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Math Help - 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:

    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 { {k-np} \over {\sqrt {npq}} }, we can scale the above formula to the standard normal curve: {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 :

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

    But why is the integral given as:
    {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; December 9th 2007 at 11:54 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:

    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 { {k-np} \over {\sqrt {npq}} }, we can scale the above formula to the standard normal curve: {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 :

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

    But why is the integral given as:
    {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 dn to dz in the integral.

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

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