Question

If a random variable $X$ has Poisson distribution with mean $2$, then $P(X>1.5)$equals

• A. $\frac{3}{e^2}$
• B. $1-\frac{3}{e^2}$
• C. $0$
• D. $\frac{2}{e^2}$

Answer 1

The correct option is B

$1-\frac{3}{e^2}$

Explanation for the correct options:

Find the value of $\mathit{P}\mathbf{(}\mathit{X}\mathbf{}\mathbf{>}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{)}\mathbf{}$:

We know that,

The probability density function in Poisson distribution is

$\mathit{P}\mathbf{(}\mathit{X}\mathbf{=}\mathbf{}\mathit{x}\mathbf{)}\mathbf{}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{\lambda }}\mathbf{.}\mathbf{}\mathbf{\lambda }\mathbf{x}}{\mathbf{x}\mathbf{!}}\mathbf{}$

Given that mean $=2=\lambda$

Now, $P(X>1.5)=P(X=2)+P(X=3)+\dots .$

$$\begin{gathered} =1-\left[P\right(X=0)+P(X=1\left)\right] \\ =1-[e^{-2}+2e^{-2}] \\ =1-3e^{-2} \\ =1-\frac{3}{e^2} \end{gathered}$$

Hence ,the correct option is B.