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Trying to understand Central Limit Theorem via example.

My question is,

In a simple random sample of $1000$ physicists taken among all universities in a country, the number of papers published by the sampled physicists in the past year had a mean of $1.1$ and a standard deviation of $1.8$.

In this example, does the Central Limit Theorem say that the distribution of the number of papers published by the sampled faculty in the past year is roughly normal?

I know that the sample mean is close to the population mean and can be considered as the population mean. The sample size is also quite large, and I assume it's large enough to say the means are distributed roughly normal by CLT. However, the standard deviation is large. I know that on a normal the standard deviation are $+ - 1$, but i don't know how to use the standard deviation given.

If it's not normal then what shape is it?

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  • $\begingroup$ what does SD stand for? $\endgroup$ – Berci Mar 21 '17 at 23:45
  • $\begingroup$ standard deviation $\endgroup$ – Jayant.M Mar 22 '17 at 0:02
  • $\begingroup$ The size of the standard deviation does not really matter very much for the normal approximation, mostly because it is "scaled out" of the problem when you take the normal approximation. Specifically the normal approximation tells you that $\frac{\overline{X}-\mu}{\sigma \sqrt{n}}$ is approximately normally distributed with mean $0$ and variance $1$. In your example, $\mu=1.1,\sigma=1.8,n=1000$. $\endgroup$ – Ian Mar 22 '17 at 0:05
  • $\begingroup$ The standard deviation of a standard normal distributed is $\sigma=+1$. It´s a unique number. The following is not right:"...on a normal the standard deviation are +−1". $\endgroup$ – callculus Mar 22 '17 at 0:09
  • $\begingroup$ Since the number of papers cannot be negative, it seems "the number of papers" cannot have a "roughly normal" distribution. $\endgroup$ – GEdgar Mar 22 '17 at 11:37
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There's no telling what distribution the number of papers per person has. That is not what the theorem is about. Here's the point. Since each value in the data set is a random variable, the sample mean $\overline{X}_n$ is itself a random variable, with an approximately normal distribution provided the sample is large. More precisely, when the sample size $n$ goes to infinity, $$\sqrt{n}\,\frac{\overline{X}_n-\mu}\sigma\to N\left( 0,\,1\right)$$

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  • $\begingroup$ Please leave a comment if you downvote. I neutralized the downvote by upvoting. $\endgroup$ – callculus Mar 22 '17 at 0:14
  • $\begingroup$ I'm confused, if we suppose the number of papers published by the sampled faculty in the past year are i.i.d. and we let $S_n$ be the sum of all or i.i.d events, $X$, then the total number of papers published by the sampled faculty in the past year is roughly normal with mean = $1000 \cdot 1.1$ and standard deviation $\sqrt{1000} \cdot 1.8$. Is this not true by Central Limit Theorem? $\endgroup$ – Jayant.M Mar 22 '17 at 5:25
  • $\begingroup$ @callculus "Neutralizing" downvotes (or upvotes) by upvoting (or downvoting) is a bad idea, as common sense indicates and as has been discussed several times on meta. $\endgroup$ – Did Mar 22 '17 at 7:39
  • $\begingroup$ @Did Sorry, I haven´t known that. But what I know is that it has been discussed the topic downvoting without leaving a comment. Could it be changed or it is technical impossible ? $\endgroup$ – callculus Mar 22 '17 at 7:53
  • $\begingroup$ @callculus Yes it was discussed, and the suggestion reappears periodically, with each time the conclusion that it would be impractical. In fact applying it would be a sure way to destroy the SE vote system. (As an aside, let me ask you this: would you be adamant as well that upvoters should leave a comment explaining each of their upvotes? If not, why?) $\endgroup$ – Did Mar 22 '17 at 7:57

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