Maths › Further Statistics 1 › The Central Limit Theorem
The Central Limit Theorem
Average enough independent observations from almost any distribution and the average behaves normally, with a spread that shrinks as the square root of the sample size.
Builds on The normal distribution and The Poisson distribution.
IN THIS TOPIC
- State the Central Limit Theorem and the distribution it gives the sample mean.
- Compute probabilities for a sample mean or a sample total from a parent distribution meeting the theorem's conditions, for large enough n.
- Say what the theorem does not claim.
COMMON MISCONCEPTION
The sample mean can only be treated as normal when the population it comes from is itself normal.
Averages turn normal
Take a random sample of n independent observations from a population with mean μ and finite variance σ². For large n the sample mean is approximately normal, whatever shape the parent has:
The booklet supplies both parameters. Under Sampling distributions it prints the sample mean as an unbiased estimator of μ with variance σ²/n, and it prints the standardised form as N(0, 1). What it cannot print is the theorem itself, so be ready to state in words that the approximation holds for large n whatever the parent looks like. The parent may be skewed, discrete, or bounded. Averaging smooths all of that away. The variance of the mean is σ²/n, so the standard error σ/√n falls with the square root of the sample size. Quadruple n and you halve the spread.
The same result runs on totals. Adding n independent observations gives a total that is approximately N(nμ, nσ²), which is often the quicker route when a question asks about a combined weight or a combined time.
WORKED EXAMPLE
A probability for a sample mean
A population has mean 50 and standard deviation 12, with unknown shape. For a sample of 36, find the probability the sample mean exceeds 53.
By the Central Limit Theorem, the mean is approximately N(50, 144/36), that is N(50, 4) with standard error 2.
z = (53 − 50)/2 = 1.5.
P(Z > 1.5) = 0.0668. The parent's shape never entered the calculation, which is the whole point of the theorem.
What the theorem does not say
It says nothing about the population. A skewed parent stays skewed however large the sample grows, and a histogram of 500 raw observations will show that skew perfectly clearly. Only the sampling distribution of the mean turns normal.
It is also an approximation with no fixed threshold. Textbooks offer n above 30 as a working rule, but a wildly skewed parent needs more and a nearly symmetric one needs less. And the conclusion collapses if the observations are not independent, or if the sample is not random, since a biased sampling method shifts the centre and no amount of data will pull it back.
GUIDED PRACTICE
Spotting the misuse
A student writes: 'The sample of 40 lifetimes came from an exponential population, so by the Central Limit Theorem the lifetimes are approximately normal.' Correct the statement.
Show the working
The lifetimes themselves stay exponential and stay skewed. Sample size does not change the parent.
What is approximately normal is the mean of the 40 lifetimes, with variance σ²/40, and the total of the 40, with variance 40σ².
A probability about one lifetime needs the exponential model; only a probability about the average may use the normal one.
Where the square root bites
Because the standard error carries √n and not n, precision is expensive. Halving the spread of the sample mean costs four times the data. That single fact governs how large surveys are designed, and why sample sizes climb so steeply as the required margin narrows.
GUIDED PRACTICE
Sizing a sample
A population has standard deviation 20. How large must a sample be for the standard error of the mean to be at most 2?
Show the working
Standard error = 20/√n ≤ 2, so √n ≥ 10.
n ≥ 100.
Reducing the standard error to 1 would need n ≥ 400. Halving the error quadruples the sample.
ASSESSMENT FOCUS
- Quote the conclusion as a distribution, with mean μ, variance σ²/n, approximately normal.
- Divide the variance by n, not the standard deviation. The standard error is σ/√n.
- Say 'approximately' and say that n is large. The theorem is an approximation, and a complete answer says so.
- For a total rather than a mean, use N(nμ, nσ²) and check which the question asked for.
CHECK YOURSELF
A distribution has mean 30 and variance 100. Write down the approximate distribution of the mean of a sample of 25, and of the total of the 25.
Show a hint
Divide the variance by n for the mean; multiply by n for the total.
Show the answer
The mean is approximately N(30, 4), with standard error 2. The total is approximately N(750, 2500), with standard deviation 50.
For large n from a random sample, the sample mean is approximately N(μ, σ²/n) and the total approximately N(nμ, nσ²), whatever the parent.
The parent itself never turns normal, and the standard error σ/√n needs four times the data for half the spread.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the the central limit theorem questions page.
CHECK YOUR PROGRESS
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- State the Central Limit Theorem and the distribution it gives the sample mean.
- Compute probabilities for a sample mean or a sample total from a parent distribution meeting the theorem's conditions, for large enough n.
- Say what the theorem does not claim.
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