Question

Flights 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 reserva-tion will not show up for the flight. Also assume that the airline accepts 182 reservations for the 168 seats that are avail-able. Find the probability that when 182 reservations are accepted for American Airlines Flight 201, there are more pas-sengers showing up than there are seats available. Table A-1 (the Binomial Probabilities table) cannot be used and calcu-lations with the binomial probability formula would be extremely time-consuming and tedious. The best approach is to use statistics software. (See Section 5-3 in the textbook for instructions describing the use of StatCrunch, Excel, STATDISK) .

A) Is the proba-bility 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?

B) 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. Now use trial and error by changing the “n” to determine the maximum number of tickets that you can “safely” sell, but still make as much profit as possible. Give a detailed account of the procedure you used to arrive at your conclusion. The maximum number of tickets I can “safe-ly” sell is?

C)Use proper notation to show how you calculated this probability. P(        ) is the number of successes I computed the probability of because (answer). I figured out my “n” by (answer in question form)   

Answer 1

A. There are 168 seats, and Airlines company accepts 182 reservations. The probability of a passenger with reservation will not show up is 0.0995.

Suppose, X denotes the number of passengers who will not show up.

Then P(X>=182-168) = P(X>=14)

= 1 - P(X<=13)

Here, X follows a binomial distribution with parameters n=182 and p=0.0995.

This can be calculated using R-code

1 - pbinom(13, 182,0.0995)

= 0.8759

This value indicates the probability that no passenger is compensated, and this probability is not very high. So the airlines company should check their policy to reduce the number.

B. For this problem, we need to find maximum of n such that the probability of not showing up of n passengers out of (168+n) passengers is 0.05 Or less. We can use R code, and make trial and error.

Use the following R-code:

p<-NULL

for( i in 1:14) {

p[i] = pbinom(i, 168+i, 0.0995) }

p

## From the values of p, we can decide n = 10.

C. This has been done already.