If a normal interval is centred at the mean, split the probability outside it equally between the two tails. Convert the upper endpoint to a left-tail probability, find its standard normal quantile, then divide the distance from the mean by that quantile.
The 2027 A-Level H2 Mathematics 9758 syllabus covers normal-curve symmetry and relationships between probabilities, cutoffs and distribution parameters. This guide finds an unknown standard deviation from a known central probability.
- Confirm the interval is centred at the mean.
- Subtract the central probability from 1 and halve the remainder.
- Use the area to the left of the upper endpoint to find its z-score.
- Divide the endpoint’s distance from the mean by the positive z-score.
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Turn a central probability into a quantile
Suppose the mean is μ and the interval goes from μ − d to μ + d, where d is positive. If the interval contains probability p, the area outside is 1 − p. Symmetry gives each tail probability (1 − p)/2.
The upper endpoint therefore has cumulative probability 1 − (1 − p)/2 = (1 + p)/2. For an inverse normal function that expects a left-tail area, use this value with mean 0 and standard deviation 1.
The resulting z-score equals d/σ. Rearranging gives σ = d/z. Here d is the half-width of the interval, measured in the same units as X.
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Worked example | Central 90% from 42 to 58
The mass X of a model component is normally distributed with mean 50 g and unknown standard deviation. Given P(42 ≤ X ≤ 58) = 0.90, find the standard deviation.
| Region | Probability |
|---|---|
| Below 42 g | 0.05 |
| Between 42 g and 58 g | 0.90 |
| Above 58 g | 0.05 |
| At or below 58 g | 0.95 |
The 95th percentile of the standard normal distribution is approximately 1.6448536. Since 58 − 50 = 8 g, 8/σ = 1.6448536. Hence σ ≈ 4.86 g to three significant figures.
Check by standardising both endpoints with the unrounded value. They are approximately −1.645 and +1.645, with about 0.05 in each tail. Together the two tails leave 0.90 in the middle.
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Practise with a different central percentage
For a normal distribution with mean 100, the interval from 88 to 112 contains 80% of the probability. Each tail contains 10%, so the upper boundary corresponds to a left-tail area of 0.90 and z ≈ 1.28155.
The half-width is 12, giving σ ≈ 12/1.28155 ≈ 9.36. A quick check is to multiply the standard deviation by the positive z-score and add the result to 100. You should recover the upper endpoint of 112, allowing for rounding.
If an interval has unequal distances from the known mean, the tails are generally unequal. Set up the probability as the difference between two cumulative areas and use the information supplied in the question.
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CHAPTER 4 OF 4 · Back to contents
Frequently asked questions
Why does the calculator need 0.95 in the 90% example?
The upper endpoint has the 90% centre and the 5% lower tail to its left. Together those areas give 0.95.
Can I use the lower endpoint instead?
Yes. For the worked example, divide the signed distance −8 by the lower z-score −1.6448536. The two negative signs give the same positive standard deviation.
What should I report if the question asks for variance?
Square the unrounded standard deviation and round the result as directed. In the worked example the variance is about 23.7 g².
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