The proprietor serves an average of 800 students a day, with inter-arrival times following an exponential distribution. Most students check out with very few items, but there are some that do all their grocery shopping there, which results in service times that are non-exponential. Based on recorded data, the checkout clerk takes an average of 1.2 minutes to ring up a student, with a standard deviation of 3.5 minutes. How many students, on average, are waiting to be checked out?
To find the average number of students waiting to be checked out, we can use the M/M/1 queuing model. M/M/1 stands for Markovian arrival process/Markovian service process/single server.
In this case, the inter-arrival times follow an exponential distribution, and the service times are non-exponential but can be approximated as exponential. Let's assume that the service times in this case also follow an exponential distribution.
To calculate the average number of students waiting to be checked out, we need to know two key parameters: arrival rate (λ) and service rate (μ).
The arrival rate (λ) is the average number of students arriving per unit of time. In this case, we are given that the proprietor serves an average of 800 students a day. Assuming the business operates for 24 hours a day, the arrival rate can be calculated as follows:
λ = 800 students / 24 hours = 33.33 students per hour
The service rate (μ) is the average number of students served per unit of time. In this case, we are given that the checkout clerk takes an average of 1.2 minutes (0.02 hours) to ring up a student.
μ = 1 / average service time = 1 / 0.02 hours = 50 students per hour
Now, we can use Little's Law to find the average number of students in the system (i.e., waiting to be checked out).
Average number of students in the system (L) = λ / (μ - λ)
L = 33.33 / (50 - 33.33) = 33.33 / 16.67 = 2 students
Therefore, on average, there are 2 students waiting to be checked out.