Question

A random variable X has Poisson distribution with mean 2. Then P $(x>1.5)$ equals

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

Answer 1

The correct option is C $1-\frac{3}{e^2}$

We know, $P(x=k)=e^{-\lambda }.\frac{{\lambda }^k}{K!}$
Here $\lambda =2$
$\therefore P(x>1.5)=P(x\ge 2)$

$=1-p(x=0)-P(x=1)$

$=1-e^{-\lambda }-e^{-\lambda }\left(\frac{\lambda }{1!}\right)$

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