Find normal percentiles and central probability limits
Convert upper-tail or central-area requirements into cumulative probabilities before using an inverse normal value.
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
- Read cumulative and critical normal tables.
- Standardise a normal measurement.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
If P(X > c) = 0.10, which cumulative probability locates c?
What you will learn
- Find a percentile using x = μ + zσ.
- Translate upper tails and central coverage into the correct cumulative quantiles.
A percentile is a cumulative boundary
For X ~ N(100,225), σ = 15. The 90th percentile c has P(X ≤ c) = 0.90, so use the MF19 critical value z ≈ 1.282. Then (c − 100)/15 ≈ 1.282, giving c ≈ 119.23. A requirement that only 10% exceed c uses the same cumulative probability 0.90.
The bottom 10th percentile instead has z ≈ −1.282 and c ≈ 100 − 1.282(15) = 80.77. Check the percentile’s side of the mean before calculating. Probability inputs strictly between zero and one give finite normal quantiles; cumulative zero and one do not describe finite cut-offs in an unbounded normal model.
A central probability leaves two tails
A central 95% interval leaves 5% outside, split as 2.5% in each tail. Its cumulative boundaries are 0.025 and 0.975, so the quoted critical values are −1.960 and 1.960. For Y ~ N(80,16), the limits are 80 ± 1.960(4), approximately 72.16 and 87.84.
Do not use z = 1.645 for a central 95% interval: that is the cumulative 95th-percentile boundary, leaving 5% on one side. Inverse values quoted to three decimals yield approximate limits, not exact identities for the underlying curve. Keep the chosen percentile, sign, units and standardisation visible in the working.
Pause and explain
Which upper cumulative probability belongs to a central 95% normal interval?
Worked example
X ~ N(100,225). Find c so that P(X > c) = 0.10 using z₀.₉₀ ≈ 1.282.
Show the worked solution
- The cumulative probability below c is 0.90.
- Write (c − 100)/15 ≈ 1.282.
- Solve c ≈ 100 + 15(1.282) = 119.23.
Answer c ≈ 119.23.
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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.
X ~ N(100,225). Find the 90th percentile using z₀.₉₀ ≈ 1.282.
Give me a hint
Multiply the quoted standard normal quantile by SD and add the mean.
Compare my reasoning
- σ = 15 and (c − 100)/15 ≈ 1.282.
- Thus c ≈ 100 + 19.23.
- The 90th-percentile boundary is approximately 119.23, above the mean.
Approximately 119.23.
Look for these in your work
- Use cumulative probability 0.90.
- Show standardisation and the original measurement units.
X ~ N(100,225). Find the bottom 10th percentile using the same positive critical value 1.282.
Give me a hint
Reflect the standard normal quantile before returning to X.
Compare my reasoning
- The lower tenth-percentile z value is approximately −1.282.
- (c − 100)/15 ≈ −1.282.
- Solve c ≈ 100 − 19.23 = 80.77.
Approximately 80.77.
Look for these in your work
- Use a negative z for a lower percentile.
- Check the answer is below the mean.
Y ~ N(80,16). Find central 95% limits using z₀.₉₇₅ ≈ 1.960.
Give me a hint
Leave 2.5% in each tail and use SD four.
Compare my reasoning
- The cumulative boundaries are 0.025 and 0.975.
- Their z values are approximately −1.960 and 1.960.
- The measurement limits are 80 ± 7.84, giving 72.16 and 87.84.
Approximately 72.16 to 87.84.
Look for these in your work
- Split the excluded probability into equal tails.
- Use the variance convention and state approximate limits.
Common mistakes
- Convert an upper-tail probability before choosing a cumulative quantile.
- Split the excluded probability between both tails for central coverage.
- Quoted critical values yield approximate measurement limits.
What can you explain now?
Does central 95% use the cumulative 95th-percentile z value?
Compare with the explanation
No; its upper cumulative boundary is 97.5%.
The remaining 5% is divided equally between the lower and upper tails.
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Make it stick
Convert the required area to a cumulative quantile, check its sign, and solve the full standardisation for the original boundary.
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.