# Poisson distribution ?

1. Mar 1, 2012

### XodoX

1. The problem statement, all variables and given/known data
Telephone calls enter a college switchboard according to a Poisson process on the average of three calls every 4 minutes (i.e., at a rate of λ=0.75 per minute). Let W denote the waiting time in minutes until the second call. Compute P(W>1.5 minutes).

2. Relevant equations

3. The attempt at a solution

I don't get it. No idea how to do it. I guess 1.5 means here that at the most 1 event can occur here.

2. Mar 1, 2012

### Ray Vickson

You DO know how to do it. Your guess is correct: do you see why?

RGV

3. Mar 1, 2012

### XodoX

No, I don't. The poisson distribution is not in this chapter. It's Weibull, Gompertz, extreme value, gamma, chi-square, and logonormal. I don't know which one of those it is.

4. Mar 1, 2012

### Ray Vickson

Well, it's related to the Gamma.

However, let me ask you: what does the Poisson distribution represent? Never mind if it is not in that chapter; it is either in another chapter or else in another book or else in thousands of web pages. So, you have a Poisson distribution with m = 1.5*0.75 = 9/8 = 1.125; that would be the expected number of calls to occur in a 1.5 minute period. You can find the probability distribution of the number of calls in a 1.5-minute period by using the Poisson distribution formula for mean m. Now ask: if you need to wait > 1.5 min for the second call, how many have arrived before time 1.5? What is the probability of that?

RGV

Last edited: Mar 2, 2012