Business Simulation Exam

ADM 3305 – Practice Problems Solution Problem 2

There are eight employees who wash cars by hand at a car-wash center that opens at 10AM and closes at 6PM. The arrival of cars to the car wash follows a Poisson process with rate 12/hour. Two people work together on each car, and the time to clean a car by a team of two follows a normal distribution with a mean of 20 minutes and a standard deviation of 2 minutes.

a) The manager would like to have a 10-minute team meeting. What is the probability that no cars arrive in the next 10 minutes?

The c.d.f. of an exponential distribution with arrival rate l is F(t) = 1 – e-lt. The arrival rate is 12 per hour, and 10 minutes is 1/6 of an hour, so we want:

1 – F(1/6) = 1- ( 1 – e-12(1/6) ) = e-2 ≈ .135 Or, if you want to think of it in minutes, the arrival rate is 1/5 per minute, so we want: 1 – F(10) = 1- ( 1 – e-1/5(10) ) = e-2 ≈ .135

b) The manager had to delay the meeting to take a phone call for 20 minutes. During this time, surprisingly no customers arrived to the car wash. If the manager now decides to have the 10-minute meeting, compared to the answer in part a), she faces: (1) an increased chance, (2) a decreased chance, (3) the same chance of a customer arriving during the meeting. Choose one of the options and justify your answer.

By the memoryless property of the exponential distribution, the distribution for the time until the next arrival, conditional on the fact that 20 minutes have passed without an arrival, is the same as the original distribution. Therefore, the chance remains the same of an arrival occurring in the next 10 minutes.

You could also answer this by the properties of the Poisson process, which say that the number of arrivals in non-overlapping intervals are independent, and the number of arrivals in a time interval are independent of the start time of that interval.