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Math Help - Normal Distribution approximation of binomial distribution problem

  1. #1
    Junior Member beebe's Avatar
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    Normal Distribution approximation of binomial distribution problem

    This problem is meant to be solved using the normal distribution. It is claimed that a bag M&M's contains 24% blue candies. Assuming that this is true, find the probability that at least 27 blue M&M's are in a given package containing 100 total M&M's.

    What I did:

    n=100, p=.24, q=.76

    \mu =np=(100)(.24)=24

    \sigma =\sqrt{npq}=\sqrt{(100)(.24)(.76)}=4.2708

    z = \frac{x-\mu }{\sigma}=\frac{27-24}{4.2708}=.7024

    On a "from the left" z-table I get .7580, so my probability is 1-.7580=.2420

    The answer given in the textbook is .2776, or technically .2748 if one uses a binomial distribution. Where did I go wrong? Thanks for taking the time to read this.
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  2. #2
    MHF Contributor
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    Re: Normal Distribution approximation of binomial distribution problem

    Hey beebe.

    Have you ever come across continuity correction?

    Continuity correction - Wikipedia, the free encyclopedia
    Thanks from beebe
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  3. #3
    Junior Member beebe's Avatar
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    Re: Normal Distribution approximation of binomial distribution problem

    Yes. I'm surprised I forgot about that. Thank you very much.
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