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Math Help - Airline Report Problem

  1. #1
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    Airline Report Problem

    Hello everybody,
    I have solved the way my professor tought us in class but I'm getting a different answer. If somebody can help me I will really Appreciate.....

    An airline reports that it has been experiencing a 15% rate of no-shows on advanced reservations. Among 150 advanced reservations, find the probability that there will be fewer than 20 no-shows. Use the normal distribution to approximate the binomial distribution.

    Thanks....
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  2. #2
    Senior Member Twig's Avatar
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    hi

    hi

    The normal approximation

    Binomial distribution - Wikipedia, the free encyclopedia

     N(np, np(1-p)) (1)

     0.15 \cdot 150 = 22.5

    Now use the formula given, (1). I gotta go know, in a hurry, Iīll check in later. GL
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  3. #3
    Senior Member Twig's Avatar
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     np = 150 \cdot 0.15 = 22.5
     np(1-p) = 19.125

     \frac{Z-\mu}{\sigma} = \frac{30 - 22.5}{4.37} \approx 1.72

     P(Z\leq 1.72) \approx 0.9573

    To be honest, I have never done this before, but I hope itīs right, otherwise let me know!
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  4. #4
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    Quote Originally Posted by Twig View Post
     np = 150 \cdot 0.15 = 22.5
     np(1-p) = 19.125

     \frac{Z-\mu}{\sigma} = \frac{<b>30</b> - 22.5}{4.37} \approx 1.72

     P(Z\leq 1.72) \approx 0.9573

    To be honest, I have never done this before, but I hope itīs right, otherwise let me know!
    The answer does not match with the ones I have; I see you have used 30; I dont know how you got that? I tried to use 20 as given in the question, still it doesn't give the right answer.
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  5. #5
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    Quote Originally Posted by Allenge View Post
    Hello everybody,
    I have solved the way my professor tought us in class but I'm getting a different answer. If somebody can help me I will really Appreciate.....

    An airline reports that it has been experiencing a 15% rate of no-shows on advanced reservations. Among 150 advanced reservations, find the probability that there will be fewer than 20 no-shows. Use the normal distribution to approximate the binomial distribution.

    Thanks....
    Let X be the random variable number of no shows.

    X ~ Binomial(n = 150, p = 0.15).

    Calculate Pr(X < 20) using the normal approximation.

    Please say where you're stuck if you are still stuck.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    Let X be the random variable number of no shows.

    X ~ Binomial(n = 150, p = 0.15).

    Calculate Pr(X < 20) using the normal approximation.

    Please say where you're stuck if you are still stuck.
    Still stuck; tried to use the normal approximation but I'm not getting the right answer.

    p(x < 20) => p(z< {20-22.5}/4.37) = -0.5721

    p(z< -0.5721) = 0.2843

    This is what I get but it is not one of the answers....
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  7. #7
    Senior Member Twig's Avatar
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    Mr fantastic, where is the fault in my reasoning?

    thanks
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  8. #8
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    Any more Suggestions please...?
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  9. #9
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    Quote Originally Posted by Allenge View Post
    Still stuck; tried to use the normal approximation but I'm not getting the right answer.

    p(x < 20) => p(z< {20-22.5}/4.37) = -0.5721

    p(z< -0.5721) = 0.2843

    This is what I get but it is not one of the answers....
    Did you use the continuity correction?

    Quote Originally Posted by Twig View Post
    Mr fantastic, where is the fault in my reasoning?

    thanks
    Quote Originally Posted by Twig View Post
     np = 150 \cdot 0.15 = 22.5

     np(1-p) = 19.125

     \frac{Z-\mu}{\sigma} = \frac{{\color{red}30} - 22.5}{4.37} \approx 1.72

     P(Z\leq 1.72) \approx 0.9573

    To be honest, I have never done this before, but I hope itīs right, otherwise let me know!
    Where did the 30 come from?
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