Question

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

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

Answer 1

Answer: (D)

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