1. ## Normal Approximation confusion

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.

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

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

$k=2: {10 \choose 2}p^2 q^8 = 0.3020$
... and so on.

The normal approximation is given as:

$K=0: 0.0904, 0.2307, 0.3154$...... 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: $a_k = h*{ 1 \over {\sqrt{2 \pi}}} e^{{-0.5(kh)}^2}$

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

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

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.

2. 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.

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

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

$k=2: {10 \choose 2}p^2 q^8 = 0.3020$
... and so on.

The normal approximation is given as:

$K=0: 0.0904, 0.2307, 0.3154$...... 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: $a_k = h*{ 1 \over {\sqrt{2 \pi}}} e^{{-0.5(kh)}^2}$

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

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

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.
Doing it this way you should have:

$
a_0=\frac{1}{\sqrt{2\pi}\sqrt{10\times 0.2 \times 0.8}}e^{-\frac{2}{1.6}}\approx 0.0904
$

But to do this correctly you really need:

$
p(0)\approx P(0.5;np,npq)-P(-0.5;np,npq)
$

where $P(x;\mu,\sigma^2)$ is the cumulative normal distribution with mean $\mu$ and variance $\sigma^2$.

RonL

3. much appreciated.

But what is the formula you applied to get:

$e^{-2 \over {1.6}}$

How does K=0 come in?

4. Originally Posted by chopet
much appreciated.

But what is the formula you applied to get:

$e^{-2 \over {1.6}}$

How does K=0 come in?
The exponetial is of -(k-np)^2/(2npq), put k=0, n=10, p=0.2 and q=0.8

RonL