American Airlines Flight 201 from New York\'s JFK airport to LAX airport in Los

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 Airlines 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 wil not show up for the flight. 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. 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? 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

Here the probability that a passenger with a reservation will not show up for the flight = 0.0995

so, if 182 reservations are accepted,

Expected number of passenger that will show for the flight if 182 reservations are accepted = 182 * (1 - 0.0995) = 163.891

Standard deviation of passengers that will show for the flight if 182 reservations are accepted = sqrt [0.0995 * 0.9005 * 182] = 4.0382

Pr(X > 168 ; 163.891; 4.0382)

taking continuity correction

Pr(X > 168 ; 163.891; 4.0382) = NORM (X > 168.5;163.891; 4.0382)  

Z = (168.5 - 163.891)/ 4.0382 = 1.14

Pr(Z > 1.14) = 0.1269

Here the probability is more than 0.05, so it seem too high so that changes must be made to make it lower.

Now, we have 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. If the number is X then

expected number of pessanger to show up = 0.9005X

Standard deviation of pessanger to show up = sqrt [0.9005 * 0.0995 * X]

Pr(X > 168; 0.9005X ; sqrt [0.9005 * 0.0995 * X] ) < 0.05

Z - value = 1.645

using coninuity factor

(168.5 - 0.9005X) / sqrt [0.9005 * 0.0995 * X] = 1.645

(168.5 - 0.9005X) = 1.645 * sqrt [0.9005 * 0.0995 * X]

(168.5 - 0.9005X) 2 = 0.2425 X

28392.25 + 0.8109X2 - 303.711X = 0

X = 179.78 or say 180

so they must book maximum 180 seats to have overbooking chances less than 0.05

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