Question

Air-USA has a policy of booking as many as 21 persons on an airplane that can seat only 19. (Past studies have revealed that only 85% of the booked passengers actually arrive for the flight.)

Find the probability that if Air-USA books 21 persons, not enough seats will be available. prob = ?

Is this probability low enough so that overbooking is not a real concern for passengers if you define unusual as 5% or less?

yes, it is low enough not to be a concern

no, it is not low enough to not be a concern

What about defining unusual as 10% or less?

yes, it is low enough not to be a concern

no, it is not low enough to not be a concern

Answer 1

Solution:

Air-USA has a policy of booking as many as 21 persons on an airplane that can seat only 19. (Past studies have revealed that only 85% of the booked passengers actually arrive for the flight.)

Find the probability that if Air-USA books 21 persons, not enough seats will be available. prob = ?

We are given

n = 21,

p = 0.85,

We have to find P(X>19)

P(X>19) = 1- P(X≤19)

P(X≤19) = 0.844962

(by using excel command =binomdist(19,21, 0.85, 1))

P(X>19) = 1- P(X≤19)

P(X>19) = 1- 0.844962

P(X>19) = 0.155038

Required probability = 0.155038

Is this probability low enough so that overbooking is not a real concern for passengers if you define unusual as 5% or less?

No, it is not low enough to not be a concern, because above probability 0.155038 is greater than 0.05 or 5%.

What about defining unusual as 10% or less?

No, it is not low enough to not be a concern, because above probability 0.155038 is greater than 0.10 or 10%.