Three distributions, and only one of them goes normal

Keep three objects separate:

1. The population distribution: the values of every individual. It has a fixed mean $\mu$ and standard deviation $\sigma$, and a shape that sampling cannot alter.
2. The distribution of one sample: the $n$ values actually collected. Its histogram resembles the population's shape, more closely as $n$ grows, and it has its own $\bar{x}$ and $s$.
3. The sampling distribution of a statistic: the values the statistic would take across all possible samples of size $n$. This one is not a distribution of individuals at all; it is a distribution of summaries.

The vocabulary that follows depends on knowing which is meant. "Larger samples have less variability" is false about the third object only if you get lucky with wording: a larger sample's own standard deviation $s$ estimates $\sigma$ and does not shrink, while the sampling distribution of $\bar{x}$ genuinely does narrow. And a skewed population sampled at $n = 100$ still produces a skewed-looking sample; it is the sampling distribution of $\bar{x}$ that goes normal.

For a simple random sample of size $n$ from a population with mean $\mu$ and standard deviation $\sigma$:

$$\mu_{\bar{x}} = \mu, \qquad \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}.$$

The first says $\bar{x}$ is unbiased: across all possible samples the sample means center on the population mean, with no systematic drift. The second says the spread of those sample means shrinks as $n$ grows, because an extreme individual gets diluted by the other $n - 1$ values.

Bottles are filled with mean 12.0 oz and standard deviation 0.4 oz. For samples of 16 bottles, $\sigma_{\bar{x}} = \frac{0.4}{\sqrt{16}} = 0.1$ oz. The square root is what governs the improvement: cutting $\sigma_{\bar{x}}$ in half takes four times the sample, so $n = 64$ gives $\frac{0.4}{8} = 0.05$ oz.

The formula assumes independent observations, which sampling without replacement violates. The 10% condition covers it: if the sample is no more than a tenth of the population, $\frac{\sigma}{\sqrt{n}}$ is a good approximation. Note also that $\sigma$ belongs to the population, so $\frac{\sigma}{\sqrt{n}}$ is the spread of the statistic, never of the data.

The Central Limit Theorem: for a sufficiently large sample size, the sampling distribution of $\bar{x}$ is approximately normal, regardless of the population's shape. The common working threshold is $n \ge 30$, though a strongly skewed population needs more and a nearly normal population needs far less. If the population is itself normal, $\bar{x}$ is exactly normal for every $n$.

Three things the theorem does not say, all of them common readings:

1. The population does not become normal. It is fixed. A skewed population stays skewed no matter how large the sample.
2. The data in the sample do not become normal. A large sample's histogram looks more like the skewed population, not less.
3. It says nothing about the center or spread. Those come from $\mu_{\bar{x}} = \mu$ and $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$, which hold at every $n$. The CLT supplies only the shape.

Its importance is practical: once the shape is known to be approximately normal, a probability about $\bar{x}$ becomes a normal calculation, which is the machinery every confidence interval and significance test in the rest of the course runs on.

A probability question about a sample mean uses the sampling distribution's own standard deviation:

$$z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}.$$

For the bottles ($\mu = 12.0$, $\sigma = 0.4$, $n = 16$), the probability a sample of 16 averages more than 12.2 oz is

$$z = \frac{12.2 - 12.0}{0.1} = 2.0, \qquad P(\bar{x} > 12.2) \approx 0.023.$$

The same 12.2 for a single bottle gives $z = \frac{12.2 - 12.0}{0.4} = 0.5$ and a probability of about 0.309. The two questions differ by more than a factor of ten, and the only difference in the arithmetic is whether $\sigma$ or $\frac{\sigma}{\sqrt{n}}$ went underneath. Read the question for whether it asks about one individual or about the average of $n$, and pick the standard deviation that matches.

The same structure returns in the next unit for a sample proportion, with $\mu_{\hat{p}} = p$ and $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$: a statistic, its center, its spread, and a normal shape once the sample is large enough.