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Math Help - Binomial to Normal Approximation

  1. #1
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    Binomial to Normal Approximation

    Assume an airline manages to sell 160 seats for a plane that is only capable of seating 155. Assume that 0.05 of ticket holders cancel their reservation. What is the probability that the place will be overbooked?

    I found these http://www.mathhelpforum.com/math-he...obability.html and http://www.mathhelpforum.com/math-he...rs-flight.html that deal with the exact same question, although they focus on solving the problem through the binomial method, which I have no trouble doing, nevertheless I would like to know how to solve it using the Normal Approximation. I am not sure what variables I have to use to get the result...

    I would think that my variance would be npq = 160*0.05*0.95=7.6

    then I would have:

    P\left(\frac{np-nr}{\sqrt{npq}}= \frac{8-7.75}{\sqrt{7.6}} = 0.90685\right)

    so I would have:

    P(Z\leq 0.90685) = 0.8186

    but the solution in the back of the book gives 0.8980 which would be P(Z\leq 1.27) = 0.8980
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  2. #2
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    Quote Originally Posted by lllll View Post
    Assume an airline manages to sell 160 seats for a plane that is only capable of seating 155. Assume that 0.05 of ticket holders cancel their reservation. What is the probability that the place will be overbooked?

    I found these http://www.mathhelpforum.com/math-he...obability.html and http://www.mathhelpforum.com/math-he...rs-flight.html that deal with the exact same question, although they focus on solving the problem through the binomial method, which I have no trouble doing, nevertheless I would like to know how to solve it using the Normal Approximation. I am not sure what variables I have to use to get the result...

    I would think that my variance would be npq = 160*0.05*0.95=7.6

    then I would have:

    P\left(\frac{np-nr}{\sqrt{npq}}= \frac{8-7.75}{\sqrt{7.6}} = 0.90685\right)

    so I would have:

    P(Z\leq 0.90685) = 0.8186

    but the solution in the back of the book gives 0.8980 which would be P(Z\leq 1.27) = 0.8980
    The mean number of passengers that arrive is 152, the variance of this is 7.6 and so the standard deviation is \approx 2.757.

    Now to be overbooked we need to calculate the probability that more than 155 passengers turn up. The z-score corresponding to this is:

    z=\frac{155.5-152}{2.757}\approx 1.269

    We use 155.5 in this calculation as a continuity correction, 154.5 to 155.5 passengers round to 155 when we are forced to an integer.

    CB
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