Question

You work at a bank as a teller and have been asked to create a contingency plan should more or less than the mean number of customers show up over a hour span. The current Poisson distributed mean customers arriving per hour is 7. However, in order to go above and beyond, you want to present on the likelihood that any two customers arrive less than 2 minutes apart. Given that, what is the probability that any two random customers arrive at the bank less than 2 minutes apart?

79.1889%

9.2387%

20.8111%

18.2231%

Answer 1

Mean arrival rate = 7 per hour = (7/60) per minute = 0.1166667 per minute

Let X be the number of customers arrive within 2 minutes.

Mean arrival rate in 2 minutes = 2 * 0.1166667 = 0.2333334 per 2 minutes

X ~ Poisson( = 0.2333334)

Probability that any two random customers arrive at the bank less than 2 minutes apart = 1 - probability that any two random customers arrive at the bank more than 2 minutes apart

= 1 - probability that no customers arrive at the bank within next 2 minutes

= 1 - P(X = 0)

= 1 - exp(-0.2333334) * 0.2333334⁰ /0!

= 1 - exp(-0.2333334)

= 0.208111

= 20.8111%