For a Poisson distribution, if mean $=l$, variance $=m$ and $l+m=8$, then $e^4[1-P(X2)]=$

Question

For a Poisson distribution, if mean $=l$, variance $=m$ and $l+m=8$, then $e^4[1-P(X>2)]=$

Aicharya · Accepted Answer

For Poisson distribution, mean = variance $l=m=4$ So, $e^4[1-P(X>2)]$ $\begin{aligned} & =e^4[P \leq 2] \ & =e^4[P(X=0)+P(X=1)+P(X=2)]] \ & =e^4\left[\frac{e^{-4} 	imes 4^0}{0 !}+\frac{e^{-4} 	imes 4^1}{1 !}+\frac{e^{-4} 	imes 4^2}{2 !}ight] . \ & =e^4 	imes e^{-4}(1+4+8)=13\end{aligned}$

For a Poisson distribution, if mean $=l$, variance $=m$ and $l+m=8$, then $e^4[1-P(X>2)]=$

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