In a Poisson distribution,if frac{P(X=5)}{P(X | vedclass

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

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$\frac{1}{5}$

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(B) For a Poisson distribution with parameter $\lambda$,the probability mass function is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.Given $\frac{P(X=5)}{P(X=2)} = \frac{1}{7500}$,we have $\frac{\frac{e^{-\lambda} \lambda^5}{5!}}{\frac{e^{-\lambda} \lambda^2}{2!}} = \frac{\lambda^3}{5 	imes 4 	imes 3} = \frac{\lambda^3}{60} = \frac{1}{7500}$.Thus,$\lambda^3 = \frac{60}{7500} = \frac{6}{750} = \frac{1}{125}$.Taking the cube root,$\lambda = \frac{1}{5}$.Checking with the second condition: $\frac{P(X=5)}{P(X=3)} = \frac{\frac{e^{-\lambda} \lambda^5}{5!}}{\frac{e^{-\lambda} \lambda^3}{3!}} = \frac{\lambda^2}{5 	imes 4} = \frac{\lambda^2}{20}$.Substituting $\lambda = \frac{1}{5}$,we get $\frac{(1/5)^2}{20} = \frac{1/25}{20} = \frac{1}{500}$.This matches the given condition. Therefore,the mean of the distribution is $\lambda = \frac{1}{5}$.

Outcome	Probability
$\omega_{1}$	$0.1$
$\omega_{2}$	$0.01$
$\omega_{3}$	$0.05$
$\omega_{4}$	$0.03$
$\omega_{5}$	$0.01$
$\omega_{6}$	$0.2$
$\omega_{7}$	$0.6$

In a Poisson distribution,if $\frac{P(X=5)}{P(X=2)}=\frac{1}{7500}$ and $\frac{P(X=5)}{P(X=3)}=\frac{1}{500}$,then the mean of the distribution is

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