Central limit theorem

For large enough samples, the sampling distribution of the mean is approximately normal with mean μ and standard error σ/√n, whatever the shape of the population.

\xbar \;\dot\sim\; N\Big(\mu, \frac{\sigma^2}{n}\Big) \quad \text{as } n \text{ grows}

Approximate sampling distribution of the mean

$\mu$: population mean
$\sigma$: population SD (must be finite)
$n$: sample size

\phat \;\dot\sim\; N\Big(p, \frac{p(1-p)}{n}\Big) \quad \text{when } np \ge 10,\; n(1-p) \ge 10

The same result for a proportion

$p$: population proportion

When to use

Justifies z and t procedures for means and proportions when the population is not normal, provided $n$ is large enough.

Assumptions

Independent observations from a population with finite variance. "Large enough" depends on shape: $n \approx 15$ –30 is often fine for moderately skewed data; strongly skewed or heavy-tailed populations (claims, incomes) can need hundreds. A normal population gives an exactly normal $\xbar$ for any $n$ .

How to interpret

The theorem is about the sampling distribution of the mean, not about the data. The data keep their shape however large $n$ is.
The spread shrinks like $1/\sqrt{n}$ while the shape becomes more normal.

Common mistakes

"With n ≥ 30 the data are normal." No: the mean is approximately normal.
Treating 30 as a universal threshold. Simulate or check the skewness.
Applying it to dependent data (time series, clustered samples) without adjustment.

Example

Waiting times are exponential with mean 4 and SD 4 minutes. For samples of $n = 36$ , $\xbar$ is approximately $N(4, 0.667^2)$ , so $P(\xbar > 5) \approx P(Z > 1.5) = 0.0668$ , even though individual waits are strongly skewed.