Can you explain the queueing question/answer in this MBA exam?

Question

In the paper that describes how ChatGPT could pass an MBA exam there is a question and associated answer relating to queueing that seems intuitively wrong. I copy the question and the answer the author of the paper gives (NOT ChatGPT) in full below, in short the question is: given a polling station with a max rate of 12 people/hour and an expected flow rate of 10 people/hour what is the average length of queue experienced, with an answer of 5 people. With max rate higher than the expected rate I would expect an average closer to zero than five, and I would expect the average rate to depend on how long the station is open, as the first person is sure to have no queue.

I looked at the wiki page on queuing theory, and I could not find anything that looks like "Average Waiting Time = Average Processing Time x Utilization / (1-Utilization)".

Question (written by exam authors)

The Pennsylvania Department of State is implementing a new electronic voting system. Voters will now use a very simple self-service computer kiosk for casting their ballots. If that kiosk is busy, voters will patiently queue up and wait until it is there turn.

It is expected that voters will spend on average 5 minutes at the kiosk. This time will vary across voters with a standard deviation of 5 minutes. Voters are expected to arrive at a demand rate of 10 voters per hour. These arrivals will be randomly spread out over the hour (you can assume that the number of voters arriving in any time period follows a Poisson distribution).

What is the average amount of time that a voter will have to wait before casting their vote?

Answer (written by paper authors)

To find the right answer, one must look at a standard equation from queuing theory. The equation for the average waiting time states that:

Average Waiting Time = Average Processing Time x Utilization / (1-Utilization).

Plugging in an average processing time of 5 minutes and an average utilization of 5/6, we get:

Average Waiting Time = 5 x (5/6) / (1 - 5/6) = 25 minutes.

So, the correct answer is 25 minutes waiting in line. If we add the 5 minutes at the kiosk, we obtain a total of 30 minutes.

Answer 1

Not being a queueing theorist I give it a go:

• The system is an $M/M/1$ queue with mean interarrival time of $1/\lambda$ being six minutes ($10$ voters per hour) and mean exponential service time of $1/\mu$ being five minutes.

• With the notation $\rho=\lambda/\mu\approx 0.83333$ and by Little's law the expected waiting time of a voter in the system (queue plus casting the vote) is then $$\tag{1} \mathbb E[S]=\frac{1/\mu}{1-\rho}=\frac{1}{\mu-\lambda}=30\text{ minutes }. $$ The mean waiting time (in the queue) is $$ \mathbb E[W]=\mathbb E[S]-1/\mu=25 \text{ minutes }. $$ Formula (1) has an intuitive explanation here.