# Thread: central limit theorem and n being "large enough"

1. ## central limit theorem and n being "large enough"

Let X ~ Binom(n, pi). Then by the Central Limit Theorem, if n is large enough, X is approximately equal to Norm(pi, sqrt(npi(1 - pi))). I need to show that if pi > .5 and npi >= 10, then 3sqrt(pi(1-pi)/n) < pi.

What i tried was rewriting n as n >= 10/pi, and then plugging this in to the left half of the expression. I got it reduced to 3sqrt((1-pi)/10) < 1, but I am not sure where to go from there.

Or maybe there is some other way to do this that I am not thinking of.

2. IS pi a probability?
and the mean of a binomial is n times the probability.

3. Originally Posted by ecc5
Let X ~ Binom(n, pi). Then by the Central Limit Theorem, if n is large enough, X is approximately equal to Norm(pi, sqrt(npi(1 - pi))).
The second parameter in $\displaystyle N(m,\sigma^2)$ (which is the usual notation for the normal distribution) is usually the variance not the standard deviation.

I need to show that if pi > .5 and npi >= 10, then 3sqrt(pi(1-pi)/n) < pi.

What i tried was rewriting n as n >= 10/pi, and then plugging this in to the left half of the expression. I got it reduced to 3sqrt((1-pi)/10) < 1, but I am not sure where to go from there.

Or maybe there is some other way to do this that I am not thinking of.
So you want to show that:

$\displaystyle \dfrac{p(1-p)}{n}<\dfrac{p^2}{9}$

or:

$\displaystyle \dfrac{9p(1-p)}{np^2}<1$

So we start:

$\displaystyle \dfrac{9p(1-p)}{np^2}=\dfrac{9p(1-p)}{(np)p}\le \dfrac{9p(1-p)}{10p}=\dfrac{9(1-p)}{10}$

etc

4. How did you get to this step?

5. How did you get the first step?

6. Originally Posted by ecc5
How did you get to this step?
Since every thing is positive, ou square 3sqrt(p(1-p)/n) < p and for our purposes this is equivalent to what you wanted to prove.

CB