Mathematics 9709 · syllabus 6.4
Cambridge A Level Mathematics 9709 — Sampling and estimation topical past papers
Random samples, the distribution of the sample mean and the Central Limit Theorem, unbiased estimates, and confidence intervals for a mean or a proportion.
- Questions
- 71
- Years
- 2013–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 50
Anyone can browse these pages. Opening the questions and their mark schemes takes a free Momentum account.
What Sampling and estimation covers
Section 6.4 of Paper 6 starts with the difference between a population and a sample, why a sample should be random, and how random numbers can be used to choose one. You may need to explain, in context, why a proposed method, such as surveying the first people through a door, is not random. Named schemes such as stratified or quota sampling are outside the syllabus.
The sample mean X̄ is itself a random variable, with E(X̄) = μ and Var(X̄) = σ²/n. If the population is normal, X̄ is exactly normal; if not, the Central Limit Theorem says X̄ is approximately normal when n is large. From a sample you calculate unbiased estimates of the population mean and variance, and the variance estimate divides by n − 1 rather than n.
Confidence intervals use these results. An interval for μ is x̄ ± z × σ/√n, using the population variance if it is known, or the unbiased estimate from a large sample, and an approximate interval for a population proportion is p̂ ± z√(p̂(1 − p̂)/n). The same ideas about the distribution of a sample mean lead straight into hypothesis tests.
In the syllabus
The headings this topic has in the Cambridge International 2026–2027 syllabus, as the board prints them.
- 6.4 Sampling and estimation
What you should be able to do
- Explain why a sampling method is or is not random, in context
- Use E(X̄) = μ and Var(X̄) = σ²/n to find probabilities about a sample mean
- Say when the Central Limit Theorem is needed and when X̄ is exactly normal
- Calculate unbiased estimates of the population mean and variance from raw or summarised data
- Find and interpret a confidence interval for a population mean
- Find an approximate confidence interval for a population proportion from a large sample
Key ideas
- E(X̄) = μ and Var(X̄) = σ²/n
- Central Limit Theorem: for large n, X̄ is approximately N(μ, σ²/n), whatever the population's distribution
- Unbiased estimate of σ²: s² = (Σx² − (Σx)²/n)/(n − 1) = n/(n − 1) × (Σx²/n − x̄²)
- Confidence interval for μ: x̄ ± z × σ/√n, with z = 1.96 for 95%
- Approximate confidence interval for a proportion: p̂ ± z√(p̂(1 − p̂)/n)
How Sampling and estimation has been examined
From 2022 to 2026, Sampling and estimation was on Paper 6 · Probability & Statistics 2 in 13 of 14 exam series, worth about 16% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 6 · Probability & Statistics 2 | 50 | 16% | 13 of 14 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 5.
Filed in Momentum under: Sampling.
A worked example
A random sample of 50 observations of a variable X gives Σx = 620 and Σx² = 7945. Calculate unbiased estimates of the population mean and variance, and find a 95% confidence interval for the population mean.
- Unbiased estimate of μ: x̄ = 620/50 = 12.4.
- Unbiased estimate of σ²: s² = (7945 − 620²/50)/49 = (7945 − 7688)/49 = 257/49 = 5.245.
- n = 50 is large, so by the Central Limit Theorem X̄ is approximately normal, with variance estimated by s²/50 = 0.1049.
- 95% interval: 12.4 ± 1.96 × √0.1049 = 12.4 ± 1.96 × 0.3239 = 12.4 ± 0.635.
Answer: Mean 12.4; variance 5.24 (257/49); 95% confidence interval (11.8, 13.0)
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Using σ instead of σ/√n as the standard deviation of the sample mean.
- Dividing by n instead of n − 1 for the unbiased estimate of the population variance.
- Saying the Central Limit Theorem was needed when the population is already normal: then X̄ is normal for any sample size.
- Interpreting a 95% interval as '95% chance that μ is in this interval': μ is fixed, and it is 95% of intervals made this way that contain it.
Exam technique
- State which result you are using: 'the population is normal, so X̄ is normal', or 'n is large, so by the Central Limit Theorem X̄ is approximately normal'.
- Show the unbiased variance calculation in full and keep it unrounded for the confidence interval.
- Give a confidence interval's limits to 3 significant figures, written as an interval.
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.