For a Poisson variate X,if P(X=2)=3 P(X=3),th | vedclass

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(A) The probability mass function of a Poisson distribution is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$,where $\lambda$ is the mean of th

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$1$

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(A) The probability mass function of a Poisson distribution is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$,where $\lambda$ is the mean of the distribution. Given the condition $P(X=2) = 3 P(X=3)$. Substituting the formula: $\frac{e^{-\lambda} \lambda^2}{2!} = 3 \cdot \frac{e^{-\lambda} \lambda^3}{3!}$ Dividing both sides by $e^{-\lambda} \lambda^2$ (assuming $\lambda 
eq 0$): $\frac{1}{2} = 3 \cdot \frac{\lambda}{3 \cdot 2 \cdot 1}$ $\frac{1}{2} = \frac{3 \lambda}{6}$ $\frac{1}{2} = \frac{\lambda}{2}$ $\lambda = 1$. Thus,the mean of the Poisson distribution is $1$.

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

For a Poisson variate $X$,if $P(X=2)=3 P(X=3)$,then the mean of $X$ is:

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