# Math Help - Normal Approximation: what happens to the 2pi?

1. ## 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?

2. Originally Posted by chopet
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?
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

3. Thank you for normalising my day!