11 QUESTION 5 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? It is e asy to see t hat , o n average, the operation has a sufficient amount of capacity. 10 voters show up per hour, each voter requires on average 5 minutes . So , in th e averag e hour, the kiosk will be utilized for 50 minu tes a nd be idle f or 1 0 m inutes. We refer to 50/60 as the capacity utilization of the kiosk. How did Chat GPT hand le this que stion? Let’s take a look. ANSWER 5 To find the average amount of time that a voter will have to wait before casting their vote, we need to calculate the average time that the kiosk is occupied. If the average time that a voter spends at the kiosk is 5 minutes, and the demand rate is 10 voters per hour, then the average time that the kiosk is occupied is 50 minutes per hour (5 minutes per voter * 10 voters per hour). The average time that the kiosk is not occupied is therefore 60 minutes per hour - 50 minutes per hour = 10 minutes per hour.