Question

In: Statistics and Probability

American Airlines Flight 201 from New York's JFK Airport to LAX airport in Los Angeles uses...

American Airlines Flight 201 from New York's JFK Airport to LAX airport in Los Angeles uses a Boeing 767 which has 168 seats available for passengers. Because some people will reservations don't show up, airlines routinely overbook. Tha airline loses revenue if there are empty seats but if more passengers show up than there are seats, the airline must pay compensation to bumped passagers.

Use the start crunch Binomial calculator to answer the following based on the number of people with a reservation of 190.

Suppose American Airlines accepts 190 reservations for this flight.

Show calculation of n=190
a) What is the probability that exactly 168 passengers will show up?
Download and attach the Binomial calculator from StatCrunch into the discussion with your answer.( Use "option" in the left corner of graph and then use the Download "option)

b) What is the probability that more than 168 passengers will sjow up (some people will get bumped)?

c) Looking at the graph of the binomial calculator, what number of passengers is the most likely to show up?

Use trial and error in changing the value of "n" with the Binomial Calculator to answer the question below. try using values between 170 and 190 for n.

What is the number of reservations, n, the airline can accept so that the probability of more than 168 passengers showing up is as close as possible to 2% (but not higher than 2%) Download and attach the Binomial calculator from statcrunch which shows your answer.

Solutions

Expert Solution

I checked up this problem from the source. There is a key statement missing in the post:

"Assume that there is a 0.0995 probability that a passenger with a reservation wil not show up for the flight. "

This is the probability of success (or failure depending on how you see it) in one trial of the Binomial distribution of the number of passenger with a reservation who do not not show up for the flight

So I shall be using this data, which was missing in your post.

Let us denote X as the random variable, for the number of passengers who show up. Then probability for sucess in a single trial is 1-0.095 = 0.905

All vlues below have been obtained using a probability calculator

a) If 168 passengers show up, the above formula gives

b) We need the probability that X is greater than or equal to 169, all the way up to 190.

c) The number of passengers who are most likely to show up, is the average or expected value of the random variable, which is obtained by the formula

For the last part, trial and error may take a long time as we are not sure how to reach the desired value quickly, and given that Binomial distribution has quite a complex formula.

Another strategy is to apply the Normal approximation for Binomial distribution since Binomial probability does not give us a single solvable equation

For a 2% probability or less, the critical Z value is 2.054. Hence, we want this Z score to exceed 2.054, when X = 168

Solving the inequality gives that the least integer value of n is at most 177. I used Desmos for getting this value of n

This can again be verified by reapplying the Binomial probability to this value of n, i.e.

The obtained probability is actually only 1.08% instead of our required 2%, which is fine because if we actually took n = 178 the probability jumps above 2% to 2.22%. Thus, n = 177 is the optimum bookings that the airline should allow

orchestra answered 1 year ago
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