What is not clearing the calculator's old data?

Leftover values from a previous question quietly corrupt the mean and standard deviation. Clear the list every time.

NSWMaths Standard 2Syllabus dot point

How are measures of spread used to describe how widely data is spread out, and how do they let you compare two data sets?

Calculate measures of spread, including the range, quartiles and interquartile range, and the population standard deviation using technology

A focused answer to the HSC Maths Standard 2 dot point on measures of spread. The range, the quartiles and interquartile range, the five-number summary, the population standard deviation from a calculator, and how to compare the spread of two data sets, with worked Australian examples.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-21

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

Have a quick question? Jump to the Q&A page

What this dot point is asking

NESA wants you to measure how spread out a set of data is, not just where its centre sits. Two classes can share the same average mark yet look completely different - one bunched tightly around the mean, the other scattered from fails to top marks. A measure of spread puts a number on that scatter. You need three of them: the range (the crudest), the interquartile range or IQR (the spread of the middle half), and the standard deviation (the average distance of the values from the mean, read straight off a calculator). You also need to assemble the five-number summary and to use these measures to compare two data sets. The arithmetic for range and IQR is light; the marks are won by quoting the right figure for the right purpose and by reading the population standard deviation, $\sigma_n$ , correctly from the calculator.

The answer

A measure of spread answers one question: how far apart are the values? There are three you must know, from crudest to most refined.

Range $=$ maximum $-$ minimum. One subtraction, but it uses only the two extreme values, so a single odd value distorts it.
Interquartile range (IQR) $= Q_3 - Q_1$ . This is the spread of the middle $50\%$ of the data, so it ignores the lowest and highest quarters and is barely affected by extreme values.
Standard deviation $(\sigma_n)$ $=$ roughly the average distance of the values from the mean. It uses every value, and in this course you read it from your calculator rather than computing it by hand.

The first two are built from the quartiles, so start there.

Quartiles and the five-number summary

The median splits ordered data into a lower half and an upper half. The quartiles split it into quarters:

$Q_1$ (the lower quartile) is the median of the lower half - a quarter of the way through the data.
$Q_2$ is the median itself - halfway through.
$Q_3$ (the upper quartile) is the median of the upper half - three quarters of the way through.

The method this page uses throughout (the standard HSC approach): first find the median. If there is an odd number of values, leave the middle value out of both halves. If there is an even number, split straight down the middle. Then $Q_1$ is the median of the lower half and $Q_3$ is the median of the upper half. Used consistently, this gives the quartiles your calculator's statistics mode reports.

Putting the pieces together, the five-number summary lists, in order:

\text{minimum}, \quad Q_1, \quad \text{median}, \quad Q_3, \quad \text{maximum}.

These five numbers carve the data into four quarters and are the exact inputs a box plot is drawn from. The number line below shows them for the ordered set $3, 5, 6, 8, 9, 11, 12, 14, 15, 17, 19$ .

The range

The range is the simplest measure of spread:

\text{range} = \text{maximum} - \text{minimum}.

For the set above, range $= 19 - 3 = 16$ . It is quick and tells you the full width of the data, but because it relies only on the two extreme values it is very sensitive: one freak value sends it shooting up, even if every other value is tightly packed. Use it for a rough sense of spread, and reach for the IQR when extremes might mislead.

The interquartile range

The interquartile range measures the spread of the central half of the data:

\text{IQR} = Q_3 - Q_1.

For the set above, IQR $= 15 - 6 = 9$ . Because it chops off the bottom quarter and the top quarter, the IQR ignores extreme values almost entirely - which is its great strength. When a data set has an unusually high or low value, the IQR describes the typical spread far better than the range does. It is also the basis of the $1.5 \times \text{IQR}$ outlier test you meet on the next dot point.

The standard deviation (population, $\sigma_n$ )

The standard deviation measures spread by averaging how far each value sits from the mean. Unlike the range and IQR, it uses every value in the data. A small standard deviation means the values cluster close to the mean; a large one means they are widely scattered. The smallest it can ever be is $0$ , which happens only when every value is identical.

In Mathematics Standard you are not asked to compute it by hand from the formula - you read it from the calculator's statistics mode. There are two versions, and you must pick the right one:

Population standard deviation $\sigma_n$ (sometimes shown as $\sigma x$ ) - this is the HSC convention and the one you quote unless a question says otherwise.
Sample standard deviation $s_{n-1}$ (sometimes $sx$ ) - slightly larger; not the value used in this course.

Finding the standard deviation on a calculator

The exam expects you to drive your calculator's statistics mode confidently. The steps are the same on every approved model, with only the key names differing:

Enter statistics mode (often MODE then STAT or SD), choosing single-variable (1-VAR) data.
Clear any previous data before you start - an uncleared list is the most common cause of a wrong answer.
Type each value, pressing the data-entry key (commonly DT, M+, or = inside a list) after each one. For repeated values, many calculators let you enter the value then a frequency.
Read off the statistics. Find the key for the population standard deviation, labelled $\sigma_n$ or $\sigma x$ . The mean is labelled $\bar{x}$ .
Round as asked, usually to two decimal places, and quote $\sigma_n$ (not $s_{n-1}$ ).

A quick sanity check: the standard deviation is never negative and is usually smaller than the range. If your calculator shows a value larger than the range, you have read the wrong key or mis-entered the data.

