Infer both normal parameters from two percentiles
Turn two cumulative statements into simultaneous equations, solve for centre and spread, and check probability ordering.
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
- Use signed normal critical values.
- Solve simultaneous linear equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which equation corresponds to P(X < 40) = 0.10 using z ≈ −1.282?
What you will learn
- Infer μ and σ using two distinct percentile conditions.
- Check consistency, positivity and the precision of quoted quantiles.
Each percentile supplies one equation
Suppose P(X < 40) = 0.10 and P(X < 60) = 0.90. The quoted z values are −1.282 and 1.282, so 40 = μ − 1.282σ and 60 = μ + 1.282σ. Adding gives μ = 50; subtracting gives 20 = 2.564σ and σ ≈ 7.8003.
The variance is approximately 60.8449, calculated from the unrounded SD. These are approximate parameter estimates from rounded critical values. One percentile alone supplies only one relationship between the two unknowns; do not claim it uniquely identifies both μ and σ without further information.
Use unequal quantiles and check ordering
If P(X < 30) = 0.75 and P(X < 42) = 0.95, use z ≈ 0.674 and 1.645. Then 30 = μ + 0.674σ and 42 = μ + 1.645σ. Subtraction gives σ ≈ 12/0.971 = 12.3584, followed by μ ≈ 21.6704. The mean is not automatically the midpoint of unequal percentiles.
Cumulative probabilities must increase with their boundary. Claims P(X < 70) = 0.90 and P(X < 50) = 0.95 are incompatible: the smaller event is assigned the larger probability. Reject such a pair before fitting parameters. For a valid pair, substitute the solved μ and σ into both standardisations and ensure σ > 0.
Pause and explain
Can P(X < 50) = 0.95 and P(X < 70) = 0.90 both hold?
Worked example
P(X < 40) = 0.10 and P(X < 60) = 0.90 for a normal variable. Infer μ and σ using critical values ±1.282.
Show the worked solution
- Write 40 = μ − 1.282σ and 60 = μ + 1.282σ.
- Add to get μ = 50, then subtract to get σ ≈ 20/2.564.
- The inferred SD is approximately 7.8003, so σ ≈ 7.80.
Answer μ ≈ 50; σ ≈ 7.80 using the quoted critical values.
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.
A normal variable has tenth and ninetieth percentiles 40 and 60. Infer μ and σ using critical values ±1.282.
Give me a hint
Add the two equations for the centre and subtract them for the spread.
Compare my reasoning
- 40 = μ − 1.282σ and 60 = μ + 1.282σ.
- Adding gives μ = 50; subtracting gives 20 = 2.564σ.
- σ ≈ 20/2.564 = 7.8003, approximately 7.80.
μ ≈ 50; σ ≈ 7.80.
Look for these in your work
- Use both percentile equations.
- Check the positive SD and approximate quantile precision.
P(X < 30) = 0.75 and P(X < 42) = 0.95. Infer μ and σ using critical values 0.674 and 1.645.
Give me a hint
Subtract the equations before substituting back for μ.
Compare my reasoning
- 30 = μ + 0.674σ and 42 = μ + 1.645σ.
- σ ≈ 12/(1.645 − 0.674) = 12.3584.
- μ ≈ 30 − 0.674(12.3584) = 21.6704.
μ ≈ 21.67; σ ≈ 12.36.
Look for these in your work
- Use the difference of the distinct quantiles.
- Do not assume the mean is the midpoint of unequal percentiles.
Assess the claims P(X < 70) = 0.90 and P(X < 50) = 0.95 for the same normal variable.
Give me a hint
Compare the containment of the two events before calculating parameters.
Compare my reasoning
- Every outcome below 50 is also below 70.
- Therefore P(X < 50) ≤ P(X < 70).
- The claimed 0.95 > 0.90 violates that ordering, so no valid model can satisfy both.
Incompatible cumulative ordering; no valid normal model.
Look for these in your work
- Identify the nested events.
- Reject the inconsistent pair without changing its values.
Common mistakes
- Two unknown parameters need two independent relationships.
- Unequal percentiles do not place the mean at their midpoint.
- Reject reversed cumulative ordering instead of forcing a fitted model.
What can you explain now?
Does one known percentile determine both mean and SD?
Compare with the explanation
No; it gives only x = μ + zσ.
Different positive spreads can be paired with different means to satisfy one such relationship.
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Make it stick
Convert each percentile to x = μ + zσ, solve the independent relationships, then check positive spread and increasing cumulative 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.