Originally Posted by

**chopet** 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!