A random variable X has Poisson distribution with mean 2.

Question

A random variable $X$ has Poisson distribution with mean $2$.

Then $$P\left( {X > 1.5} \right)$$ equals :

Accepted Answer

In a position distribution,

$$P(X = r) = {{{e^{ - \lambda }}{\lambda ^r}} \over {r!}}$$ ($$\lambda$$ = mean).

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

$$ = 1 - \{ P(0) + P(1)\} $$

$$ = 1 - \left( {{e^{ - 2}} + {{{e^{ - 2}} \times 2} \over 1}} \right) = 1 - {3 \over {{e^2}}}$$

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