Question

A flight scheduler makes a mistake and schedules two airplanes to land at the exact same time at an airport with a single runway. Assume that once an airplane approaches the airport, time it takes to land the airplane and clear the runway for another plane is an exponentially distributed random variable with mean 40 minutes. The first plane approaches the airport on time, but the other one is 10 minutes late. The second airplane can’t start landing until the first plane clears. Find the expected value of the time between first airplane approaching and landing/clearance of the second plane.

Answer 1

As exponential distribution follows memoryless property and we are given here that the second airplane can’t start landing until the first plane clears, therefore the time between first airplane approaching and landing/clearance of the second plane can be thought of as a sum of 2 exponential random variables. The waiting time for the second plane starts only after the first plane has been cleared already.

Therefore the total waiting time here could be modelled as:

Y = 2X where X is the distribution of the waiting time of 1 plane.

Now we are given the mean of X here as 40 minutes. Therefore E(X) = 40, which means that E(Y) = 2E(X) = 80

Therefore 80 minutes is the required expected value of time between first airplane approaching and landing / clearance of the second plane.