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(C) For a Poisson distribution,the mean $\lambda$ is equal to the variance. Given variance $= 2$,so $\lambda = 2$. The probability mass function is $P

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$(e^2 - 5)/e^2$

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(C) For a Poisson distribution,the mean $\lambda$ is equal to the variance. Given variance $= 2$,so $\lambda = 2$. The probability mass function is $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$. We need to find $P(X \geq 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]$. $P(X=0) = \frac{e^{-2} 2^0}{0!} = e^{-2}$. $P(X=1) = \frac{e^{-2} 2^1}{1!} = 2e^{-2}$. $P(X=2) = \frac{e^{-2} 2^2}{2!} = \frac{4e^{-2}}{2} = 2e^{-2}$. Sum $= e^{-2} + 2e^{-2} + 2e^{-2} = 5e^{-2} = \frac{5}{e^2}$. Therefore,$P(X \geq 3) = 1 - \frac{5}{e^2} = \frac{e^2 - 5}{e^2}$.

If $X$ follows a Poisson distribution with variance $2$,then $P(X \geq 3) = $

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