If X is a Poisson variate with mean 2,then Pl | vedclass

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(C) For a Poisson distribution with mean $\lambda = 2$,the probability mass function is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} = \frac

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) For a Poisson distribution with mean $\lambda = 2$,the probability mass function is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} = \frac{e^{-2} 2^k}{k!}$.We need to find $P(X > \frac{3}{2})$. Since $X$ takes only non-negative integer values,$X > \frac{3}{2}$ is equivalent to $X \geq 2$.Using the complement rule,$P(X \geq 2) = 1 - P(X Calculating the individual probabilities:$P(X = 0) = \frac{e^{-2} 2^0}{0!} = e^{-2} = \frac{1}{e^2}$.$P(X = 1) = \frac{e^{-2} 2^1}{1!} = 2e^{-2} = \frac{2}{e^2}$.Therefore,$P(X \geq 2) = 1 - [\frac{1}{e^2} + \frac{2}{e^2}] = 1 - \frac{3}{e^2} = \frac{e^2 - 3}{e^2}$.Thus,the correct option is $C$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X = x)$	$k$	$\frac{k}{3}$	$\frac{k}{4}$	$\frac{k}{2}$	$\frac{k}{2}$

$X$	$-3$	$-1$	$0$	$1$	$3$	$5$	$7$	$9$
$F(X)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

If $X$ is a Poisson variate with mean $2$,then $P\left(X>\frac{3}{2}\right)=$

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