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How is naturally varying data distributed?

The shape of the normal distribution, symmetry about the mean, and the 68, 95, 99.7 percent rule.

A focused answer to AQA GCSE Statistics on the normal distribution, covering the bell-shaped curve, symmetry about the mean, the role of the standard deviation, and the 68, 95 and 99.7 percent rule.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The shape of the normal distribution
  3. The role of the standard deviation
  4. The 68, 95, 99.7 percent rule

What this dot point is asking

AQA wants you to recognise the shape of the normal distribution, know it is symmetrical about the mean, understand how the standard deviation controls its width, and use the 6868, 9595, 99.799.7 percent rule to estimate the proportion of data in a given range.

The shape of the normal distribution

Because it is symmetrical, the mean, median and mode all lie at the centre, directly under the peak. The curve never quite touches the horizontal axis: in principle values can lie any distance from the mean, but the chance of extreme values becomes vanishingly small. The normal distribution arises whenever a quantity is the result of many small, independent influences adding together, which is why heights, manufacturing variation and measurement errors so often look normal.

The role of the standard deviation

Two normal distributions can share a mean but look quite different: a sample of carefully machined parts has a small standard deviation (a tall, thin curve), while a sample of handmade parts has a large one (a short, wide curve). Total area under any normal curve is 11 (it is a probability), so a taller curve must be narrower to keep the area fixed.

The 68, 95, 99.7 percent rule

By symmetry, the data outside a band is split equally between the two tails. Outside ±1σ\pm 1\sigma is 32%32\%, so 16%16\% lies in each tail; outside ±2σ\pm 2\sigma is 5%5\%, so 2.5%2.5\% in each tail; outside ±3σ\pm 3\sigma is 0.3%0.3\%, so 0.15%0.15\% in each tail. Splitting the band from the mean outward gives the useful half-band figures: mean to 1σ1\sigma is 34%34\%, 1σ1\sigma to 2σ2\sigma is 13.5%13.5\%, and 2σ2\sigma to 3σ3\sigma is 2.35%2.35\%.

A reliable method for any range question is to convert each endpoint into a number of standard deviations from the mean, mark the half-band percentages on a quick sketch of the curve, and then add or subtract the regions you need. The exact figures are slightly higher than the round numbers (the bands are really 68.27%68.27\%, 95.45%95.45\% and 99.73%99.73\%), but the rounded 6868, 9595, 99.799.7 values are what AQA expects. Note that the rule only gives percentages at whole numbers of standard deviations; for endpoints in between, such as 1.4σ1.4\sigma, you would need a standardised score and a table, which is the standardised-scores topic.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20194 marksThe masses of apples are normally distributed with mean 160160 g and standard deviation 1515 g. (a) Between what two masses do about 95%95\% of the apples lie? (b) Estimate the percentage of apples with mass greater than 175175 g.
Show worked answer →

(a) 95%95\% lie within 22 standard deviations: 1602(15)=130160 - 2(15) = 130 g to 160+2(15)=190160 + 2(15) = 190 g.

(b) 175175 g is 17516015=1\frac{175 - 160}{15} = 1 standard deviation above the mean. About 68%68\% lie within 11 standard deviation, so 32%32\% lie outside, and by symmetry half of that, 16%16\%, is above 175175 g.

Markers reward the ±2σ\pm 2\sigma band for part (a), and identifying 175175 g as 1σ1\sigma above the mean then halving the 32%32\% tail for part (b).

AQA 20213 marksA normal distribution has mean 5050 and standard deviation 44. Estimate the percentage of values between 4242 and 5454.
Show worked answer →

4242 is 42504=2\frac{42 - 50}{4} = -2 standard deviations (so 2σ2\sigma below), and 5454 is 54504=+1\frac{54 - 50}{4} = +1 standard deviation above.

From the mean to 2σ2\sigma below covers half of 95%95\%, that is 47.5%47.5\%; from the mean to 1σ1\sigma above covers half of 68%68\%, that is 34%34\%.

Total =47.5%+34%=81.5%= 47.5\% + 34\% = 81.5\%.

Markers reward expressing each endpoint in standard deviations and summing the correct half-band percentages.

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