The same idea works for averages. Take many random samples, find each sample's mean $\bar{x}$, and look at how those means vary. The answer is one of the most important facts in statistics, the central limit theorem, and it parallels the sampling distribution of a proportion.
Key idea
For random samples of size $n$ from a population with mean $\mu$ and SD $\sigma$:
$$\mu_{\bar{x}} = \mu, \qquad \sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}}.$$
Central limit theorem: when $n$ is large, the sampling distribution of $\bar{x}$ is approximately normal, whatever the shape of the population. (If the population itself is normal, $\bar{x}$ is normal for every n.)
Why it works
The sample total is a sum of $n$ independent values, so its mean is $n\mu$ and, because variances add, its variance is $n\sigma^2$. Dividing the total by $n$ gives $\bar{x}$: mean $\mu$ and variance $n\sigma^2/n^2 = \sigma^2/n$, so SD $\sigma/\sqrt{n}$. The normal shape is the deep part: when many independent pieces are averaged, high and low values offset each other, and the lopsidedness of any one value washes out. How large n must be depends on the skew; $n \ge 30$ is a common rule of thumb for moderately skewed populations.
Worked example: Center and spread
Commute times in a city are skewed right with mean 30 minutes and SD 12 minutes. For random samples of 36 workers, describe the sampling distribution of $\bar{x}$.
- Center: $\mu_{\bar{x}} = 30$ minutes.
- Spread: $\sigma_{\bar{x}} = \dfrac{12}{\sqrt{36}} = 2$ minutes.
- Shape: $n = 36$ is fairly large, so by the CLT $\bar{x}$ is approximately normal.
Answer: About normal, mean 30, SD 2.
Worked example: A probability about the mean
Find the probability that the mean commute of 36 workers is more than 33 minutes. ($P(Z < 1.50) = 0.9332$.)
- $z = \dfrac{33 - 30}{2} = 1.5$.
- $$P(\bar{x} > 33) = 1 - 0.9332 = 0.0668$$
- We could not do this for a single worker: one commute is skewed, not normal. The CLT applies to the mean.
Answer: About $0.067$.
Worked example: Too small a sample
Could you use the normal model for the mean of just 4 commutes?
- With n = 4 from a strongly skewed population, $\bar{x}$ is still noticeably skewed.
- Its mean (30) and SD ($12/2 = 6$) are still correct, but normal-curve probabilities would be off.
Answer: No: the shape is not yet close to normal.
Common mistake
Thinking the central limit theorem makes the data normal. A large sample from a skewed population looks skewed, like the population. The theorem is about the distribution of $\bar{x}$ over many samples. Check: are you describing individual values (population shape) or sample means (approximately normal for large n)?
The econometrics probability chapter lets you watch averages of lopsided data pile up into a bell, and builds every regression test on this theorem.