ScotlandApplications of MathematicsSyllabus dot point

How do you use averages and measures of spread, including the five-figure summary, the semi-interquartile range and standard deviation, to compare two data sets?

Using a combination of statistics to compare data sets, calculating the mean, median, mode and range, the five-figure summary and semi-interquartile range, and the standard deviation, then comparing an average with a measure of spread.

A focused answer to the SQA National 5 Applications of Mathematics statistics content on comparing data, covering the mean, median, mode and range, the five-figure summary and semi-interquartile range, the standard deviation, and comparing two data sets using both an average and a measure of spread.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-15

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

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What this dot point is asking
Averages and the five-figure summary
Standard deviation
Comparing two data sets
Examples in context
Try this

What this dot point is asking

The SQA wants you to calculate averages (mean, median, mode) and measures of spread (range, five-figure summary, semi-interquartile range and standard deviation), then use a combination of an average and a spread to compare two data sets and state a conclusion in context.

Averages and the five-figure summary

The three averages each summarise a data set with one number. The five-figure summary then describes the spread by splitting ordered data into quarters.

Find the five-figure summary

We want to find all five key values: the minimum, $Q_1$ , the median, $Q_3$ and the maximum. The data must be in order before anything else is done.

Step 1: Confirm order and identify the minimum and maximum

The data $4, 7, 8, 9, 12, 15, 18$ is already in ascending order. The minimum is $4$ and the maximum is $18$ .

Step 2: Find the median

With $7$ values, the median is the middle one: position $\frac{7+1}{2} = 4$ . Counting along the ordered list, the 4th value is $9$ . So the median is $9$ .

Step 3: Find the quartiles from the two halves

Split the data at the median. The lower half is the three values below it: $4, 7, 8$ . The middle value of those three is $7$ , so $Q_1 = 7$ . The upper half is the three values above the median: $12, 15, 18$ . The middle value is $15$ , so $Q_3 = 15$ .

Final answer: Five-figure summary: $4,\ 7,\ 9,\ 15,\ 18$ .

The semi-interquartile range measures the spread of the middle half of the data and ignores extreme values, so it is not distorted by one unusually large or small value. This makes it a good partner for the median, which is also resistant to extreme values; together they describe a data set that contains an outlier far better than the mean and range would.

When the number of values is even, the median is the mean of the two middle values, and each half (for the quartiles) is taken as the values on that side of the median. Always count the values carefully and order them first, because a single misplacement changes every quartile.

Standard deviation

The standard deviation measures, on average, how far each value lies from the mean. Unlike the SIQR, it uses every value, so it is the most sensitive measure of spread.

Calculate the standard deviation of 2, 4, 6

The standard deviation measures how far, on average, each value sits from the mean. We work through 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 how many there are.

\bar{x} = \dfrac{2 + 4 + 6}{3} = \dfrac{12}{3} = 4

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

Subtract the mean from each value to get the deviation, then square it to remove the minus sign.

$x$	$x - \bar{x}$	$(x - \bar{x})^2$
$2$	$-2$	$4$
$4$	$0$	$0$
$6$	$2$	$4$

The sum of the squared deviations is $4 + 0 + 4 = 8$ .

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

We divide by $n - 1 = 2$ rather than $n = 3$ because this gives the sample standard deviation, which is what the National 5 formula requires.

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

Final answer: The standard deviation is $2$ .

A small standard deviation means the data is tightly clustered around the mean; a large one means it is widely spread.

Comparing two data sets

A fair comparison always quotes both an average and a measure of spread, then states a conclusion in context. The average says which set is typically higher; the spread says which is more consistent. Choose matching measures for both sets: compare medians with semi-interquartile ranges, or means with standard deviations, but do not mix a median from one set with a mean from the other. The conclusion should answer the question asked, for example which class did better on average and which was more consistent.

Examples in context

Comparisons drive real decisions. Two delivery firms with the same mean time but different spreads differ in reliability; two batteries with the same average life but different standard deviations differ in consistency; two classes' marks are compared by median and SIQR. The SQA rewards quoting both an average and a spread and concluding in context, the core skill here.

Try this

Q1. Find the mean of $5, 8, 11, 16$ . [2 marks]

Cue. $\dfrac{5 + 8 + 11 + 16}{4} = 10$ .

Q2. For the summary $Q_1 = 6$ , $Q_3 = 18$ , find the semi-interquartile range. [2 marks]

Cue. $\dfrac{18 - 6}{2} = 6$ .

Q3. Two sets have the same mean but standard deviations $3$ and $9$ . Which is more consistent? [1 mark]

Cue. The set with standard deviation $3$ .

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 N5 Apps style4 marksFor the data

3, 5, 5, 8, 11, 14, 16

, find the median and the semi-interquartile range.

Show worked answer →

With $7$ ordered values, the median is the middle (4th) value, which is $8$ (1 mark). The lower quartile is the median of the lower half ( $3, 5, 5$ ), which is $5$ ; the upper quartile is the median of the upper half ( $11, 14, 16$ ), which is $14$ (1 mark for both quartiles). The semi-interquartile range is $\dfrac{Q_3 - Q_1}{2} = \dfrac{14 - 5}{2}$ (1 mark). Evaluate: $\dfrac{9}{2} = 4.5$ (1 mark). Markers reward the median, both quartiles, the SIQR formula, and the value.

SQA N5 Apps style3 marksTwo classes sit the same test. Class A has mean

62

and standard deviation

4

; Class B has mean

62

and standard deviation

11

. Compare the performance of the two classes.

Show worked answer →

Both classes have the same mean of $62$ , so on average they performed equally well (1 mark). Class A has a much smaller standard deviation ( $4$ against $11$ ), so its marks are more consistent and closer to the mean (1 mark). Class B's marks are more spread out, so its results are more variable, with more very high and very low marks (1 mark). Markers reward comparing the averages and the spreads, both stated in context. A comparison needs both, not just the means.

Related dot points

Sources & how we know this

SQA National 5 Applications of Mathematics Course Specification — SQA (2023)
SQA National 5 Applications of Mathematics - Course overview and resources — SQA (2023)