SOLUTION: Suppose that the probability that a passenger will miss a flight is 0.0999. Airlines do not like flights with empty seats, but it is also not desirable to have overbooked flight
Algebra ->
Probability-and-statistics
-> SOLUTION: Suppose that the probability that a passenger will miss a flight is 0.0999. Airlines do not like flights with empty seats, but it is also not desirable to have overbooked flight
Log On
Question 1204520: Suppose that the probability that a passenger will miss a flight is 0.0999. 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 57 passengers.
(a) If 59 tickets are sold, what is the probability that 58 or 59 passengers show up for the flight resulting in an overbooked flight?
(b) Suppose that 63 tickets are sold. What is the probability that a passenger will have to be "bumped"?
(c) For a plane with seating capacity of 250 passengers, what is the largest number of tickets that can be sold to keep the probability of a passenger being "bumped" below 1%? Answer by ikleyn(52834) (Show Source):
You can put this solution on YOUR website! .
Suppose that the probability that a passenger will miss a flight is 0.0999.
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 57 passengers.
(a) If 59 tickets are sold, what is the probability that 58 or 59 passengers show up
for the flight resulting in an overbooked flight?
(b) Suppose that 63 tickets are sold. What is the probability that a passenger
will have to be "bumped"?
(c) For a plane with seating capacity of 250 passengers, what is the largest number of tickets
that can be sold to keep the probability of a passenger being "bumped" below 1%?
~~~~~~~~~~~~~~~~~~~~~~~~~~~
Question (a)
It is a binomial experiment.
There are 59 potential passengers. The probability that every given passenger will show up is 1 - 0.0999 = 0.9001.
The events for individual passengers are INDEPENDENT.
The question is: find the probability that of 59 passengers that bought tickets, 58 or 59 will show up
(it is the event "the flight is overbooked").
n = 59 (the number of trials);
p = 0.9001 (the probability of "success" to each individual trial);
k >= 58 (the event of overbooking flight).
P(n=59; p=0.9001; k>=58) = P(n=59; p=0.9001; k=58) + P(n=59; p=0.9001; k=59) = + =
= = 0.01316 (rounded). ANSWERQuestion (b)
It is a binomial experiment very similar to part (a).
There are 63 potential passengers. The probability that every given passenger will show up is 1 - 0.0999 = 0.9001.
The events for individual passengers are INDEPENDENT.
The question is: find the probability that of 63 passengers that bought tickets, at least 58 will show up
(it is the event "the flight is overbooked").
n = 63 (the number of trials);
p = 0.9001 (the probability of "success" to each individual trial);
k >= 58 (the event of overbooking flight).
We want to find the probability P = P(n=63; p=0.9001; k>=58)
Go to web-site https://stattrek.com/online-calculator/binomial and use online (free of charge) calculator there.
Its interface is very simple, so even a beginner student can easily work with it.
It facilitates using many-addend formula, so calculations are really easy and fast.
The ANSWER is
P = P(n=63; p=0.9001; k>=58) = 0.3893.
Alternatively, you may use standard function binomcfd on your regular TI-83/84 calculator.
About this function and how to use it read from this web-page https://www.statology.org/binompdf-vs-binomcdf/
The regular calculator will provide the same answer.
Solved.
--------------------
My post is just very long, so I prefer to stop here.
