The number of customers entering a 24-hour convenience store every 10-minutes can be modeled by the Poisson distribution with a a mean of λ=4.5 customers. You are to look at the amount of time passing between successive customers entering the convenience store, represented by X.
if 2.22 minutes pass between successive customers entering the store, what is the probability at most 2.5 minutes will pass between one customer entering the store to the next?
What I have tried so far;
λ = 4.5, x value 1 = 0 minutes, x value 2 = 2.5 minutes.
looking at the distribution of x between 0 and 2.5 the value is 0.1736. Is this the correct way to find the solution?
Best Answer
The key observation to make is that if $$N \sim \operatorname{Poisson}(\lambda = 4.5)$$ counts the random number of customers entering the store in a given $10$-minute period, then the interarrival time random variable $X$ that measures the random time between the arrivals of successive customers is $$X \sim \operatorname{Exponential}(\mu = 10/\lambda),$$ with cumulative distribution function $$F_X(x) = \Pr[X \le x] = 1 - e^{-x/\mu}, \quad x > 0.$$ This is the fundamental relationship in a (homogeneous) Poisson process.