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Math Help - Function approaching standard normal distribution

  1. #1
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    Function approaching standard normal distribution

    I want to show that this function:

    (P^ - p) / (root)((p(1-p))/n)

    will approximate the standard normal distribution. Here P^ is an unbiased estimator for p, and p is the probability for an event. n is the number of trials.

    The first thing I notice is that this formula looks very similiar to the formula used in the central limit theorem, so I am guessing that there is a way to manipulate this one into that one. My textbook doesn't explain or proove these things, so any help to make me understand this will be very appreciated!
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  2. #2
    MHF Contributor matheagle's Avatar
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    It's the same as my previous post.

    \hat P={\sum_{i=1}^nY_i\over n} where the Y's are iid Bernoulli's with mean p.

    So \hat P is a sample mean.

    {\bar X-\mu\over \sigma/\sqrt{n}} approaches a st normal as n goes to infinity.
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    All right, so in that case I see that P^ - p will equal X_ - .

    But I can't get the (root)((p(1-p))/n) to quite equal (sigma)/(root)n. Let's see... the variance in a binomial distribution is np(1-p), so:

    (root)((p(1-p))/n) =

    (root)((np((1-p))/n^2) =

    (root)((np((1-p))) / (root)n^2 =

    (sigma) / n

    I seem to be missing a root. Maybe it will still be standard normally distributed?
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  4. #4
    MHF Contributor matheagle's Avatar
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    If you want me to help, you will need to type in TeX.
    YOU can learn by hitting QUOTE and copying from my work.
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  5. #5
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    Will this do?
    Attached Thumbnails Attached Thumbnails Function approaching standard normal distribution-111.jpg  
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  6. #6
    MHF Contributor matheagle's Avatar
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    The mean of a BERNOULLI is p and it's variance is p(1-p) since n=1 for a BERNOULLI.

    SO  {\bar X-\mu\over \sqrt{\sigma^2/n}} ={\hat P-p\over \sqrt{p(1-p)/n}}
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  7. #7
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    Haha, now I see it! Thanks alot!!!
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