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EnglandStatisticsSyllabus dot point

How do you measure spread using every value in the data?

Variance and standard deviation, calculating standard deviation from a list and a frequency table, and interpreting it.

A focused answer to AQA GCSE Statistics on standard deviation, covering variance, calculating standard deviation from a list and from a frequency table, and interpreting standard deviation as spread around the mean.

Generated by Claude Opus 4.810 min answer

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

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  1. What this dot point is asking
  2. Variance and standard deviation
  3. Calculating from a list
  4. Calculating from a frequency table
  5. Interpreting standard deviation

What this dot point is asking

AQA wants you to understand variance and standard deviation, calculate standard deviation from a list of values and from a frequency table, and interpret it as a measure of spread about the mean. Standard deviation is the spread measure that pairs with the mean, and it feeds directly into standardised scores and the normal distribution later in the course.

Variance and standard deviation

Deviations from the mean are squared so that values above and below the mean do not cancel out (the plain sum of deviations is always zero). Squaring also weights large deviations more heavily, which is why standard deviation is sensitive to outliers. Taking the square root at the end returns the measure to the original units, so a standard deviation of 0.040.04 seconds is directly comparable with the data.

The two forms are algebraically identical. The first (deviation) form makes the meaning clear; the second (sum-of-squares) form is faster because you avoid subtracting the mean from every value before squaring.

Calculating from a list

Calculating from a frequency table

The method mirrors the list version: build an fxfx column to get the mean, and an fx2fx^2 column for the sum of squares. For grouped data, use the class midpoints as the values of xx, exactly as for the estimated mean, and the result is then an estimated standard deviation.

Interpreting standard deviation

A larger standard deviation means the data is more spread out around the mean; a smaller one means the values cluster tightly. Because deviations are squared, standard deviation reacts strongly to outliers, more so than the interquartile range, so a single extreme value can inflate it noticeably. When comparing two data sets with the same mean, the one with the smaller standard deviation is the more consistent, a point examiners love to test in context (manufacturing tolerances, exam marks, race times).

Standard deviation is also the bridge to later topics. It is the spread measure that pairs with the mean, just as the interquartile range pairs with the median, so a comparison using means should quote standard deviations. It is the unit in which the normal distribution measures position: the 6868, 9595, 99.799.7 percent rule counts whole standard deviations from the mean. And it is the denominator of the standardised score, which expresses how many standard deviations a value lies from the mean so that results on different scales can be compared. Mastering the calculation here therefore unlocks the normal distribution and standardised scores in the final module.

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 20204 marksThe reaction times (seconds) of five athletes are 0.12,0.15,0.18,0.20,0.250.12, 0.15, 0.18, 0.20, 0.25. Calculate the standard deviation, giving your answer to 33 significant figures.
Show worked answer β†’

Mean: xˉ=0.12+0.15+0.18+0.20+0.255=0.905=0.18\bar{x} = \frac{0.12 + 0.15 + 0.18 + 0.20 + 0.25}{5} = \frac{0.90}{5} = 0.18.

Sum of squares: βˆ‘x2=0.122+0.152+0.182+0.202+0.252=0.0144+0.0225+0.0324+0.0400+0.0625=0.1718\sum x^2 = 0.12^2 + 0.15^2 + 0.18^2 + 0.20^2 + 0.25^2 = 0.0144 + 0.0225 + 0.0324 + 0.0400 + 0.0625 = 0.1718.

Apply the formula: Οƒ=0.17185βˆ’0.182=0.03436βˆ’0.0324=0.00196β‰ˆ0.0443\sigma = \sqrt{\frac{0.1718}{5} - 0.18^2} = \sqrt{0.03436 - 0.0324} = \sqrt{0.00196} \approx 0.0443 s.

Markers reward the mean, the sum of squares, the correct substitution into βˆ‘x2nβˆ’xΛ‰2\sqrt{\frac{\sum x^2}{n} - \bar{x}^2}, and the rounded answer 0.04430.0443.

AQA 20222 marksTwo factories make rods with the same mean length. Factory A has standard deviation 0.40.4 mm and Factory B has standard deviation 1.11.1 mm. Explain which factory produces more consistent rods.
Show worked answer β†’

Factory A is more consistent because it has the smaller standard deviation (0.40.4 mm versus 1.11.1 mm).

A smaller standard deviation means the rod lengths lie closer to the mean, so there is less variation about the target length.

Markers reward naming Factory A and linking the smaller standard deviation to less spread about the mean, in context.

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