To find a score from a percentile, read the table backwards to get z, then use X = μ + zσ. Sketch the curve first, because the sketch tells you whether z is positive or negative.
This lesson is part of choosing the right probability model. It builds on standardising a normal random variable.
Which way does the calculation run?
In the forward direction you start with a score, find z = (X − μ) ÷ σ, then read an area. In the reverse direction you start with an area, read z from the table, then rebuild the score.
The table lists the right-tail area Q(z) for positive z. So a stated “top 10%” is a right-tail area of 0.10, and z can be read directly.
Worked example: the top 10%
Here is an original example. In a fictional school test, scores are normally distributed with mean 60 and standard deviation 8. The top 10% of scores are above k. Find k.
- Sketch the curve and shade the right tail with area 0.10. It sits above the mean, so z > 0.
- Find z with Q(z) = 0.10. The table gives Q(1.28) = 0.1003, so z ≈ 1.28.
- Rebuild the score: k = 60 + 1.28 × 8 = 60 + 10.24 = 70.24.
- Check: (70.24 − 60) ÷ 8 = 1.28, which returns the same z.
So about a tenth of the fictional group scored above 70.24 on that test. This is a statement about that group and that test only.
Worked example: the bottom 20%
Use the same test. The lowest 20% of scores are below m. Find m.
The shaded area is on the left, below the mean, so z will be negative. The table has only right tails, so use symmetry: the z with left tail 0.20 is the negative of the z with right tail 0.20.
Q(z) = 0.20 gives z ≈ 0.84 (the table has Q(0.84) = 0.2005). So the left-tail z is −0.84.
m = 60 − 0.84 × 8 = 60 − 6.72 = 53.28.
The mistake that costs marks
The common slip is to use the positive z for a lower-tail question, writing m = 60 + 0.84 × 8 = 66.72. A score of 66.72 is above the mean, yet the question asked about the bottom 20%.
| Step | Wrong | Right |
|---|---|---|
| Where is the shaded tail? | (skipped) | Left of the mean |
| Sign of z | +0.84 | −0.84 |
| Score m | 66.72 | 53.28 |
| Quick check | Above the mean, so it cannot be bottom 20% | Below the mean |
The fix is the sketch. Before you touch the table, mark the mean and shade the tail. If the tail is left of the mean, the score must be less than the mean.
A note on percentiles and grades
A percentile in a question describes a position inside a made-up group. It does not tell you what mark earns a particular SPM grade, and this lesson does not try to predict any real result. The lesson only teaches how to move from an area back to a score.
Check yourself
Scores in a fictional test are normally distributed with mean 50 and standard deviation 10. The top 2.5% score above k. Find k.
Answer
The shaded area is the right tail with area 0.025, so z > 0.
Q(z) = 0.025 gives z = 1.96.
k = 50 + 1.96 × 10 = 50 + 19.6 = 69.6.
Check: (69.6 − 50) ÷ 10 = 1.96.
What to study next
After you can recover a score, the next skill is checking an answer against a picture. Continue with reconciling a probability calculation with the shaded diagram. A related exam skill is finding unknown normal-distribution parameters.
For a teacher to run the reverse direction on your own questions, see online one-to-one Additional Mathematics tuition.