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

2 / e^{2}

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

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\frac{3}{e^{2}}

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Solution

The correct option is D $1 - \frac{3}{e^{2}}$
For Poisson distribution of mean = $μ = 2$
P $(x; μ) = \frac{e^{- μ} μ^{x}}{x!}$
$P (X > 1.5) = 1 - [P (X = 0) + P (X = 1)]$
P $(0; 2) = \frac{e^{- 2} 2^{0}}{0!} = e^{- 2}$
P $(1; 2) = \frac{e^{- 2} 2}{1!} = 2 e^{- 2}$
$P (X > 1.5) = 1 - [P (X = 0) + P (X = 1)]$
$= 1 - [e^{- 2} + 2 e^{- 2}] = 1 - 3 e^{- 2} = 1 - \frac{3}{e^{2}}$

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