The Central Limit Theorem

The second route to a normal X̄

Most business populations are not normal. Days-to-pay is right-skewed — most customers pay near the due date, a few take months. Order sizes, claim amounts, call durations: skewed, all of them. L8.3's rule does not apply. The Central Limit Theorem (CLT) is what does:

> For a random sample of size $n$ from any population with mean $\mu$ and finite variance ${\sigma }^2$, the sampling distribution of $\bar{X}$ is approximately $N(\mu ,{\sigma }^2/n)$ when $n$ is large.

The centre and the spread are L8.2's, exact for every $n$. What the theorem adds is the shape, and it adds it only approximately and only for large $n$. The convention the course and the paper use for "large" is

$$n>30$$

— a convention, not a law. A mildly skewed population's mean is nearly normal by $n=15$; a severely skewed one may need 50 or more. State the convention as such: "$n=36>30$, so by the CLT $\bar{X}$ is approximately normal."

Worked example — days to pay

Harbourview Clearing's customers take a right-skewed number of days to pay, with $\mu =38$ days and $\sigma =12$ days. An analyst samples $n=36$ invoices. Since $36>30$, the CLT gives

$$\bar{X}\approx N\left(38,\frac{144}{36}\right)=N(38,4),\mathrm{SE}=\frac{12}{6}=2\text{ days}$$

Probability the sample averages more than 41 days.

$$z=\frac{41-38}{2}=1.50,P(\bar{X}>41)=1-\Phi (1.50)=1-0.9332=0.0668$$

Probability the sample average is between 36 and 42 days. $z=-1.00$ and $2.00$:

$$P(36<\bar{X}<42)=\Phi (2.00)-\Phi (-1.00)=0.9772-0.1587=0.8185$$

What the CLT does not say

Probability a single invoice takes more than 41 days. This is $P(X>41)$ for one skewed value. The CLT says nothing about it — the theorem describes the average, and the individual invoices keep whatever skewed shape they had. With only $\mu$ and $\sigma$ and no distribution, this probability cannot be computed with the course's tools. On the paper, saying so is the mark; standardising 41 with $\sigma =12$ and reading the normal table is the error.

The theorem also does not rescue a small sample from a skewed population. Eight invoices from Harbourview's skewed days-to-pay: $n=8$ is not large, the population is not normal, and neither rule applies — no probability about $\bar{X}$ can be computed. That is the honest answer, and the paper asks for it.

The decision, in one table

Every question about a sample mean starts by placing itself in a row of this table, and every later module's "conditions" line is this table restated.