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I have the following questions as homework where I have to decide whether they are true or false. a) The standard deviation of the distribution of the sum of independent random variables is also called the central limit theorem. My guess for this question is that it should be true as the variance is the sum of all the variances of the number of variables involved and thus the standard deviation is the square root of the variance.

b) When a great many n random variables are drawn from a population that is normally distributed, the distribution of the sample mean will be normal regardless of the sample size n. For this part I don't really know how to work it out and decide if its true or false.

Can someone help me? Thank you.

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  • $\begingroup$ Where are you stuck? What kind of help you need? What have you tried already? $\endgroup$
    – Tim
    Nov 15, 2020 at 20:55
  • $\begingroup$ For the first part I don't know if the logic I followed is correct meaning with the variance and covariance implication I wrote above. For the second part I dont really know how to interpret the question $\endgroup$
    – mathslover
    Nov 15, 2020 at 20:57

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Strictly speaking the Central Limit Theorem states that means $\bar X_n$ of large samples from any population, in which the variance $\sigma^2$ exists, has approximately a normal distribution for large $n,$ and that the limiting distribution as $n \rightarrow \infty$ is exactly normal.

To be specific about the approximate distribution of $\bar X_n,$ one often says $X_n \stackrel{aprx}{\sim}\mathsf{Norm}(\mu, \sigma/\sqrt{n}),$ where $\mu$ is the population mean and $\sigma$ is the population standard deviation.

(a) It is true that $SD(\bar X_n) = \sigma/\sqrt{n},$ for all $n = 1,2,3,\dots.$ This fact may be useful in computations for sufficiently large $n$ to obtain results about a nearly-normal $\bar X_n,$ but it is not a limit theorem ("Central" or otherwise).

(b) The statement is true, but if one is sampling from a normal population with mean $\mu$ and standard deviation $\sigma,$ then the exact distribution of $\bar X_n$ is $\mathsf{Norm}(\mu, \sigma/\sqrt{n}).$ Again here, you have a true statement that is not a limit theorem.

The CLT has to do with the behavior of $\bar X_n$ for a sample from a non-normal distribution as $n$ becomes infinite. I believe the point of the question is to get you to focus on the fact that the CLT is a limit theorem.

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  • $\begingroup$ thank you very much so it turns out that part a is false and b is true right? $\endgroup$
    – mathslover
    Nov 15, 2020 at 21:54
  • $\begingroup$ Both statements are true. Neither one is equivalent to the CLT. $\endgroup$
    – BruceET
    Nov 15, 2020 at 22:02
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    $\begingroup$ Thank you very much sir. $\endgroup$
    – mathslover
    Nov 15, 2020 at 22:13

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