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September 23rd, 2015, 09:01 AM   #1
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Normal approximation and Central Limit Theorem

Hi,

I have the following theorem of Central Limit Theorem:

$\displaystyle Z_n = \frac{\bar{X}_n - \mu}{\frac{\sigma}{\sqrt{n}}}$

I know that $\displaystyle \bar{X}_n $ is the mean of all n outcomes of each n trials:
$\displaystyle \bar{X}_n = \frac{X_1 + X_2 + ... + X_n}{n}$

I know also that $\displaystyle \mu$ is the mean or expectation of each single trial:
$\displaystyle \mu = E(X) = \frac{x_1 + x_2 + ... + x_n}{n}$

I know that sigma is the standard deviation but however, what I don't understand is the denominator in the theorem $\displaystyle \frac{\sigma}{\sqrt{n}}$, what it is? What is that square root of n?

thanks!
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September 23rd, 2015, 10:18 AM   #2
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is the standard deviation for each individual trial. is the standard deviation for the average of all n trials.
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