Applying the Theorem

Sample means are a normal variable

The central limit theorem says that when the sample size n is large enough, the sample mean x̄ is approximately normal, with mean $\mu$μ and standard deviation $\frac{\sigma }{\sqrt{n}}$σ/√n, the standard error. In symbols, x̄ ~ $N\left(\mu ,\frac{{\sigma }^2}{n}\right)$N(μ, σ²/n) approximately, whatever the shape of the population.

From then on x̄ is treated like any other normal variable: standardize it, read the table, or run the table backward. The only change is the spread. A z-score for a sample mean divides by the standard error, not by $\sigma$σ: $z=\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$z = (x̄ − μ)/(σ/√n).

An example

The lifetimes of a type of bulb have mean 800 hours and standard deviation 120 hours. They are not normal: most bulbs last a while, and a few fail very early. A shop tests a sample of 36 bulbs. What is the probability that their mean lifetime is less than 770 hours?

The standard error is $\frac{120}{\sqrt{36}}=\frac{120}{6}=20$120/√36 = 120/6 = 20 hours, so x̄ ~ $N\left(800,20^2\right)$N(800, 20²) approximately. Standardize 770: $z=\frac{770-800}{20}=-\frac{30}{20}=-1.5$z = (770 − 800)/20 = −30/20 = −1.5. Then P(x̄ < 770) = P(Z < −1.5) = 1 − 0.9332 = 0.0668.

Running the table backward works too. The middle 95% of sample means lie within 1.96 standard errors of 800: from 800 − 1.96 × 20 = 760.8 to 800 + 1.96 × 20 = 839.2 hours.

The mean lifetime of 36 bulbs, $N\left(800,20^2\right)$N(800, 20²). The shaded area to the left of 770 is 0.0668.

A total instead of a mean

The total T of the n values is n times their mean, so it is approximately normal too. Its mean is $n\mu$nμ, and since the variances of independent values add, its variance is $n{\sigma }^2$nσ² and its standard deviation $\sigma \sqrt{n}$σ√n.

For the 36 bulbs, T has mean 36 × 800 = 28,800 hours and standard deviation 120 × 6 = 720 hours. The mean is less than 770 exactly when the total is less than 36 × 770 = 27,720, and $z=\frac{27,720-28,800}{720}=-\frac{1080}{720}=-1.5$z = (27,720 − 28,800)/720 = −1080/720 = −1.5, the same z as before.

Thirty is a rule of thumb

The theorem says the means get closer and closer to normal as n grows. It names no sample size at which they become normal, because there is none: the approximation improves gradually.

Take a population piled up at small values with a long tail to the right, like the times between calls to a help line, with mean 1 minute and standard deviation 1 minute. The normal approximation says that x̄ lands more than 1.5 standard errors above the mean with probability 0.0668, and the same below it. The exact probabilities, worked out from this population, are in the table below.

At n = 4 the approximation is poor: the upper tail is 0.0818 and the lower tail only 0.0190. By n = 36 the two tails are 0.0742 and 0.0559, and by n = 100 they are 0.0716 and 0.0607. Nothing changes suddenly at 30; around there the approximation is usually good enough to use.

The exact chance that x̄ lands more than 1.5 standard errors above, or below, the mean of a population with a long right tail. Both columns close in on the normal value 0.0668 as n grows.

The means of 600 samples of size n from a population with mean 1 and standard deviation 1 and a long right tail. At n = 4 the pile of means leans to the right. Drag n up: the pile turns symmetric and narrows, its spread $\frac{\sigma }{\sqrt{n}}$σ/√n.

How lopsided the population is

How large n must be depends on the population. A symmetric population needs very few: for values spread evenly between 0 and 1, the chance that the mean of only 4 of them lands more than 1.5 standard errors above the mean is 0.0688, already close to 0.0668.

A population with a long tail needs more, as the table shows, and a badly lopsided one, such as incomes or insurance claims, where a few values are enormous, can need hundreds. If the population is itself normal, x̄ is exactly normal for every n, and no rule of thumb is needed.

The usual mistakes

Treating 30 as a switch. A sample of 29 is not useless and a sample of 31 is not perfect; the approximation improves gradually, and how fast depends on the shape of the population.

Applying the theorem to one value. It is the mean of the 36 bulbs that is approximately normal. One bulb’s lifetime keeps the population’s shape, so P(one bulb lasts less than 770 hours) cannot be found from a normal table.

Standardizing a mean with $\sigma$σ. The spread of x̄ is $\frac{\sigma }{\sqrt{n}}=20$σ/√n = 20 hours, not $\sigma =120$σ = 120 hours: $\frac{770-800}{120}=-0.25$(770 − 800)/120 = −0.25 would be the z-score of a single bulb.

Thinking the population must be normal. That is the case the theorem does not need: the means become normal whatever the population looks like.

An application

In the application below, one parcel’s mass is not normal, but the total mass of 50 parcels is approximately normal, with mean 50 times the mean and variance 50 times the variance.