A poisson variate X is such that P(X=2)= 9 P(X= 4) + 90. P(X= 6) then mean and standard deviations are

Given X follows Poisson distribution.

And given that $\mathrm{P}(\mathrm{X}=2)=9 \mathrm{P}(\mathrm{X}=4)+90 \mathrm{P}(\mathrm{X}=6)$

$$\begin{gathered} \Rightarrow \frac{{\mathrm{e}}^{-\mathrm{\lambda }}·{\mathrm{\lambda }}^2}{2!}=9 \left(\frac{{\mathrm{e}}^{-\mathrm{\lambda }}·{\mathrm{\lambda }}^4}{4!}\right)+90 \left(\frac{{\mathrm{e}}^{-\mathrm{\lambda }}·{\mathrm{\lambda }}^6}{6!}\right) \\ \Rightarrow \frac{{\mathrm{\lambda }}^2}{2}=9 \left(\frac{{\mathrm{\lambda }}^4}{24}\right)+\frac{90\times {\mathrm{\lambda }}^6}{720} \\ \Rightarrow \frac{{\mathrm{\lambda }}^2}{2}=\frac{3{\mathrm{\lambda }}^4}{8}+\frac{{\mathrm{\lambda }}^6}{8} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{{\mathrm{\lambda }}^2}{2}=\frac{3{\mathrm{\lambda }}^4+{\mathrm{\lambda }}^6}{8} \\ \Rightarrow 4{\mathrm{\lambda }}^2=3{\mathrm{\lambda }}^4+{\mathrm{\lambda }}^6 \\ \Rightarrow {\mathrm{\lambda }}^4+3{\mathrm{\lambda }}^2-4=0 \\ \Rightarrow {\mathrm{\lambda }}^4+4{\mathrm{\lambda }}^2-{\mathrm{\lambda }}^2-4=0 \\ \Rightarrow ({\mathrm{\lambda }}^2-1) ({\mathrm{\lambda }}^2+4)=0 \\ \Rightarrow {\mathrm{\lambda }}^2=1 \\ \Rightarrow \mathrm{\lambda }=1 \end{gathered}$$

$\therefore$ mean $\left(\mathrm{\lambda }\right)=1$ 

and standard deviation $=\sqrt{\mathrm{\lambda }}=1$