Q80 / 160 · AP EAMCET 2022 (05 Jul Shift 1) · Mathematics

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

  • A
    $8$
  • Correct answer
    B
    $13$
  • C
    $9$
  • D
    $12$

Solution

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

Hence, the correct answer is option: 'B'.

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