A random variable X has a Poisson distributio | vedclass

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(A) For a Poisson distribution,the probability mass function is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$,where $\lambda = 2$.We need to

Vedclass · Accepted Answer

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

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(A) For a Poisson distribution,the probability mass function is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$,where $\lambda = 2$.We need to find $P(X > 1.5)$. Since $X$ takes only non-negative integer values,$P(X > 1.5) = P(X \ge 2)$.This can be calculated as $P(X \ge 2) = 1 - P(X = 0) - P(X = 1)$.For $k = 0$: $P(X = 0) = \frac{e^{-2} 2^0}{0!} = \frac{e^{-2} \cdot 1}{1} = \frac{1}{e^2}$.For $k = 1$: $P(X = 1) = \frac{e^{-2} 2^1}{1!} = \frac{e^{-2} \cdot 2}{1} = \frac{2}{e^2}$.Therefore,$P(X > 1.5) = 1 - (\frac{1}{e^2} + \frac{2}{e^2}) = 1 - \frac{3}{e^2}$.

$A$ random variable $X$ has a Poisson distribution with mean $\lambda = 2$. Then $P(X > 1.5)$ is equal to:

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