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Customers arrive randomly and independently at a service window, and the time between arrivals has an exponential distribution with a mean of 12 minutes. Let X equal the number of arrivals per hour. What is P[X = 10]?

Now the solution to this problem uses this logic:

If the time between arrivals is exponential with mean 12 minutes and arrival times are independent then the number of arrivals in any single minute is Poisson with mean 1/12. Since the sum of independent Poisson variables is also Poisson, the number of arrivals in an hour will be Poisson with mean: 5

My question is how can an exponential distribution be related to a Poisson PMF? Thank you

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Let $N_t \sim Pois(\lambda t)$ be the number of events in $(0,t],$ for $t > 0.$ Then the probability of seeing no events in $(0,t]$ is $P(N_t = 0) = e^{-\lambda t}.$

Starting at time $t=0,$ let $X$ be the waiting time for the first event. Then another way to say there are no events in $(0,t]$ is $P(X > t) = e^{-\lambda t}.$

So $F_X(t) = P(X \le t) = 1 - P(X > t) = 1 - e^{-\lambda t},$ which is the CDF of $X \sim Exp(\lambda).$

If you want the PDF of $X,$ then $f_X(t) = F_X^\prime(t) = \lambda e^{-\lambda t},\,$ for $t > 0.$

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Broadly speaking, the exponential distribution is a continuous distribution that tells you how long to wait until the next arrival, and the Poisson distribution is a discrete distribution that tells you how many arrivals happen within a given interval of time.

More specifically: If interarrival times are independent and exponentially distributed with mean $1/\lambda$, then the number of arrivals during a given interval of time of length $t$ obey a Poisson distribution with mean $\lambda t$.

This is a basic principle of stochastic processes and queueing theory. Proving that these two distributions have this relationship is straightforward and can be found on the web (if you would like to see a demonstration, I can put one here). Although it's worth seeing once, you don't need to know how to prove it; it's sufficient to know the relationship holds.

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