American Airlines Flight 201 from New York’s JFK airport to LAX airport in Los A

Question

American Airlines Flight 201 from New York’s JFK airport to LAX airport in Los Angeles uses a Boeing 767-200 with 168 seats available for passengers. Because some people with reservations don’t show up, American can overbook by accepting more than 168 reservations. If the flight is not overbooked, the airline will lose revenue due to empty seats, but if too many seats are sold and some passengers are denied seats, the airline loses money from the compensation that must be given to the bumped passengers. Assume that there is a 0.0995 probability that a passenger with a reservation will not show up for the flight (based on data from the IBM research paper “Passenger-Based Predictive Modeling of Airline No-Show Rates,” by Lawrence, Hong, and Cherrier). Also assume that the airline accepts 182 reservations for the 168 seats that are available. Find the probability that when 182 reservations are accepted for American Airlines Flight 201, there are more passengers showing up than there are seats available. The best approach is to use statistics software or a TI-83/84 Plus calculator. Is the probability of overbooking small enough so that it does not happen very often, or does it seem too high so that changes must be made to make it lower? Now use trial and error to find the maximum number of reservations that could be accepted so that the probability of having more passengers than seats is 0.05 or less.

Explanation / Answer

Seats available = 168

Acceping reservations = 182

Prob for cancellation = 0.0995

Each cancellation is independent, and there are two outcomes.

X- no of cancellations is binomial, with n = 182 and p = 0.0995

P(passengers >168)

= P(X<182-168)

= 0.12413

p value >0.05, hence we can say that normal average is 168 only.

No need to revise.

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P(X>14) <=0.05

n is to be found out given the above.

Using binomial table we check for this

For 204 reservations we have the prob that having more passengers is 0.05 or less.

For = P(X<14)

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