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
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 $t > 0.$