Question

In: Statistics and Probability

Suppose that the probability that a passenger will miss a flight is 0.0976. Airlines do not...

Suppose that the probability that a passenger will miss a flight is 0.0976. Airlines do not like flights with empty​ seats, but it is also not desirable to have overbooked flights because passengers must be​ "bumped" from the flight. Suppose that an airplane has a seating capacity of 59 passengers.

​(a) If 61 tickets are​ sold, what is the probability that 60 or 61 passengers show up for the flight resulting in an overbooked​ flight?

​(b) Suppose that 65 tickets are sold. What is the probability that a passenger will have to be​ "bumped"?

​(c) For a plane with seating capacity of 58 ​passengers, how many tickets may be sold to keep the probability of a passenger being​ "bumped" below 5​%?

Solutions

Expert Solution

a) The number of passengers who would show up for the flight is modelled here as binomial distribution given as:

The required probability here is computed as:
P(X = 60) + P(X = 61)

This is computed in EXCEL as:
=binom.dist(60,61,0.9024,FALSE)+binom.dist(61,61,0.9024,FALSE)

0.0145 is the output here.

Therefore 0.0145 is the required probability here.

b) Probability that passenger will be bumped here is computed as:

P(X >= 60) = 1 - P(X <= 59)

This is computed for n = 65 in EXCEL here as:
=1-binom.dist(59,65,0.9024,TRUE)

0.3817 is the output here.

Therefore 0.3817 is the required probability here.

c) This is sold using hit and trial method here.

For n = 62, we have here:
P(X <= 58)

This is computed as:
=binom.dist(58,62,0.9024,TRUE)

the output here is 0.8666 but we can take it to 0.95 as we want only 0.05 probability of a passenger being bumped.

For n = 60
P(X <= 58)

This is computed as:
=binom.dist(58,60,0.9024,TRUE)

0.9842 is the output here.

For n = 61,

This is computed as:
=binom.dist(58,61,0.9024,TRUE)

The output here is 0.9448 < 0.95

Therefore n = 60 is the required number of tickets which should be sold here.


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