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(A) The number of customers visiting the shop follows a Poisson distribution with parameter $\lambda = 4$. The probability mass function is given by $

Vedclass · Accepted Answer

$\frac{e^4-13}{e^4}$

Vedclass · Answer

(A) The number of customers visiting the shop follows a Poisson distribution with parameter $\lambda = 4$. The probability mass function is given by $P(X=x) = \frac{\lambda^x e^{-\lambda}}{x!}$. We need to find the probability that more than $2$ customers visit, which is $P(X > 2)$. $P(X > 2) = 1 - P(X \leq 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$. Substituting the values: $P(X=0) = \frac{4^0 e^{-4}}{0!} = e^{-4}$. $P(X=1) = \frac{4^1 e^{-4}}{1!} = 4e^{-4}$. $P(X=2) = \frac{4^2 e^{-4}}{2!} = \frac{16e^{-4}}{2} = 8e^{-4}$. Therefore, $P(X > 2) = 1 - [e^{-4} + 4e^{-4} + 8e^{-4}] = 1 - 13e^{-4}$. This can be written as $1 - \frac{13}{e^4} = \frac{e^4 - 13}{e^4}$.

$X = x$	$0$	$1$	$2$	$3$	$4$
$P(X = x)$	$k$	$2k$	$4k$	$2k$	$k$

If on an average $4$ customers visit a shop in an hour, then the probability that more than $2$ customers visit the shop in a specific hour is

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