Question

A ticket outlet operator is concerned about customer satisfaction levels dropping when people are waiting for long periods in a queue. The number of minutes between people arriving at a ticket outlet is shown to follow a Poisson distribution, average of 10 persons arriving every 5 minutes. The ticket outlet operator requests information on the probability that one or more persons arriving at the ticket outlet within a 3 minute period. You calculate this to be:

Answer 1

The number of minutes between people arriving at a ticket outlet is shown to follow a Poisson distribution, average of 10 persons arriving every 5 minutes.

That is we expect 10/5 = 2 person per minute.

Next we want to find  the probability that one or more persons arriving at the ticket outlet within a 3 minute period.

Mathematically, P( X >= 1) = 1 - P( X = 0) .....( 1 )

Where X = number of persons arrive in 3 minutes.

We expect 3*2 = 6 persons in 3 minutes.

So X follows Poisson random variable with mean = 6

The formula of Poisson distribution is as follows:

Plug x = 0, and = 6 in P(X = x), we get:

P(X = 0) = = 0.002479

Plug this value in equation ( 1 ), we get:

P( X >= 1) = 1 - P( X = 0) = 1 - 0.002479 = 0.997521 ( This is the finsl answer).