Calls for dial-in connections to a computer centre arrive at an average rate of four per minute. The calls follow a Poisson distribution. If a call arrives at the beginning of a 1-minute interval, what is the probability that a second call will not arrive in the next 20 seconds?
I'm having trouble putting this into the Poisson probability distribution formula. It seems different from some of the other problems I've done. How do you solve this?
Jack
OK. I think I got it, but could someone confirm this:
I let Y be the time between the arrival of two dial-in connections in minutes. We have P(Y>1/3)
(lambda)t=4 and (1/lambda)=(1/4)
Thus, f(y)=(1/0.25)e^(-y/0.25)=4e^(-4y)
P(Y>1/3)=integration of 4e^(-4y) from (1/3) to infinity. This equals e^(-4/3).
Is this the probability?