# Thread: Function approaching standard normal distribution

1. ## 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!

2. 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.

3. 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?

4. If you want me to help, you will need to type in TeX.
YOU can learn by hitting QUOTE and copying from my work.

5. Will this do?

6. The mean of a BERNOULLI is p and it's variance is p(1-p) since n=1 for a BERNOULLI.

SO ${\bar X-\mu\over \sqrt{\sigma^2/n}} ={\hat P-p\over \sqrt{p(1-p)/n}}$

7. Haha, now I see it! Thanks alot!!!