Results 1 to 2 of 2

Math Help - Probability/proportion

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    12

    Probability/proportion

    Hey guys, here's a question that is confusing me:

    An airline knows that some people who make reservations do not show for their flight. So, for example, it sells 52 seats for a 50 seat flight. If, on average, 4% of those who make reservations are no-shows, and we assume independence for the n=52 trials,
    (a) for what proportion of flights overbooked this way will there not be enought seats?
    (b) why is the assumption of independence not realistic?
    (c) why is the proportion of all 50 seat flights not having enought seats les than the proportion computed in part (a)?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by kelli_rie View Post
    Hey guys, here's a question that is confusing me:

    An airline knows that some people who make reservations do not show for their flight. So, for example, it sells 52 seats for a 50 seat flight. If, on average, 4% of those who make reservations are no-shows, and we assume independence for the n=52 trials,
    (a) for what proportion of flights overbooked this way will there not be enought seats?
    (b) why is the assumption of independence not realistic?
    (c) why is the proportion of all 50 seat flights not having enought seats les than the proportion computed in part (a)?

    Thanks!
    Hi kelli_ri,

    a) Let X be the number of no-shows. If we assume independence of events, then X has a Binomial distribution with n = 52 and p = 0.04. A flight will be overbooked if X=0 or X=1, and then
    P(X = 0) = \binom{52}{0} p^0 (1-p)^{52}
    P(X = 1) = \binom{52}{1} p^1 (1-p)^{51}
    P(\text{overbooking}) = P(X = 0) + P(X = 1).
    I'll let you do the computation.

    b) There are many common factors that might influence multiple passengers. For example, if the weather is bad then many people might not make it to the airport on time.

    c) Whoever wrote this question believes something like bad weather is likely, so several passengers might not show up. However, it is by no means certain that the actual probability is less than that calculated in part a). Lack of independence can also make it more likely that people will show up, increasing the probability of overbooking. For example, suppose the weather is unusually good for a change. Then you have the reverse effect of bad weather and people are more likely to make it to the airport on time. So lack of independence can increase or decrease the probability of overbooking, and without additional evidence we can't really say which is more likely.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Inverse proportion/Direct proportion question
    Posted in the Statistics Forum
    Replies: 3
    Last Post: September 26th 2011, 08:22 AM
  2. Replies: 3
    Last Post: July 6th 2009, 08:25 PM
  3. Data Analysis -- Proportion Problem and Probability
    Posted in the Advanced Statistics Forum
    Replies: 11
    Last Post: April 24th 2009, 07:34 AM
  4. Probability Distributions???Grades and Proportion
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: June 26th 2008, 08:28 PM
  5. Proportion
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 4th 2006, 06:42 AM

Search Tags


/mathhelpforum @mathhelpforum