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

Vedclass · Accepted Answer

$\frac{1}{5}$

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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 $\lambda = \frac{1}{5}$.

$X = x_i$	$1$	$2$	$3$	$5$
$P(X = x_i)$	$2k^2$	$k$	$k$	$k^2$

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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