Question

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

• A. 4
• B. 2
• C. $\frac{1}{2}$
• D. $\sqrt{2}$

Answer 1

Answer: (D)

Given,
$2P(X=1)=5P(X=5)+2P(X=3)$
Now, poission's variate is $p(X=r)=\frac{e^{-\lambda }{\lambda }^r}{r!}$
$\Rightarrow \frac{2e^{-\lambda }\lambda }{1!}=\frac{5e^{-\lambda }{\lambda }^5}{5!}+\frac{2e^{-\lambda }{\lambda }^3}{3!}$
$\Rightarrow 2=\frac{{\lambda }^4}{24}+\frac{{\lambda }^2}{3}$
$\Rightarrow 2=\frac{3{\lambda }^4+24{\lambda }^2}{72}$
$\Rightarrow 144=3{\lambda }^4+24{\lambda }^2$
$\Rightarrow 48={\lambda }^4+8{\lambda }^2$
$\Rightarrow {\lambda }^4+8{\lambda }^2-48=0$
Let ${\lambda }^2=y$
$\Rightarrow y^2+8y-48=0$
$\Rightarrow y^2+12y-4y-48=0$
$\Rightarrow y(y+12-4(y+12)=0$
$\Rightarrow (y+12)(y-4)+0$
$\Rightarrow y=4(\because y\ne -12)$
$\Rightarrow {\lambda }^2=4$
$\Rightarrow \lambda =2$
$\therefore$ Standard deviation $=\sqrt{\lambda }=\sqrt{2}$