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This is a question from an old test, taken completely out of context:

"The customers arrive at a shop according to a Poisson distribution with mean $\lambda$ per hour. Each customer takes $1/k$ hours of service. What is the expected value and the variance of the service time for customers arriving within an hour? What is the probability that the service time exceeds $t$ hours?

I am not entirely certain how to approach this question. This is supposed to be a simple Poisson-distribution question, with no Poisson processes or queuing.

Are there any ways to look at this question in a simpler way? What is the general approach here?

Apologies if I am completely misunderstanding this.

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  • $\begingroup$ I would start by defining the number of customers arriving at the shop in a given hour as $X \sim Pois(\lambda)$ and $k^{-1}X$ as the service time they require. Can you then see how to answer the first two parts using the properties of expectations and variances? Can you write down an expression for the final part? Show us your working so far and we can see where the difficulty lies. $\endgroup$ – M. Berk Oct 22 '13 at 9:13
  • $\begingroup$ Okay, if the service time is also proportional to the Poisson, we can say that the expectation is $1/kE(X)$, and the variance will be $1/k^2Var(X)$. The probability of seeing a wait time longer than $t$ is going to be $P(X>t)$ which is the integral of the distribution from $t+1$ to $\inf$. I was not certain that the wait time are also $Poisson$. $\endgroup$ – ivan-k Oct 23 '13 at 1:55
  • $\begingroup$ In the absence of further information it seems reasonable to assume that $k$ is fixed and identical for each customer. However, $k^{-1}X$ is not Poisson distributed - for starters $E[k^{-1}X] \ne Var[k^{-1}X]$. $\endgroup$ – M. Berk Oct 23 '13 at 8:26
  • $\begingroup$ I do not have enough rep to up-vote your comment. I believe it is clear to me know. Thank you kindly. $\endgroup$ – ivan-k Oct 23 '13 at 21:02

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