Central Limit Theorem (AI HL)
Take repeated random samples of size \(n\) from a population and average each one: those sample means cluster into a predictable normal shape of their own, however the original population was distributed. The central limit theorem pins down exactly what that shape is - its mean and its variance - so you can turn "what's the chance this sample average lands somewhere?" into an ordinary normal-distribution calculation. It's part of the broader Probability & Distributions topic.
25 questions on this sub-topic.
The sample mean distribution
Covered under IB syllabus reference AHL 4.15: a linear combination of \(n\) independent normal random variables is itself normally distributed, and in particular the sample mean \(\bar{X}\) of a random sample drawn from \(N(\mu,\sigma^2)\) satisfies \(\bar{X}\sim N\!\left(\mu,\dfrac{\sigma^2}{n}\right)\). The central limit theorem extends this beyond normal populations: whatever distribution the population actually has, \(\bar{X}\) approaches normality as \(n\) grows, and the IB treats \(n>30\) as large enough to assume this in an exam.
Distribution of \(\bar{X}\)
\(\bar{X} \sim N\!\left(\mu,\, \dfrac{\sigma^2}{n}\right)\)
The mean of the sample mean equals the population mean \(\mu\); its variance is the population variance \(\sigma^2\) shrunk by a factor of \(n\), so bigger samples give a tighter spread.
Standardising \(\bar{X}\)
\(Z = \dfrac{\bar{X}-\mu}{\sigma/\sqrt{n}}\)
Once you know \(\bar{X}\)'s mean and variance, any probability about it - or any sample-size question run in reverse - is read straight off the normal distribution, usually on the GDC.
Need the wider distributions toolkit and formula-booklet reference table? See Probability & Distributions.
Worked examples
A population has mean \(\mu = 50\) and standard deviation \(\sigma = 10\). A random sample of size \(n = 25\) is taken.
(a)(i) State the mean of the sample mean \(\bar{X}.\)
(a)(ii) State its variance.
(b) Find \(P(\bar{X} > 53)\).
Worked solution
(a)(i) By the Central Limit Theorem, \(\bar{X} \sim N\!\left(50,\, \dfrac{10^2}{25}\right),\) mean correct A1
(a)(ii) \(= N(50,\, 4)\), variance correct A1
(b) \(P(\bar{X} > 53) = P\!\left(Z > \dfrac{53-50}{2}\right) = P(Z > 1.5)\) M1
\(= 1 - \Phi(1.5) \approx 0.0668\) A1
A population has \(\mu = 100\) and \(\sigma = 20\). Find the minimum sample size \(n\) such that \(P(\bar{X} > 103) < 0.05.\)
Worked solution
\(\bar{X} \sim N\!\left(100, \dfrac{400}{n}\right).\) M1
We need \(P(Z > 1.645)\) to give: \(\dfrac{103-100}{20/\sqrt{n}} \ge 1.645.\) M1
\(\sqrt{n} \ge \dfrac{1.645 \times 20}{3} = 10.97.\) A1
\(n \ge 120.2\), so \(n = 121.\) A1
Common mistakes
- Using \(\sigma\) instead of \(\sigma^2/n\) as the variance. \(\bar{X}\) is less spread out than a single observation - dividing by \(n\) (or \(\sqrt{n}\) if you're working with the standard deviation) is essential, not optional, and skipping it inflates every probability that follows.
- Confusing \(X\) with \(\bar{X}\). A single observation is still \(N(\mu,\sigma^2)\); only the sample mean of \(n\) observations gets the shrunken variance \(\sigma^2/n\). Read the question carefully to see which one is actually being asked about.
- Forgetting the theorem needs a large \(n\) for a non-normal population. If the population itself is stated to be normal, \(\bar{X}\)'s distribution is exact for any \(n\). Otherwise, the CLT only kicks in as an approximation once \(n\) is large - the IB treats \(n>30\) as sufficient, so quote that threshold if a question asks you to justify using it.
Ready to practise properly?
25 central limit theorem questions, marked instantly like the real exam.
Quick answers
What does the central limit theorem say about the sample mean?
If a population has mean \(\mu\) and standard deviation \(\sigma\), then the sample mean \(\bar{X}\) of a random sample of size \(n\) is distributed \(N\!\left(\mu,\dfrac{\sigma^2}{n}\right)\) - exactly if the population itself is normal, and approximately for any population once \(n\) is large (the IB treats \(n>30\) as large enough).
Why does the variance of the sample mean divide by n?
Averaging \(n\) independent observations cancels out random fluctuation: the variance of a sum of \(n\) independent variables adds, but dividing that sum by \(n\) to form a mean divides the variance by \(n^2\), leaving \(\sigma^2/n\). Larger samples give a sample mean that clusters more tightly around \(\mu\).