Originally Posted by

**chopet** There is an example on the normal approximation for binomial distribution.

Given n =10, p = 0.2.

We calculate binomial distribution from k=0 to k=6.

$\displaystyle k=0: {10 \choose 0}p^0 q^10 = 0.1074$

$\displaystyle k=1: {10 \choose 1}p^1 q^9 = 0.2684$

$\displaystyle k=2: {10 \choose 2}p^2 q^8 = 0.3020$

... and so on.

The normal approximation is given as:

$\displaystyle K=0: 0.0904, 0.2307, 0.3154$...... and so on.

Ok, I've been trying to figure out how to arrive at normal approximations given above, but to no avail.

Do I use:$\displaystyle a_k = h*{ 1 \over {\sqrt{2 \pi}}} e^{{-0.5(kh)}^2}$

where h=$\displaystyle {1 \over {\sqrt{2 \pi npq}}} $

For example,

$\displaystyle a_0 = { 1 \over {\sqrt{2 \pi}}} {1 \over {\sqrt{2 \pi 10*0.2*0.8}}} = 0.1258$

which is different from the answer given.

I thank you for your comments. This is a particularly vexing problem, as I want to thoroughly understand the inner workigs of Normal Approximations.