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Thread: Normal Approximation: what happens to the 2pi?

  1. #1
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    Normal Approximation: Please check my reasoning

    After going through all the painstaking proofs, I have finally come to this step:

    $\displaystyle p_k= {n \choose k} p^k q^{n-k} \sim {1 \over { \sqrt{2 \pi npq}}} e^{ -0.5 ({ {k-np} \over {\sqrt {npq}} })^2 }$

    If we denote z as $\displaystyle { {k-np} \over {\sqrt {npq}} }$, we can scale the above formula to the standard normal curve: $\displaystyle {1 \over { \sqrt{2 \pi npq}}} e^{ -0.5 z^2 }$

    When we calculate the area under 2 points on the z-axis, say, A to B, we do :

    $\displaystyle {1 \over { \sqrt{2 \pi npq}}} \sum_{z=A}^{B} e^{ -0.5 z^2 } $

    But why is the integral given as:
    $\displaystyle {1 \over { \sqrt{2 \pi}}} \int_{A}^{B} e^{ -0.5 z^2 } dz$

    What happens to the "npq" term?
    Please help to alleviate my doubt if you can. Thanks!
    Last edited by chopet; Dec 9th 2007 at 10:54 PM.
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  2. #2
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    Quote Originally Posted by chopet View Post
    After going through all the painstaking proofs, I have finally come to this step:

    $\displaystyle p_k= {n \choose k} p^k q^{n-k} \sim {1 \over { \sqrt{2 \pi npq}}} e^{ -0.5 ({ {k-np} \over {\sqrt {npq}} })^2 }$

    If we denote z as $\displaystyle { {k-np} \over {\sqrt {npq}} }$, we can scale the above formula to the standard normal curve: $\displaystyle {1 \over { \sqrt{2 \pi npq}}} e^{ -0.5 z^2 }$

    When we calculate the area under 2 points on the z-axis, say, A to B, we do :

    $\displaystyle {1 \over { \sqrt{2 \pi npq}}} \sum_{z=A}^{B} e^{ -0.5 z^2 } $

    But why is the integral given as:
    $\displaystyle {1 \over { \sqrt{2 \pi}}} \int_{A}^{B} e^{ -0.5 z^2 } dz$

    What happens to the "npq" term?
    Please help to alleviate my doubt if you can. Thanks!
    It disappears when you convert $\displaystyle dn$ to $\displaystyle dz$ in the integral.

    That is you approximate the sum over the relevant range of $\displaystyle n$ by an integral over a range of $\displaystyle n$, then you change the variable from $\displaystyle n$ to $\displaystyle z$.

    ZB
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  3. #3
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    Thank you for normalising my day!
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