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Thread: 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:

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

    so I would have:

    $\displaystyle P(Z\leq 0.90685) = 0.8186$

    but the solution in the back of the book gives 0.8980 which would be $\displaystyle P(Z\leq 1.27) = 0.8980$
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  2. #2
    Grand Panjandrum
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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:

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

    so I would have:

    $\displaystyle P(Z\leq 0.90685) = 0.8186$

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

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

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

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

    CB
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