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Math Help - Normal approximation to binomial distribution.

  1. #1
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    Normal approximation to binomial distribution.

    Hey, I'm stuck on a homework problem that I've been working on...

    Transportation officials tell us that 80% of drivers wear seat belts while driving. What is the probability of observing 518 or fewer drivers wearing seat belts in a sample of 700 drivers?

    Hint: Use normal distribution to approximate the binomial distribution

    What I have so far is

    xbar = 518
    μ = mean = (n * p) = (700 * .80) = 560
    σ = standard deviation = sqrt(n * p * q) = sqrt(700 * .80 * .20) = 10.583

    Any help will be greatly appreciated. Thanks
    Last edited by mr fantastic; June 21st 2010 at 04:19 PM. Reason: Re-titled.
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  2. #2
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    Quote Originally Posted by lilyungn View Post
    Hey, I'm stuck on a homework problem that I've been working on...

    Transportation officials tell us that 80% of drivers wear seat belts while driving. What is the probability of observing 518 or fewer drivers wearing seat belts in a sample of 700 drivers?

    Hint: Use normal distribution to approximate the binomial distribution

    What I have so far is

    xbar = 518
    μ = mean = (n * p) = (700 * .80) = 560
    σ = standard deviation = sqrt(n * p * q) = sqrt(700 * .80 * .20) = 10.583

    Any help will be greatly appreciated. Thanks
    Have you reviewed any examples in your class notes or textbook? Where are you stuck here?
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    How do I calculate the probability? I dont understand what it means by use normal distribution to approximate the binomial distribution
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    Quote Originally Posted by lilyungn View Post
    How do I calculate the probability? I dont understand what it means by use normal distribution to approximate the binomial distribution
    What do your classnotes and textbook say? Probably similar to what is said here, for example: http://courses.wcupa.edu/rbove/Beren...section6_5.pdf
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    When I do it in the calculator I get a binomial of 0.000072 ... is that the right answer?
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    Quote Originally Posted by lilyungn View Post
    When I do it in the calculator I get a binomial of 0.000072 ... is that the right answer?
    The answer is correct but how you got the answer is wrong. You have to use the normal approximation.

    In an exam, if the question prescribes the method but you do it a different way you will get zero marks.
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  7. #7
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    How can I use the normal approximation, I'm allowed to use a TI-83 in class.
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    Quote Originally Posted by lilyungn View Post
    How can I use the normal approximation, Mr F says: Did you read the link I posted earlier?

    I'm allowed to use a TI-83 in class. Mr F says: As I have already said, IF the question prescribes the method, then you have to do it that way. If the question does not precribe the method, then you can do it any way you please.
    ..
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    Is it possible you can show me showing the work how you use it?
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  10. #10
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    You want
    P(X \leq 518) where  X \sim Bi(700,0.8)

    We'll get an approximate answer by using another variable, Y
    Y \sim N(\mu,\sigma^2)

    You have already worked out that
    \mu = np =560
    \sigma^2 = np(1-p) = 112
    Do you understand why these values are used? We are making the mean and variance of the normal distribution equal to the mean and variance of the binomial distribution we want to approximate.


    Now:
    P(X \leq 518) \approx P(Y \leq 518.5) (continuity correction)
    P(Y \leq 518.5) = P \left( \frac{Y-560}{\sqrt{112}} < \frac{518.5-560}{\sqrt{112}} \right) (standardise)

    P(Y \leq 518.5) = \Phi \left( \frac{518.5-560}{\sqrt{112}} \right)
    P(Y \leq 518.5) = 0.000044
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