State the central limit theorem and use it to compute probabilities for the sample mean

What this dot point is asking

SCSA wants you to state the theorem precisely, recognise when it applies, and use it to find probabilities about the sample mean by standardising.

The statement

The remarkable part is the phrase regardless of shape. The population might be skewed, bimodal or discrete, yet the averaging in $\bar{X}$ smooths it toward a normal shape as $n$ increases. If the population is already normal, $\bar{X}$ is exactly normal for every $n$.

The mean and standard error of the sample mean

Two exact facts underpin the theorem and hold for any sample size. The mean of the sampling distribution equals the population mean, $\mu_{\bar{X}} = \mu$, so the sample mean is an unbiased estimator of the population mean. The variance of the sample mean is $\dfrac{\sigma^2}{n}$, so its standard deviation, called the standard error, is $\dfrac{\sigma}{\sqrt{n}}$. The $\sqrt{n}$ in the denominator is the engine of inference: quadrupling the sample size only halves the standard error, which is why precision improves slowly with sample size. These two results are exact; the central limit theorem adds the further, approximate claim that the shape becomes normal.

Why it matters

The theorem is what makes inference about a mean possible with normal-distribution tools. Without it we would need to know the exact population distribution; with it, the sample mean is approximately normal and we can use $z$-values and confidence intervals.

How large is large enough

The size of $n$ needed for the approximation to be good depends on how far the population is from normal. For a population that is already roughly symmetric and bell-shaped, even $n = 10$ or so gives a good approximation. For a strongly skewed or heavy-tailed population a larger sample is needed, and a common rule of thumb is that $n \ge 30$ is sufficient for most populations met in practice. If the population is exactly normal then $\bar{X}$ is exactly normal for every $n$, however small, so no approximation is involved at all. SCSA questions usually supply a sample size that is comfortably large, and they expect you to cite the central limit theorem as the justification for treating $\bar{X}$ as normal.

Distinguishing the three distributions

A frequent source of confusion is keeping three distributions apart. The population distribution describes a single observation $X$ and has mean $\mu$ and standard deviation $\sigma$; it can be any shape. The sampling distribution of the mean describes $\bar{X}$ over all possible samples of size $n$; it has mean $\mu$, standard deviation $\dfrac{\sigma}{\sqrt{n}}$, and is approximately normal by the central limit theorem. A single sample, finally, is just $n$ observed data values from which $\bar{x}$ is computed. When a question asks for a probability about the sample mean, you work in the sampling distribution and standardise with the standard error, never with $\sigma$ alone.

Standardising the sample mean

To find a probability about $\bar{X}$, convert to the standard normal:

The denominator is the standard error, not $\sigma$. Then probabilities such as $P(\bar{X} > c)$ become standard-normal probabilities.