Year 12 · Mathematics · Statistics
The normal distribution and z-scores
Recognise normally distributed data, use the 68–95–99.7 rule to estimate proportions, and use z-scores to compare values from different distributions.
Awaiting educator review. This material was drafted by our content team with AI assistance and is waiting for review by a qualified educator. Please check it suits your child before using it. Created 10 October 2026.
Learning goals
Learners will
- Describe the shape and key features of a normal distribution.
- Use the 68–95–99.7 rule to estimate percentages of data.
- Calculate and interpret z-scores.
Success looks like
- I can sketch a normal curve and mark the mean and one, two and three standard deviations either side.
- I can estimate the percentage of data between or beyond given values.
- I can calculate a z-score and use it to compare results from different tests.
The big idea
Many measurements, such as adult heights or the masses of packaged food, form a normal distribution when graphed. The graph is a symmetric bell shape with its peak at the mean. Half the data lies below the mean and half above.
The standard deviation (σ) describes the spread. A small standard deviation gives a tall, narrow curve; a large one gives a low, wide curve.
The 68–95–99.7 rule gives approximate percentages for any normal distribution:
- About 68% of values lie within 1 standard deviation of the mean.
- About 95% lie within 2 standard deviations, and about 99.7% within 3.
Because the curve is symmetric, these percentages split evenly either side of the mean. For example, about 34% of values lie between the mean and 1 standard deviation above it, and about 47.5% lie between the mean and 2 standard deviations above it. About 2.5% lie more than 2 standard deviations above the mean.
A z-score tells you how many standard deviations a value is from the mean: z = (x − μ) / σ. A positive z-score is above the mean, a negative one is below, and z = 0 is exactly at the mean. Rearranging gives x = μ + zσ, which finds the value for a given z-score.
z-scores let us compare results from different distributions. A mark with a higher z-score is better relative to the group, even if the raw mark is lower.
Worked examples
Calculating a z-score
Heights in a group are normally distributed with mean 170 cm and standard deviation 8 cm. Find the z-score for a height of 186 cm.
- z = (x − μ) / σ = (186 − 170) / 8.
- z = 16 / 8 = 2.
Answer: z = 2 (two standard deviations above the mean)
Using the 68–95–99.7 rule
For the same heights (mean 170 cm, standard deviation 8 cm), estimate the percentage of people taller than 186 cm.
- 186 cm has z = 2.
- About 95% of heights lie between z = −2 and z = 2, so about 5% lie outside this range.
- By symmetry, half of that 5% is above z = 2.
Answer: About 2.5%
Comparing results
Sam scored 75 in a Maths test (mean 60, standard deviation 10) and 78 in an English test (mean 70, standard deviation 4). In which test did Sam do better relative to the group?
- Maths: z = (75 − 60) / 10 = 1.5.
- English: z = (78 − 70) / 4 = 2.
- The English z-score is higher.
Answer: English (z = 2 compared with z = 1.5)
Practice check
Have a go, then check your answers. Each answer comes with an explanation.
Answer guide for parents
In a normal distribution, approximately what percentage of values lie within 1 standard deviation of the mean?
68% — By the 68–95–99.7 rule, about 68% of values lie within 1 standard deviation of the mean.
Scores have mean 70 and standard deviation 5. What is the z-score of a score of 85?
3 — z = (85 − 70) / 5 = 15 / 5 = 3.
Data are normally distributed with mean 50 and standard deviation 10. What percentage of values are above 50?
50% — The normal curve is symmetric about the mean, so half of the values (50%) are above it.
Data are normally distributed with mean 100 and standard deviation 15. Approximately what percentage of values are above 130?
2.5% — 130 is 2 standard deviations above the mean. About 95% lie within 2 standard deviations, leaving 5% outside, half of which (2.5%) is above.
What does a negative z-score tell you?
The value is below the mean — A negative z-score means x − μ is negative, so the value is below the mean.
Data have mean 40 and standard deviation 6. Find the z-score of 31.
−1.5 — z = (31 − 40) / 6 = −9 / 6 = −1.5.
Data have mean 20 and standard deviation 3. Which value has a z-score of 2?
26 — x = μ + zσ = 20 + 2 × 3 = 26.
Data are normally distributed with mean 70 and standard deviation 5. Using the 68–95–99.7 rule, approximately what percentage of values lie between 65 and 80? Give the number only.
81.5% — 65 is 1 standard deviation below the mean (about 34% between 65 and 70) and 80 is 2 above (about 47.5% between 70 and 80). Total: 34 + 47.5 = 81.5%.
Watch out for
- Thinking the rule gives exact percentages. The 68–95–99.7 values are approximations for data that are close to normal.
- Subtracting in the wrong order, such as μ − x instead of x − μ, which flips the sign of the z-score.
- Assuming all data are normally distributed. Skewed data, such as house prices or incomes, should not be analysed this way.
- Comparing raw scores from different tests directly. Different means and spreads make raw scores misleading; z-scores account for this.
Tips for parents
- Ask your learner to sketch a bell curve and mark the mean and standard deviations for every question. Most errors disappear with a sketch.
- Discuss real examples, such as how manufacturers set package weights so that very few packets are underweight.
- Ask your learner to explain what a z-score means in a sentence, such as "this height is two standard deviations above average".
Go further
- Measure the hand spans of as many family members and friends as possible, draw a histogram, and decide whether the data look approximately normal.
- Find the mean and standard deviation of a data set using a calculator's statistics mode, then check what percentage of values lie within one standard deviation.
- Research how standardised scores are used to compare results across different groups, and discuss the advantages and limitations.