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Let $X$ be the number of customers arriving daily at a local restaurant, acconding to a Poisson distribution of parameter $\lambda>0$. Due to technical reasons, the restaurant can serve only the $5$ of them daily. Find the mean number of customers being served daily at the restaurant.

Attempt. The number $Y$ of customers being served daily at the restaurant is a discrete rv, taking values $0,1,2,3,4,5$. But the probability mass is not purely from the Poisson distribution. How do we determine its mass (to find the expected value)?

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    $\begingroup$ If $Y$ is the number of customer being served, then by the constraint of the restauraunt, you have $Y=\min\{X,5\}$, because either less than $5$ customers arrive and are served or more than $5$ arrive so that only $5$ can be served by the constraint. Can you calculate $E(Y)$, now? $\endgroup$ Nov 25, 2017 at 18:38

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Following the hint by @LoveTooNap29, $Y$ is a discrete random variable, whose values are $0,1,\ldots,5$ and the probability mass function of $Y$ is $P(Y=k)=P(X=k),~k=0,1,2,3,4$ and $P(Y=5)=P(X\geq 5)$, so its mean value is $$E(Y)=\sum_{k=0}^{5}k\,P(Y=k).$$

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