SQA SQA National 5 2019-style practice: Calculate the standard deviation of the data 2, 4, 5, 6, 8, to 1 decimal place.

The mean is 2 + 4 + 5 + 6 + 85 = 5. Using s = sum(x - x)^2n - 1, find the squared deviations: (-3)^2, (-1)^2, 0^2, 1^2, 3^2 = 9, 1, 0, 1, 9, summing to 20 (2 marks). Then s = 205 - 1 = sqrt(5) (1 mark). So s = 2.2 to 1 decimal place (1 mark). Markers reward the squared deviations, dividing by n - 1, and the rounded value.

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How do you calculate the standard deviation of a data set and use it to compare the spread of two data sets?

Calculating the standard deviation of a data set using the standard formula, and using the mean and standard deviation to compare two data sets.

A focused answer to the SQA National 5 Mathematics standard deviation content, covering how to calculate the standard deviation with the National 5 formula, what it measures, and how to compare two data sets using the mean and standard deviation together.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-14

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

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What this dot point is asking
What standard deviation measures
The formula
Calculating step by step
Comparing data sets
Examples in context
Try this

What this dot point is asking

The SQA wants you to calculate the standard deviation of a data set using the National 5 formula, understand that it measures how spread out the data is about the mean, and use the mean together with the standard deviation to compare two data sets.

What standard deviation measures

The mean tells you the typical value of a data set, but not how varied it is. Two sets can share a mean yet be very different: one tightly bunched, the other widely scattered. Standard deviation captures that spread by measuring, on average, how far the values lie from the mean. A small standard deviation means consistent, clustered data; a large one means variable, spread-out data.

The formula

National 5 uses the sample standard deviation, dividing by $n - 1$ . Two equivalent forms appear on the formula sheet; the deviation form below is the clearest to follow.

Calculating step by step

The deviation method follows a clear sequence: find the mean, find each value's deviation from it, square the deviations, add them, divide by $n - 1$ , and take the square root.

Find the standard deviation of 5, 7, 9

The standard deviation tells us how far the values typically lie from their mean. We apply the formula $s = \sqrt{\dfrac{\sum(x - \bar{x})^2}{n-1}}$ step by step.

Step 1: Find the mean

Add all the values and divide by $n = 3$ :

\bar{x} = \dfrac{5 + 7 + 9}{3} = \dfrac{21}{3} = 7

Step 2: Find each deviation, square it, and sum

Subtract the mean from each value to find how far it sits from the centre, then square to make all deviations positive.

Deviations: $5 - 7 = -2$ , $7 - 7 = 0$ , $9 - 7 = 2$ . Squared: $4$ , $0$ , $4$ . Sum of squares $= 4 + 0 + 4 = 8$ .

Step 3: Divide by $n - 1$ and take the square root

Dividing by $n - 1 = 2$ gives the sample variance; taking the square root returns to the original units:

s = \sqrt{\dfrac{8}{3 - 1}} = \sqrt{\dfrac{8}{2}} = \sqrt{4} = 2

Final answer: The standard deviation is $2$ .

A table helps with larger sets: a column for $x$ , a column for $x - \bar{x}$ , and a column for $(x - \bar{x})^2$ , then total the last column.

Find the standard deviation of 2, 4, 6, 8 to 1 decimal place

This example has four values, so we divide by $n - 1 = 3$ and round at the final step only.

Step 1: Find the mean

\bar{x} = \dfrac{2 + 4 + 6 + 8}{4} = \dfrac{20}{4} = 5

Step 2: Find each deviation from the mean, square it, and sum

Subtract $\bar{x} = 5$ from each value: $2 - 5 = -3$ , $4 - 5 = -1$ , $6 - 5 = 1$ , $8 - 5 = 3$ . Square each: $9$ , $1$ , $1$ , $9$ . The sum of squared deviations is $9 + 1 + 1 + 9 = 20$ .

Step 3: Divide by $n - 1$ and take the square root

s = \sqrt{\dfrac{20}{4 - 1}} = \sqrt{\dfrac{20}{3}} = \sqrt{6.6\overline{6}} \approx 2.6 \text{ to 1 decimal place}

Final answer: The standard deviation is $2.6$ to 1 decimal place.

Comparing data sets

A complete comparison quotes both the mean and the standard deviation, and explains what each says. The means compare the typical values; the standard deviations compare the consistency. The smaller standard deviation belongs to the more consistent set.

Examples in context

Standard deviation is the standard measure of reliability. A manufacturer comparing two production lines prefers the one with the smaller standard deviation, because its output varies less. Investors use it to judge how risky a return is, since a larger standard deviation means more uncertainty. Exam boards use it to see how varied a cohort's marks are about the average.

Try this

Q1. Find the mean of $4, 6, 8, 10, 12$ . [1 mark]

Cue. $\dfrac{40}{5} = 8$ .

Q2. Find the standard deviation of $3, 5, 7$ to 1 decimal place. [3 marks]

Cue. Mean $5$ ; squared deviations $4, 0, 4$ ; $s = \sqrt{\tfrac{8}{2}} = 2.0$ .

Q3. Set A has mean $20$ , SD $2$ ; Set B has mean $20$ , SD $7$ . Which is more consistent? [1 mark]

Cue. Set A, because its standard deviation is smaller.

Exam-style practice questions

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

SQA National 5 20194 marksCalculate the standard deviation of the data

2, 4, 5, 6, 8

, to 1 decimal place.

Show worked answer →

The mean is $\dfrac{2 + 4 + 5 + 6 + 8}{5} = 5$ . Using $s = \sqrt{\dfrac{\sum(x - \bar{x})^2}{n - 1}}$ , find the squared deviations: $(-3)^2, (-1)^2, 0^2, 1^2, 3^2 = 9, 1, 0, 1, 9$ , summing to $20$ (2 marks). Then $s = \sqrt{\dfrac{20}{5 - 1}} = \sqrt{5}$ (1 mark). So $s = 2.2$ to 1 decimal place (1 mark). Markers reward the squared deviations, dividing by $n - 1$ , and the rounded value.

SQA National 5 20223 marksSample P has mean

50

and standard deviation

3

. Sample Q has mean

50

and standard deviation

9

. Compare the two samples.

Show worked answer →

The means are equal at $50$ , so on average the two samples are the same (1 mark). Sample P has a smaller standard deviation ( $3$ against $9$ ), so its data is more consistent and clustered closer to the mean; Sample Q is more spread out (1 mark). A full comparison states both the average and the spread in context (1 mark). Markers reward comparing the means and the standard deviations with a clear conclusion about consistency.

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Sources & how we know this

SQA National 5 Mathematics Course Specification — SQA (2018)