Question

On any given flight, the goal of an airline is to fill the plane as much as possible, without exceeding the capacity of the plane. In order to achieve this, the airlines routinely overbook their flights in consideration of last minute cancellations. We assume that a customer cancels his/her ticket in the last minute with probability 0.03, independent of the other customers. We also assume that the airline is not able to sell more tickets in order to replace the canceled ones. What is the probability that a particular flight will be over capacity if the airline sells 325 tickets, for a plane that has a maximum capacity of 300 seats? In solving this problem, use the Central Limit Theorem, and in particular, use the De Moivre-Laplace normal approximation to the binomial distribution (with 1/2 correction) and be very careful when you choose the boundaries for probability computation. You will also need to use the standard normal CDF table. Use this table precisely as follows: In using the standard normal CDF table, first compute the input argument for the standard normal CDF with your calculator, then round this input argument value to two decimal digits after the decimal point, and finally locate the entry in the table which corresponds to the rounded input value. If you need the value of the standard normal CDF for arguments larger than 3.49 (not available in the table), you can use 1.0000. Your final answer for the problem should have four decimal digits after the decimal point.

Answer 1

Let X be the random variable denoting the number of seats filled.

We first calculate the mean and the standard deviation of the binomial distribution. This is given by:

To use the normal approximation, as stated in De Moivre-Laplace theorem, we use the same mean and the same standard deviation as the binomial distribution.

The flight is not over capacity if it has 300 or less passengers and is over capacity if it has 301 or more passengers. Since, the normal approximation is a continuous distribution, we take the cutoff limit as average of 300 and 301, that is we take it as x=300.5. This is the 0.5 correction as mentioned in the problem. The z-score is calculated as:

This is less than -3.49, the minimum in a standard z-table. So, probability of flight not being over capacity is 0.0000.

So, the probability of flight being over capacity is 1-0.0000 = 1.0000.