Question

Activity Three: Poisson and Exponential Distributions

An accountant notes that, on average, it takes 30 minutes to talk to two clients, with the time of visits following an exponential distribution. What is the probability that the time between the arrival of one client to the arrival of the next client will be less than ten minutes? Can you show the same answer using the Poisson formula by asking the probability that at least one client will be seen within a ten minute period?

Answer 1

Given, the rate of exponential distribution is 2 clients per 30 minutes = 2/30 per minute = 1/15 per minute

or, the parameter of exponential distribution is = 1/15 per minute

Let T be the time between the arrival of one client to the arrival of the next client. Then T ~ Exp( = 1/15)

The probability that the time between the arrival of one client to the arrival of the next client will be less than ten minutes = P(T < 10) = 1 - exp(- * 10)

= 1 - exp(-10/15) = 0.4866

We know that, if the time between the arrival of one client to the arrival of the next client follows exponential distribution, then the number of arrivals for a specific time range follows Poisson distribution. Let X be the number of arrivals of clients within the a ten minute period.

For 10 minute period, = 1/15 per minute = 10/15 per 10 minutes.

So, X ~ Poisson( = 10/15)

Probability that at least one client will be seen within a ten minute period = P(X 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - exp(-) / 0!

= 1 - exp(-)

1 - exp(-10/15) = 0.4866