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Math Help - [SOLVED] Airline Passenger Seats/Overbooking Probability

  1. #1
    achibaby1974
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    Unhappy [SOLVED] Airline Passenger Seats/Overbooking Probability

    Here's the problem
    Ok. I've typed the problem out and answered it to the best I can.
    Please help. Any feedback at all will help!

    There's a plane with 213 seats. Because some people with reservations don’t show up, the company overbooks the plane by accepting more passenger reservations than seats, in order to increase revenue.

    It is assumed there is a probability of 0.0995 that a passenger with a reservation will not show up for the flight.

    It is also assumed that the airline booked 236 reservations for the 213 seated airplane.

    1.)

    Find the probability that when 236 reservations are accepted, more passengers show up than the amount of seats available. In other words, find the probability of more than 213 people showing up with reservations, when 236 reservations were accepted.

    My work:
    Given:

    n = 236 - # of reservations made for the plane
    r = 213 - # of plane seats available
    p = .0995 – probability that a passenger wit ha reservation will NOT show up
    q = .9005 - probability that a passenger with a reservation will show up.

    P(more than 213 passengers show up)
    Since the probability deals with “more than, >” and not an exact value, solving the probability requires a cut off mark and cumulative adding. This accumulated sum must then be subtracted by 1. Thus, I will use
    binomcdf (n, q, r)

    P (more than 213 passengers show up) = 1 – binomcdf (236, .9005, 213)
    1 - binomcdf (236, .9005, 213)
    1 - .5735581699 = .4264418301 ≈ .4264


    IS this right? Did I use the right probability? Does it make sense? It doesn't seem tight at all to me.

    Follow-up:

    Is the probability of overbooking small enough so that it doesn’t happen very often, or does it seem too high so that changes must be made to make it lower?

    My work:
    It seems fine, because we assume that anything lower than .05 is unusual. .4264 > .05, not less than. So, the probability is not too small.

    Using trial and error, find the maximum number of reservations that could be accepted so that the probability of having more passengers than seats is 0.05or less.

    My work:
    This is why my answer didn't make sense for the probability question.
    In order to do this, do I use 1-binomcdf or just binomcdf? I'm really confused. Do I do 1-binomcdf(237...240...245..... or just do binomcdf(237...240...245) to manipulate it to be more than .05.


    PLEASE HELP!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by achibaby1974 View Post
    Here's the problem
    Ok. I've typed the problem out and answered it to the best I can.
    Please help. Any feedback at all will help!

    There's a plane with 213 seats. Because some people with reservations don’t show up, the company overbooks the plane by accepting more passenger reservations than seats, in order to increase revenue.

    It is assumed there is a probability of 0.0995 that a passenger with a reservation will not show up for the flight.

    It is also assumed that the airline booked 236 reservations for the 213 seated airplane.

    1.)

    Find the probability that when 236 reservations are accepted, more passengers show up than the amount of seats available. In other words, find the probability of more than 213 people showing up with reservations, when 236 reservations were accepted.

    My work:
    Given:

    n = 236 - # of reservations made for the plane
    r = 213 - # of plane seats available
    p = .0995 – probability that a passenger wit ha reservation will NOT show up
    q = .9005 - probability that a passenger with a reservation will show up.

    P(more than 213 passengers show up)
    Since the probability deals with “more than, >” and not an exact value, solving the probability requires a cut off mark and cumulative adding. This accumulated sum must then be subtracted by 1. Thus, I will use
    binomcdf (n, q, r)

    P (more than 213 passengers show up) = 1 – binomcdf (236, .9005, 213)
    1 - binomcdf (236, .9005, 213)
    1 - .5735581699 = .4264418301 ≈ .4264


    IS this right? Did I use the right probability? Does it make sense? It doesn't seem tight at all to me.

    Follow-up:

    Is the probability of overbooking small enough so that it doesn’t happen very often, or does it seem too high so that changes must be made to make it lower?

    My work:
    It seems fine, because we assume that anything lower than .05 is unusual. .4264 > .05, not less than. So, the probability is not too small.

    Using trial and error, find the maximum number of reservations that could be accepted so that the probability of having more passengers than seats is 0.05or less.

    My work:
    This is why my answer didn't make sense for the probability question.
    In order to do this, do I use 1-binomcdf or just binomcdf? I'm really confused. Do I do 1-binomcdf(237...240...245..... or just do binomcdf(237...240...245) to manipulate it to be more than .05.


    PLEASE HELP!
    The first part is OK. except you should have biniomcdf(213, 0.9005, 236), that is the cumulative probability that 213 or fewer
    turn up when 236 tickets are sold, but you may just be using a different convention fro the order of argumets here


    For the second part you want to find the smallest N such that:

    1 – binomcdf (213, .9005, N) <= 0.05

    assuming that you have the arguments for binomcdf in the right order.

    RonL
    Last edited by CaptainBlack; April 16th 2008 at 11:48 PM. Reason: changed order of arguments in binomcdf
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  3. #3
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    In the second part you need to know the maximum number m ( 213 < m < 236) for which (1-binomial(m,p,213)) <= 0.05, i.e. the probability that more than 213 people will turn up is small.

    Sarim.
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  4. #4
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    I have a question for those of you who are acquainted with current educational practices. I went to school back in the stone age, before computers or even electronic calculators were commonly available. (I actually knew, and still know, how to use a slide rule.) So back in my day, when men were men etc., this problem would have been solved using a normal approximation to the binomial. I haven't bothered to work it out, but I bet the answer thus obtained would be very close to the answer obtained using more exact methods.

    So the question I have is this: Has the use of computers and programmable calculators in the schools, especially high schools, become so ubiquitous that the normal approximation to the binomial is obsolete for problems like this? Or are students still taught how to apply the normal approximation?

    Just curious,
    jw
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  5. #5
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    as far as I know every student has a sophisticated calculator now, so they just feed it in, the program therein, is just a loop that can sum up any discrete distribution,

    So, easy life, you may think,
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by awkward View Post
    I have a question for those of you who are acquainted with current educational practices. I went to school back in the stone age, before computers or even electronic calculators were commonly available. (I actually knew, and still know, how to use a slide rule.) So back in my day, when men were men etc., this problem would have been solved using a normal approximation to the binomial. I haven't bothered to work it out, but I bet the answer thus obtained would be very close to the answer obtained using more exact methods.

    So the question I have is this: Has the use of computers and programmable calculators in the schools, especially high schools, become so ubiquitous that the normal approximation to the binomial is obsolete for problems like this? Or are students still taught how to apply the normal approximation?

    Just curious,
    jw
    Most of the time reality falls short of our ability to calculate, and the normal
    approximation is more than adequate as the model does not justify greater
    precision (and I can do the calculations in my head). But the minute you
    have to look up a p value you might as well used the binomial itself as it
    takes no more effort than the normal approximation.

    (I still have my slide rule and remember how to use it)

    RonL
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