The Central Limit Theorem

Deede

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How does one calculate the sampling distribution of X when you have the random sample of size n, mean, and standard deviation. I would appreciate any help that I can get.
 
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How does one calculate the sampling distribution of X when you have the random sample of size n, mean, and standard deviation. I would appreciate any help that I can get.
Suppose you know the mean and the standard deviation of the "population" distribution - but not necessarily the shape of the distribution. The central limit theorem states that if you take samples of size n from the population, the distribution of sample means will approach a Normal distribution as n gets large. Further, the mean of the sample means will equal the population mean, and the standard deviation of the sample means will be the population standard deviation divided by \(\displaystyle \sqrt{n}\).

Population:................... \(\displaystyle \mu,\;\;\sigma \)

Mean of sample of size \(\displaystyle n\): \(\displaystyle \mu, \;\;\dfrac{\sigma}{\sqrt{n}}\)
 
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How does one calculate the sampling distribution of X when you have the random sample of size n, mean, and standard deviation. I would appreciate any help that I can get.



Thank you Dr. Phil. I understand it now.

"In three words I can sum up everything I've learned about life. It goes on." Robert Frost
 
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