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An air flight can carry 120 passengers. A passenger with a reserved seat arrives

ID: 2923293 • Letter: A

Question

An air flight can carry 120 passengers. A passenger with a reserved seat arrives for the flight with probability 0.95. assume that the passengers behave independently.( use of computer software is expected-excel).

a) what is the minimum no of seats the airline should reserve for the probability of a full flight to be at least 0.90?

b) what is the maximum no of seats the airline should reserve for the probability that more passengers arrive than the flight can seat to be less than 0.10?

c) discuss some reasonable policies the airline could use to reserve seats based on these probabilities

Explanation / Answer

solution=

a) The claim is to find the minimum of seats the airline should reserve for the probability of a full flight to be at least 0.90.

Then, P(X120) = 1 - P(X < 120) 0.90 120 120 0 Then , for P(X = 120) = ( 120 (0.95) (1 0.95) = 0.0021 Similarly n P(X120) 121 0.0149 122 0.0534 123 0.1317 124 0.2521 125 0.4015 126 0.557 127 0.697 128 0.8081 129 0.8872 130 0.9381

Then for n = 130, the probability of a full flight to be at least 0.90

b) the claim is to find the maximum number of seats the airline should reserve for the probability that more passengers arrive than the flight can seat to be less than 0.10

Then, P(X120) < 0.10 n P(X120) 120 1 121 0.998 122 0.9858 123 0.9486 124 0.8723 125 0.7541 126 0.6063 127 0.4511 128 0.3104 129 0.1978 130 0.1171 131 0.0646

Then for n = 131, the probability that more passengers arrive than the flight can seat to be less than 0.10

c) the claim is to discuss some reasonable policies the airline could use to reserve seats based on these probabilities.

The airline should take a reserved number of seats that is between the number of passengers in (a) and in (b), such that both probabilities can be as small as possible.

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