Normal population, so X̄ is normal for any n

Shape, at last

L8.2 gave the centre and the spread of $\bar{X}$ but said nothing about its shape — and a probability question needs the shape, because the table is a normal table. There are two routes to a normal $\bar{X}$, and the course keeps them apart because the paper does. This lesson is the first route.

If the population is normal, the sample mean is normal — exactly, for every $n$. $\bar{X}$ is a linear combination of the $X_i$, and the normal family is closed under linear combination (L7.2). No approximation, no minimum sample size. With L8.2's centre and spread:

$$X\sim N(\mu ,{\sigma }^2)⟹\bar{X}\sim N\left(\mu ,\frac{{\sigma }^2}{n}\right)\text{ for any }n$$

In the course's notation the second argument is the variance ${\sigma }^2/n$; the SD you standardise with is $\sigma /\sqrt{n}$.

Worked example — nine invoices

Northfield Credit's invoice amounts are $X\sim N(120,15^2)$. A clerk pulls a sample of $n=9$ invoices. Then

$$\bar{X}\sim N\left(120,\frac{225}{9}\right)=N(120,25),\mathrm{SE}=\frac{15}{\sqrt{9}}=5$$

Nine is a small sample — far below any "30" — and the result is still exact, because the population is normal.

Probability the sample mean exceeds \$128. Standardise with the standard error:

$$z=\frac{128-120}{5}=1.60,P(\bar{X}>128)=1-\Phi (1.60)=1-0.9452=0.0548$$

About one sample of nine in eighteen averages above \$128. Compare a single invoice: $P(X>128)$ has $z=8/15=0.53$ and probability 0.2981 — more than five times likelier. One invoice has three times the spread of a mean of nine ($\sqrt{9}=3$), and a tail three times as far out in SE units is a very different tail.

Probability the sample mean lies within \$5 of the true mean. $P(115<\bar{X}<125)$: the limits are one SE either side of $\mu$, so $z=\pm 1.00$ and

$$P=\Phi (1.00)-\Phi (-1.00)=0.8413-0.1587=0.6826$$

The empirical rule's 68 % — for the sample mean, on a scale of SEs. The same calculation in Module 9 runs backwards: within how much of $\mu$ does $\bar{x}$ fall 95 % of the time? Answer: $\pm 1.96\times \mathrm{SE}$. That is a confidence interval, and this lesson is where it comes from.

The wrong SD

The error that loses the question is standardising $\bar{x}$ with $\sigma$ instead of $\sigma /\sqrt{n}$ — treating the sample mean as if it were one invoice. The signal is the word mean or average in the question: a question about an average of $n$ divides $\sigma$ by $\sqrt{n}$ before anything else happens. L8.5 makes this a three-way rule.

Why the course separates this from the CLT

IBS1 states the normality of $\bar{X}$ once, under the Central Limit Theorem's "$n\ge 30$" umbrella. The handouts — and the paper — distinguish two cases: the population is normal, so $\bar{X}$ is normal for any $n$ (this lesson); or the population is not normal, so $\bar{X}$ is approximately normal only when $n$ is large (L8.4). The exam asks which case you are in, and "$n=9<30$" is not a reason to refuse the question when the population is stated to be normal.