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Math Help - [SOLVED] Normal approximation to binomial

  1. #1
    Super Member Quacky's Avatar
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    [SOLVED] Normal approximation to binomial

    I've tried so many things but I never get the correct answer. Here's an example:

    For Y \sim B(20,0.4) Find P(3<Y<13) using a normal approximation

    This is what I do:

    \mu=np=(20)(0.4)=8
    \sigma^2=np(1-p)=20(0.4)(0.6)=4.8

    But at this point, I get in a mess.

    The correct answer is 0.9596

    Thanks in advance for any replies.
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  2. #2
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    Quote Originally Posted by Quacky View Post
    I've tried so many things but I never get the correct answer. Here's an example:

    For Y \sim B(20,0.4) Find P(3<Y<13) using a normal approximation

    This is what I do:

    \mu=np=(20)(0.4)=8
    \sigma^2=np(1-p)=20(0.4)(0.6)=4.8

    But at this point, I get in a mess.

    The correct answer is 0.9596

    Thanks in advance for any replies.
    Remember \sigma^2=np(1-p)=20(0.4)(0.6)=4.8 \implies \sigma = \sqrt{4.8}\approx 2.2


    Now employ z = \frac{Y-\mu}{\sigma}

    So P(3<Y<13) = P\left(\frac{3-8}{2.2}<z<\frac{13-8}{2.2}\right)

    Can you contiune?
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  3. #3
    Super Member Quacky's Avatar
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    Thanks for the fast reply, but seemingly not

    =P(-2.282<z<2.282)
    p(-2.282<z)
    =p(z>2.282)
    =0.0113

    p(z<2.282)
    =0.9887

    0.9887-0.113=0.9774
    Which is incorrect. Am I using the tables incorrectly? Or is there something wrong with my method?
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  4. #4
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    I would say P(-2.82<z<2.82) = 1-2P(z>2.82) = 1 - 2\times 0.0112 = 0.9776
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  5. #5
    Super Member Quacky's Avatar
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    0.9596 is the correct answer though, so I'm confused.
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  6. #6
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    All I can suggest is to use a better approximation for \sqrt{4.8}\approx 2.1909 . Otherwise the methodology seems good to me.
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  7. #7
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    I suggest you apply a "correction for continuity", so instead of finding
    P(3 < Y < 13),
    you find
    P(2.5 < Y < 13.5),
    where Y has a Normal distribution with mean 8 and standard deviation 2.19.
    This leads to
    P(-2.51 < Z < 2.51)
    where Z has a Normal(0,1) distribution, yielding a probability of 0.988.

    (Yes, I know, that's not the answer in the book. Sometimes the book has the wrong answer.)
    Last edited by awkward; March 10th 2010 at 02:04 PM. Reason: correction
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  8. #8
    Super Member Quacky's Avatar
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    Hm. Seeing as we're all getting around that, I suppose it must be right. Thanks again.
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