Check when a normal approximation to binomial is suitable
Test expected successes and failures, preserve binomial assumptions and use the correct matching normal mean and variance.
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.5. N(μ,σ²) uses variance as its second parameter, with σ > 0; probabilities are areas, not density heights. Show the full standardisation Z = (X − μ)/σ. The MF19 table gives Φ(z) = P(Z ≤ z), with negative values found by symmetry. Model sketches and shaded probabilities may be required. Probability answers use quoted four-decimal table values where supplied; calculations with non-tabulated z use unrounded standardisation and numerical normal probabilities, so table interpolation can differ slightly in the last digits. Inverse examples use the quoted three-decimal MF19 critical values and are approximate. A binomial normal approximation requires both np > 5 and n(1 − p) > 5, and a continuity correction. It remains approximate, including for integer probability thresholds. A continuous normal variable has zero probability at a single exact value; a discrete binomial count does not. Sample means, the Central Limit Theorem and hypothesis testing belong to Statistics 2 and are not covered here. Model assumptions need evidence in real applications. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Identify binomial modelling assumptions.
- Calculate binomial mean and variance.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For B(100,0.40), what are np and n(1 − p)?
What you will learn
- Check np > 5 and n(1 − p) > 5 for the syllabus approximation.
- Match B(n,p) with the approximate N(np,np(1 − p)) model.
Test both sides of the trial outcome
Cambridge section 5.5 requires n to be sufficiently large that both np > 5 and nq > 5, with q = 1 − p. For X ~ B(100,0.40), expected successes are 40 and expected failures are 60, so the criterion is met. A suitable matching normal variable is Y ~ N(40,24), with SD √24.
The binomial experiment still needs fixed n, independence, two classifications and a constant p. The normal approximation does not establish those assumptions. It approximates the distribution of the success count; it does not convert a dependent or changing-probability experiment into a binomial one.
A large n alone is insufficient
For B(200,0.01), np = 2 and nq = 198: the small success expectation fails the criterion despite 200 trials. For B(50,0.10), np = 5, which does not satisfy the stated strict inequality np > 5. Keep that boundary explicit rather than replacing it with a different rule of thumb.
For B(40,0.50), both expected counts are 20, and the matching normal is N(20,10), with SD √10. Meeting the criterion makes the approximation available under the syllabus, but does not guarantee negligible error for every tail. Use a continuity correction, label the answer approximate and consider exact binomial calculation when an exact decision threshold matters.
Pause and explain
Does B(50,0.10) satisfy both stated approximation inequalities?
Worked example
X ~ B(100,0.40). Check the normal-approximation criterion and state the matching model.
Show the worked solution
- np = 40 and n(1 − p) = 60, both above five.
- Mean = np = 40 and variance = np(1 − p) = 24.
- Use Y ~ N(40,24) as an approximation, with continuity correction when translating events.
Answer Criterion met; Y ~ N(40,24) approximately matches X.
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.
Check the normal approximation for B(100,0.40) and state its matching normal mean, variance and SD.
Give me a hint
Check both expected counts before using the moment formulas.
Compare my reasoning
- np = 40 and nq = 60, both greater than five.
- The matching mean is 40 and variance is 100(0.4)(0.6) = 24.
- Use approximate N(40,24), with SD √24 and event continuity corrections.
Criterion met; approximate N(40,24), SD √24.
Look for these in your work
- Check both strict inequalities.
- Distinguish variance from SD and preserve the approximation label.
Assess a normal approximation for B(50,0.10) using the stated Cambridge criterion.
Give me a hint
Compare np with a strict greater-than-five condition.
Compare my reasoning
- np = 5 and nq = 45.
- The failure expectation exceeds five, but the success expectation equals it.
- The stated criterion is not met; do not assert a normal approximation is justified by this rule.
Criterion not met: np = 5 is not > 5.
Look for these in your work
- Apply the strict boundary correctly.
- Check both outcome expectations.
Compare B(200,0.01) and B(40,0.50) for normal approximation; state the matching model where justified.
Give me a hint
A larger n may still have a small expected success count.
Compare my reasoning
- For B(200,0.01), np = 2 and nq = 198, so the criterion fails.
- For B(40,0.50), np = nq = 20, so the criterion is met.
- The latter matching model is approximate N(20,10), with SD √10.
B(200,0.01) fails; B(40,0.50) meets the criterion and matches N(20,10) approximately.
Look for these in your work
- Do not decide from n alone.
- Use the matching binomial mean and variance.
Common mistakes
- Test expected failures as well as successes.
- The syllabus criterion uses strict greater-than-five boundaries.
- Use np(1 − p) as variance and its square root as SD.
What can you explain now?
Does n = 200 guarantee a useful normal approximation if p = 0.01?
Compare with the explanation
No; np = 2 fails the criterion.
The distribution may remain strongly skewed when successes are rare.
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Make it stick
Verify the binomial assumptions and both expected counts, then use N(np,npq) with an explicit approximation and continuity correction.
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.