Use the Central Limit Theorem with clear limits
Approximate sufficiently large independent sample means while retaining the original population shape and finite-moment assumptions.
Section 6.4 · Sampling and estimation
Exam scope and assessment notes
Probability & Statistics 2 (Paper 6, A Level) · Mathematics 9709, 2026–2027, section 6.4. Paper 6 builds on Paper 5 and assumes Paper 3 calculus. Distinguish population parameters, sample statistics and repeated-sample random variables. Random selection needs a relevant frame and does not automatically remove undercoverage or non-response. Particular named sampling methods are not required. Sample-mean formulae here assume independent observations from one population with finite variance: E(X̄) = μ, Var(X̄) = σ²/n and SE = σ/√n. A mean of independent normal observations is normal; the CLT gives an approximate normal mean for an adequate large independent sample, not normal individual data. No universal sample-size cutoff guarantees that approximation. Unbiased sample variance uses n − 1, with n > 1. Mean intervals use the normal model with known variance or a large-sample approximation; a small sample with unknown variance does not justify the z method taught here. Proportion intervals use the large-sample normal approximation with adequate success and failure counts; extreme or sparse counts may invalidate it. Quoted two-sided critical values are 1.645 for 90%, 1.960 for 95% and 2.576 for 99%; retain other working unrounded. Confidence describes repeated-sampling coverage, not posterior probability for a fixed realised interval, individual observations or a guarantee. A narrower interval does not repair a biased sample. Hypothesis testing belongs to section 6.5. Written practice is self-checked; cumulative assessment and human subject-teacher review remain pending.
Before you begin
- Distinguish population and sampling distributions.
- Use finite mean and variance.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which distribution does the CLT approximate here?
What you will learn
- State an informal CLT conclusion.
- Identify when the normal approximation is unsupported.
The approximation concerns averages
Suppose an independent sample comes from a skewed finite-variance population with mean twenty and variance thirty-six. For an adequate large n, X̄ is approximately normal with mean twenty and variance 36/n. At n = 100, its approximate model is N(20,0.36), with SE 0.6.
The CLT does not change the original skewed population into a normal population. How quickly the average becomes close to normal depends on the underlying distribution, particularly skewness and tails; no single universal cutoff guarantees accuracy.
Check assumptions instead of reciting a size rule
Repeated or strongly dependent observations do not meet the independent sampling model. A valid heavy-tail density may have infinite variance or mean, so the finite-variance CLT used here cannot simply be invoked.
For n = 100 in the stipulated adequate-approximation setting, P(X̄ < 20.6) is approximately Φ(1) = 0.8413 using that quoted value. State the approximation explicitly. A small sample from an unspecified skewed population does not justify a normal mean merely because its mean and variance are known.
Pause and explain
Does n = 100 alone guarantee a valid normal mean approximation?
Worked example
Give a large-sample mean model for μ = 20, σ² = 36, n = 100 independently.
Show the worked solution
- Mean stays twenty.
- Variance is 36/100 = 0.36.
- Use an approximate normal model if the CLT approximation is adequate.
Answer Approximately N(20,0.36); SE 0.6.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For a suitable independent large sample with μ = 20, σ² = 36 and n = 100, state the mean model.
Give me a hint
Calculate centre and variance before describing its approximate shape.
Compare my reasoning
- E(X̄) = 20.
- Var(X̄) = 0.36 and SE = 0.6.
- The CLT gives approximately N(20,0.36), subject to adequacy.
Approximately N(20,0.36); SE 0.6.
Look for these in your work
- State approximation.
- Retain conditions.
Use that model to estimate P(X̄ < 20.6), with Φ(1) = 0.8413.
Give me a hint
Standardise with SE 0.6.
Compare my reasoning
- z = (20.6 − 20)/0.6 = 1.
- Use the lower area Φ(1).
- Carry the approximation label into the answer.
Approximately 0.8413.
Look for these in your work
- Use mean SE.
- Do not claim exact population normality.
Can the same finite-variance CLT justify an average from density 1/x² on x ≥ 1?
Give me a hint
Check its moments before invoking a theorem.
Compare my reasoning
- The first-moment integral diverges.
- A finite mean and variance are unavailable.
- The finite-variance CLT model taught here is not justified.
No; this density lacks the finite moments needed here.
Look for these in your work
- Check moment existence.
- Do not rely on size alone.
Common mistakes
- The CLT concerns means, not individual values.
- Large size does not fix dependence or infinite variance.
- Do not claim a universal cutoff.
What can you explain now?
Does the CLT make a skewed population normal?
Compare with the explanation
No.
It approximates the distribution of suitable sample means.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Use the CLT for suitable independent averages, stating finite moments and approximation limits.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.