Question: At my local post office, there is a central queue served by three assistants. From past experience, I know that the service times of twp of the assistants are exponentially distributed with mean 2 minutes and the service time of the third assistant is exponentially distributed with mean 3 minutes

One day, a customer enters the post office to discover that all three assistant are busy, but nobody else is waiting to be served

1. Find the distribution of the time she has to wait before she can move forward for service and show that her expected waiting time is 45 seconds. Calculate the probability that she has to wait more that 1.5 minutes for an assistant to be free to serve her.
2. Two minutes after entering the post office she is still waiting. Another customer enters the post office and stands behind her in the queue. (So now there are the same three people being served as there were at the start, and two people standing in the queue.) State the distribution of this latest arrival’s waiting time before his service commences, and hence or otherwise find the mean and standard deviation of his waiting time.

Price: $2.99

Solution: The downloadable solution consists of 2 pages
Deliverable: Word Document