12)=0 \\ \Rightarrow {\mathrm{\lambda }}^2=4 \\ \Rightarrow \mathrm{\lambda }=\sqrt{4}=2 \end{gathered}$$

$\therefore$ Standard deviation $=\sqrt{\mathrm{\lambda }}=\sqrt{2}$