# Function approaching standard normal distribution

• May 1st 2010, 09:29 PM
gralla55
Function approaching standard normal distribution
I want to show that this function:

(P^ - p) / (root)((p(1-p))/n)

will approximate the standard normal distribution. Here P^ is an unbiased estimator for p, and p is the probability for an event. n is the number of trials.

The first thing I notice is that this formula looks very similiar to the formula used in the central limit theorem, so I am guessing that there is a way to manipulate this one into that one. My textbook doesn't explain or proove these things, so any help to make me understand this will be very appreciated!
• May 2nd 2010, 06:29 PM
matheagle
It's the same as my previous post.

$\hat P={\sum_{i=1}^nY_i\over n}$ where the Y's are iid Bernoulli's with mean p.

So $\hat P$ is a sample mean.

${\bar X-\mu\over \sigma/\sqrt{n}}$ approaches a st normal as n goes to infinity.
• May 2nd 2010, 06:58 PM
gralla55
All right, so in that case I see that P^ - p will equal X_ - µ.

But I can't get the (root)((p(1-p))/n) to quite equal (sigma)/(root)n. Let's see... the variance in a binomial distribution is np(1-p), so:

(root)((p(1-p))/n) =

(root)((np((1-p))/n^2) =

(root)((np((1-p))) / (root)n^2 =

(sigma) / n

I seem to be missing a root. Maybe it will still be standard normally distributed?
• May 2nd 2010, 07:04 PM
matheagle
If you want me to help, you will need to type in TeX.
YOU can learn by hitting QUOTE and copying from my work.
• May 2nd 2010, 07:12 PM
gralla55
Will this do?
• May 2nd 2010, 07:39 PM
matheagle
The mean of a BERNOULLI is p and it's variance is p(1-p) since n=1 for a BERNOULLI.

SO ${\bar X-\mu\over \sqrt{\sigma^2/n}} ={\hat P-p\over \sqrt{p(1-p)/n}}$
• May 2nd 2010, 07:52 PM
gralla55
Haha, now I see it! Thanks alot!!!