In a Poisson distribution, the probability $P (X = 0)$ is twice the probability $P (X = 1)$ . The mean of the distribution is

\frac{1}{4}

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\frac{1}{3}

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\frac{1}{2}

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\frac{3}{4}

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Solution

The correct option is D $\frac{1}{2}$
Poisson's distribution implies
$P (X = n) = \frac{λ^{n} e^{- λ}}{n!}$
Where mean=variance= $λ$
Hence $P (x = 0) = 2 P (x = 1)$
$\frac{λ^{0} e^{- λ}}{0!} = 2 \frac{λ^{1} e^{- λ}}{1!}$
$2 λ = 1$
$λ = \frac{1}{2}$
Hence mean=variance= $0.5$

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