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Math Help - central limit theorem and n being "large enough"

  1. #1
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    central limit theorem and n being "large enough"

    Let X ~ Binom(n, pi). Then by the Central Limit Theorem, if n is large enough, X is approximately equal to Norm(pi, sqrt(npi(1 - pi))). I need to show that if pi > .5 and npi >= 10, then 3sqrt(pi(1-pi)/n) < pi.

    What i tried was rewriting n as n >= 10/pi, and then plugging this in to the left half of the expression. I got it reduced to 3sqrt((1-pi)/10) < 1, but I am not sure where to go from there.

    Or maybe there is some other way to do this that I am not thinking of.
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  2. #2
    MHF Contributor matheagle's Avatar
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    IS pi a probability?
    and the mean of a binomial is n times the probability.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ecc5 View Post
    Let X ~ Binom(n, pi). Then by the Central Limit Theorem, if n is large enough, X is approximately equal to Norm(pi, sqrt(npi(1 - pi))).
    The second parameter in N(m,\sigma^2) (which is the usual notation for the normal distribution) is usually the variance not the standard deviation.


    I need to show that if pi > .5 and npi >= 10, then 3sqrt(pi(1-pi)/n) < pi.

    What i tried was rewriting n as n >= 10/pi, and then plugging this in to the left half of the expression. I got it reduced to 3sqrt((1-pi)/10) < 1, but I am not sure where to go from there.

    Or maybe there is some other way to do this that I am not thinking of.
    So you want to show that:

    \dfrac{p(1-p)}{n}<\dfrac{p^2}{9}

    or:

    \dfrac{9p(1-p)}{np^2}<1

    So we start:

    \dfrac{9p(1-p)}{np^2}=\dfrac{9p(1-p)}{(np)p}\le \dfrac{9p(1-p)}{10p}=\dfrac{9(1-p)}{10}

    etc
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  5. #5
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    How did you get the first step?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by ecc5 View Post
    How did you get to this step?
    Since every thing is positive, ou square 3sqrt(p(1-p)/n) < p and for our purposes this is equivalent to what you wanted to prove.

    CB
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