Airline Problem with Poisson Approximation

[mutzy188]

Homework Statement

An ailrine always overbooks if possible. A particular plane ha 95 seats on a flight in which a ticket sells for $300. The airline sells 100 such tickets for this flight. Use a Poisson approximation only.

(a) If the probbility of an individual not showing is 0.05, assuming independence, what is the probability that the airline can accomidate all the passengers who do show up?

(b) If the airline must return the $300 price plus a penalty of $400 to each passenger that cannot get on the flight, what is the expected payout (penalty plus ticket refund) that the airline will pay? Answer = $598.56

The Attempt at a Solution

I know how to do part (a), and I got 0.560 as my answer, which agrees with the answer in the back of the book. I just don't know how to do part b.

My thoughts:

The probability that they cannot accomidate the passengers = 1 - .56 = 0.44.
(0.44)(400) + 300 = $476 . . . . but that's not correct. So I don't know what to do for this part. Any help would be greatly appreciated.

Thanks

 

[Avodyne]

You need to compute the expected number of passengers that will not show up, and multiply by $700.

 

[mutzy188]

You need to compute the expected number of passengers that will not show up, and multiply by $700.

How do I find the expected number of passengers that will not show up?

 

[Avodyne]

Sorry, that was a little too simplistic. If 5 or more fail to show up, there is no payout. If n fail to show up, with n<5, then the payout is $700 times 5-n. So if P_n is the probability that n people fail to show up, you need to compute (5-n)P_n summed over n from 0 to 4.

 

[Avodyne]

Well, I just did the calculation, using Poisson probabilities, and I got $614.14. Then I tried it using the more accurate binomial probabilities, and then I got $598.56, which is your book's answer. But the problem tells you to use Poisson. So your book screwed up.

 

[mutzy188]

Thanks for your help