SOLUTION: A small feeder airline knows that the probability is .10 that a reservation holder will not show up
for its daily 7:15 A.M. flight into a hub airport. The flight carries 10 passen
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Question 404719: A small feeder airline knows that the probability is .10 that a reservation holder will not show up
for its daily 7:15 A.M. flight into a hub airport. The flight carries 10 passengers. (a) If the flight is
fully booked, what is the probability that all those with reservations will show up? (b) If the airline overbooks by selling 11 seats, what is the probability that no one will have to be bumped? (c) That
more than one passenger will be bumped? *(d) The airline wants to overbook the flight by enough
seats to ensure a 95 percent chance that the flight will be full, even if some passengers may be
bumped. How many seats would it sell?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A small feeder airline knows that the probability is .10 that a reservation holder will not show up for its daily 7:15 A.M. flight into a hub airport.
The flight carries 10 passengers.
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Binomial Problem with n = 10, p(not show) = 0.1 ; p(show) = 0.9
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(a) If the flight is fully booked, what is the probability that all those with reservations will show up?
:::(0.9)^10 = 0.3487
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(b) If the airline overbooks by selling 11 seats, what is the probability that no one will have to be bumped?
P(no bump) = P(0<= show <=10) = 1 - binomcdf(11,0.9,10) = 0.3138
--------------------
(c) That more than one passenger will be bumped?
P(2<= bumped <=10) = 1 - binomcdf(10,0.1,1) = 0.2639
----------------------
Cheers,
Stan H.
==========
(d) The airline wants to overbook the flight by enough
seats to ensure a 95 percent chance that the flight will be full, even if some passengers may be
bumped. How many seats would it sell?
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