Question

An operations manager for a major airline carrier conducts an analysis of overbooked flights. The airline’s goal is to have every seat on the aircraft filled by a passenger on each flight. Based on historical data, the airline estimates that 80% of the individuals that have a ticket for a particular flight actually board that flight. In order to compensate for the no shows, the airline overbooks each flight. If the number of individuals with a ticket at the gate exceeds the capacity of the aircraft, the airline must offer compensation to any individual willing to surrender their ticket. The airline will continue to increase this compensation until the number of individuals with tickets equals the capacity of the aircraft. As the date of each flight approaches, the airline must determine how many tickets to sell. If a flight has 20 open seats remaining four days before the flight date, how many tickets should the airline sell in order to fill as many seats as possible while keeping the likelihood of overbooking less than 0.15?

a. What is the name of the distribution that you used to model this problem?

b.What are the values of the parameters of this distribution?

c. Find P(x>20)

.d. Based on your analysis, how many tickets should the airline sell?

Answer 1

a.

Let N be the number of tickets sold. Then out of N, X people came to board the flight where probability of individuals boarding the flight = 0.8

The problem can be modeled as a Binomial experiment where the number of trials are number of tickets sold and the probability of individuals boarding the flight is 0.8

Then X ~ Binomial(N, p = 0.8)

b.

We need to find the value of N.

Mean of X = 0.8N

Standard deviation of X =

Given, P(X > 20) < 0.15

By CLT,

P[Z > (20 - 0.8N)/] < 0.15

=> (20 - 0.8N)/ < 1.0364 (Using Z tables)

(20 - 0.8N)² < 1.0364² * 0.16N

400 + 0.64N² - 32N <  0.17186N

0.64N² - 32.17186N + 400 < 0

Solving by quadratic equation formula, we get

Either, N = 22.54

So, the value of N (tickets can be booked) is 23 (Rounding to nearest integer)

The parameters of the distribution are

N = 23

p = 0.8

c.

Mean of X = 0.8 * 23 = 18.4

Standard deviation of X = = 1.9183

Using CLT,

P(x>20) = P[Z > (20.5 - 18.4) / 1.9183] (Using continuity correction)

= P[Z > 1.0947]

= 0.1368 (Using Z tables)

d.

As, calculated in part (b), number of tickets sold is 23.