In the book Probability and Statistics for Engineering and the Sciences 8th edition by Devore, the Central Limit Theorem is the following:


This theorem says that $\mu_\bar X = \mu$ , and $\sigma_\bar X ^{2} = \sigma ^2 /n$. Right after that it says $\mu_ {T_0} = n\mu$, and $\sigma_ {T_0} ^{2} = n\sigma ^2$.

What does $T_0$ mean and why are there two different ways to find $\mu$ and $\sigma$? I was using this in a couple of problems and couldn't figure out which to use. Which situations would I use the first rather than the latter?


From their conclusion, I would assume that $T_0 := n\bar{X}$. Then \begin{align*} \mu_{T_0} &= \mathrm{E}[n\bar{X}] = n\mu \\ \sigma_{T_0}^2 &= \mathrm{Var}[n\bar{X}] = n^2 \frac{\sigma^2}{n} = n\sigma^2. \end{align*}

  • $\begingroup$ How would I know which one to use in a problem? Since they give completely different numbers. $\endgroup$
    – user185714
    Nov 18 '14 at 1:53
  • $\begingroup$ @user185714 It just depends on what statistic(s) you want to calculate. For example, if $X$ is a random variable that represents the proportion of something, then $nX$ would be the actual number of what you are measuring, if it's in a sample of size $n$. Thus, $\bar{X}$ would give information about the mean proportion, and $T_0$ would give information about the mean number. $\endgroup$
    – kokocijo
    Nov 18 '14 at 2:54

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