## Prompt

The Peachtree Airport in Atlanta serves light aircraft. It has a single runway and one air traffic controller to land planes. It takes an airplane 12 minutes to land and clear the runway (following an exponential distribution). Planes arrive at the airport at the rate of one every 15 minutes. The arrival rate follows a Poisson distribution. A plane is considered to have entered the “system” once it has notified the airport that it is in the vicinity and wants to land. For purposes of this analysis, you can ignore the planes taking off.

- Determine the average number of planes that will stack up (wait) to land.
- Find the average time a plane must “wait in line” before it can land.
- What is the probability that a plane approaching the airport will find at least three other planes already in the “system” waiting to land?

### ChatGPT-Generated Response

- The average number of planes that will stack up (wait) to land can be calculated as the ratio of the arrival rate to the service rate, which is the rate at which planes are cleared from the runway. The arrival rate is one plane every 15 minutes, or 1/15 per minute. The service rate is one plane every 12 minutes, or 1/12 per minute. So, the average number of planes waiting to land is given by:

Average number of planes waiting = Arrival rate / Service rate = 1/15 / 1/12 = (1/15) * (12/1) = 4/5- To find the average time a plane must “wait in line” before it can land, we can use Little’s Law, which states that the average number of items in a queue (N) is equal to the arrival rate (λ) multiplied by the average time spent in the system (W):
Since we already found that N = 4/5, we can solve for W:

- N = λ * W
So, the average time a plane must “wait in line” before it can land is 12 minutes.

W = N / λ = 4/5 / (1/15) = (4/5) * (15/1) = 4 * 3 = 12 minutes

- To find the probability that a plane approaching the airport will find at least three other planes already in the “system” waiting to land, we can use the Poisson distribution formula:

P(X >= 3) = 1 – P(X < 3) = 1 – (P(X = 0) + P(X = 1) + P(X = 2))

where X is the number of planes in the system. The mean of the Poisson distribution is equal to the arrival rate, so in this case, the mean is 1/15 per minute. So, the probability of finding 0, 1, or 2 planes in the system can be calculated as:

- P(X = 0) = e^(-λ) * (λ^0) / 0! = e^(-1/15) * (1/15)^0 / 0! = e^(-1/15)
- P(X = 1) = e^(-λ) * (λ^1) / 1! = e^(-1/15) * (1/15)^1 / 1! = (1/15) * e^(-1/15)
- P(X = 2) = e^(-λ) * (λ^2) / 2! = e^(-1/15) * (1/15)^2 / 2! = (1/15)^2 * e^(-1/15) / 2
So, the probability that a plane approaching the airport will find at least three other planes already in the “system” waiting to land is:

- P(X >= 3) = 1 – (P(X = 0) + P(X = 1) + P(X = 2)) = 1 – (e^(-1/15) + (1/15) * e^(-1/15) + (1/15)^2 * e^(-1/15) / 2)

*(click below to reveal the “expert analysis”)*

## Expert Analysis

- For part a, ChatGPT does not select the correct formula. For a system with Poisson arrivals and exponential service times, the average number of planes waiting in line for service is Lq=2(-), where μ is the service rate and λ is the number of planes that actually land per hour. In this problem, =60/12=5 planes per hour, and =60/15 = 4 planes per hour. The average number of waiting planes is Lq=2(-)=425(5-4)=3.2 planes. Compare this to ChatGPT’s answer of ⅘ of a plane.
- ChatGPT also does not select the correct formula for part b. For a system with Poisson arrivals and exponential service times, the average wait time is Wq=Lq=3.24=0.8 hours or 48 minutes.
- ChatGPT correctly identifies that the probability that three or more planes will be waiting can be calculated with the formula P(X >= 3) = 1 – P(X < 3) = 1 – (P(X = 0) + P(X = 1) + P(X = 2)), which states that 100% minus the probability that 0, 1, or 2 planes are waiting gives the probability that 3 or more planes are waiting. However, the appropriate formula to calculate the probability that there are
*n*planes waiting is Pn=(/)2P0, where P0=1-/. The probability that 3 or more planes are waiting is, therefore, 1-P2-P1-P0=1-(0.8)2(0.2)-(0.8)1(0.2)-0.2=0.512. Compare this to ChatGPT’s answer, 1 – (e^(-1/15) + (1/15) * e^(-1/15) + (1/15)^2 * e^(-1/15) / 2). A student would be unlikely to provide an unsimplified answer like this, and the decimal approximation is tiny, about 0.00005.