Calculate probabilities for means from a normal population
Use the exact normal sample-mean model and standardise with its standard error.
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
- Use normal tables.
- Calculate SE from population variance.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the variance of X̄ for n = 9 from N(100,144) independently?
What you will learn
- State the distribution of an independent normal sample mean.
- Calculate normal mean-event probabilities.
Normality holds without a large-sample argument
Independent observations with X ~ N(100,144) have population SD twelve. For n = 9, X̄ ~ N(100,16), with SE four. This normal shape follows from normal linear-combination closure even for small n; it is not a CLT approximation.
P(X̄ > 108) uses z = (108 − 100)/4 = 2. With quoted Φ(2) = 0.9772, the upper probability is 0.0228. Using population SD twelve would calculate a different event for an individual.
Change n before standardising
For n = 36, mean variance is 144/36 = 4 and SE is two. P(96 < X̄ < 104) uses z limits −2 and 2, giving 0.9544 from the quoted table.
If observations are perfectly repeated copies rather than independent draws, the mean distribution remains N(100,144). Identical marginal normal distributions alone do not justify dividing variance by n.
Pause and explain
Which divisor standardises X̄ > 108 for that n = 9 sample?
Worked example
Find P(X̄ > 108) for nine independent observations from N(100,144).
Show the worked solution
- X̄ ~ N(100,16), so SE = 4.
- Boundary z = 2.
- Upper area = 1 − 0.9772.
Answer 0.0228 using the quoted table.
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 n = 9 from N(100,144), find P(X̄ > 108), using Φ(2) = 0.9772.
Give me a hint
Find the mean distribution before choosing its upper tail.
Compare my reasoning
- Mean variance = 16; SE = 4.
- z = 2.
- Probability = 1 − 0.9772.
0.0228.
Look for these in your work
- Use SE.
- Choose the upper tail.
For n = 36 from the same population, find P(96 < X̄ < 104).
Give me a hint
Use the new SE and the same quoted normal value.
Compare my reasoning
- SE = 12/6 = 2.
- z limits are −2 and 2.
- Probability = 2(0.9772) − 1.
0.9544.
Look for these in your work
- Recalculate SE.
- Use both interval limits.
What is the distribution of the average of nine records all equal to one X ~ N(100,144)?
Give me a hint
The records are dependent copies of the same observation.
Compare my reasoning
- Average = X.
- Its mean is 100 and variance 144.
- Independent mean-variance reduction is unavailable.
N(100,144), not N(100,16).
Look for these in your work
- Check the joint mechanism.
- Do not infer independence from normal marginals.
Common mistakes
- Standardise with SE, not individual SD.
- State variance as the second normal parameter.
- Normal population still needs the independent sampling assumption.
What can you explain now?
Is a normal mean only justified for large n?
Compare with the explanation
No, if observations are independent and normal.
Normal closure gives its shape directly.
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
Normal independent observations give a normal average; use its smaller SE for mean probabilities.
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.