Comparing the spread of two data sets

A favourite exam task gives you two data sets, often with a similar mean, and asks which is more spread out or more consistent. The routine:

Compare the centres first (usually the mean or median) so you know whether you are genuinely comparing spread.
Compare a measure of spread - the standard deviation for the overall scatter, or the IQR if extreme values are in play. The data set with the smaller spread is the more consistent one.
Write the comparison in words, quoting the figures: "Class A's standard deviation ( $5.15$ ) is far smaller than Class B's ( $13.17$ ), so Class A's marks are more consistent and tightly clustered."

The diagram below compares two data sets with the same median but very different spread.

How exam questions ask about spread

The wording tells you which measure to reach for:

"Find the range" - subtract the smallest value from the largest.
"Find the five-number summary" - list minimum, $Q_1$ , median, $Q_3$ , maximum, in that order.
"Find the interquartile range / IQR" - find $Q_1$ and $Q_3$ , then subtract: $Q_3 - Q_1$ .
"Find the standard deviation" (with no other instruction) - read $\sigma_n$ from the calculator and round as asked.
"Which data set is more consistent / less variable / more spread out?" - compare a measure of spread (smaller spread $=$ more consistent), and quote the figures.
"Compare the two data sets" - comment on both centre (mean or median) and spread (IQR or standard deviation), in words, using the numbers.

Measures of spread across four data sets

Four tasks, one per skill: a five-number summary, an IQR from an even-sized set, a standard deviation from the calculator, and a spread comparison. Each follows the same plan - order the data, find what is asked, and state the result with its units.

Find a five-number summary

The marks (out of $20$ ) of $11$ students in a class quiz, in order, are $3, 5, 6, 8, 9, 11, 12, 14, 15, 17, 19$ . Find the five-number summary.

Find the minimum, maximum and median. The minimum is $3$ and the maximum is $19$ . With $11$ values the median is the $6$ th value:

\text{median} = 11.

Split into halves, leaving out the median. The lower half is $3, 5, 6, 8, 9$ and the upper half is $12, 14, 15, 17, 19$ .

Find the quartiles. Each half has $5$ values, so each quartile is the middle value of its half:

Q_1 = 6, \qquad Q_3 = 15.

State the summary. Minimum $3$ , $Q_1 = 6$ , median $11$ , $Q_3 = 15$ , maximum $19$ . (These are exactly the five values plotted on the number line earlier.)

Find the interquartile range

The daily numbers of customers at a market stall over $12$ days, in order, are $5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 21, 24$ . Find the interquartile range.

Split the even-sized set in half. With $12$ values the lower half is $5, 7, 8, 10, 11, 13$ and the upper half is $14, 16, 17, 19, 21, 24$ .

Find $Q_1$ and $Q_3$ . Each half has $6$ values, so each quartile is the average of its middle two:

Q_1 = \frac{8 + 10}{2} = 9, \qquad Q_3 = \frac{17 + 19}{2} = 18.

Subtract for the IQR.

\text{IQR} = 18 - 9 = 9 \text{ customers}.

So the middle half of the days saw a spread of $9$ customers. (For comparison the range is $24 - 5 = 19$ , more than double the IQR, because the busy and quiet extremes stretch it.)

Find the standard deviation from a calculator

A machine fills six bags with masses (in grams) $2, 4, 6, 8, 10, 12$ . Find the mean and the population standard deviation, correct to two decimal places.

Find the mean. The six values total $42$ , so

\bar{x} = \frac{42}{6} = 7 \text{ g}.

Enter the data in statistics mode. Put the calculator into single-variable statistics mode, clear the old list, then enter $2, 4, 6, 8, 10, 12$ , pressing the data-entry key after each value.

Read off $\sigma_n$ . The population standard deviation key gives

\sigma_n \approx 3.42 \text{ g}.

State the answer. The mean mass is $7$ g and the population standard deviation is $3.42$ g (to $2$ dp). Quote $\sigma_n$ , not $s_{n-1}$ , and note $\sigma_n$ ( $3.42$ ) is comfortably smaller than the range ( $12 - 2 = 10$ ), which is the sanity check that you read the right key.

Compare the spread of two data sets

Two baristas time, in seconds, how long seven coffees take. Barista P: $6, 7, 8, 9, 10, 11, 12$ . Barista Q: $3, 5, 7, 9, 11, 13, 15$ . (Each list has $7$ values.) Both have a mean of $9$ seconds. Compare their spread.

Confirm the centres match. Both lists have $7$ values totalling $63$ , so each mean is $63 \div 7 = 9$ s. With equal means, this is a fair comparison of spread.

Compare a measure of spread. The ranges are $12 - 6 = 6$ s for P and $15 - 3 = 12$ s for Q. Entering each list in the calculator gives population standard deviations

\sigma_P \approx 2.00 \text{ s}, \qquad \sigma_Q \approx 4.00 \text{ s}.

Write the comparison. Both baristas average $9$ seconds, but Barista Q's times are twice as spread out: Q's range ( $12$ s) and standard deviation ( $4.00$ s) are each double P's ( $6$ s and $2.00$ s). So Barista P is the more consistent, with times bunched close to the mean, while Barista Q is far more variable.

Final answers: the five-number summary is $3, 6, 11, 15, 19$ ; the IQR is $9$ customers; the bag-filling standard deviation is $3.42$ g; and Barista P is the more consistent, with half the spread of Barista Q.

Common traps

Forgetting to order the data first. Quartiles, the median and the range all assume the values are in ascending order. Sort before you do anything else.

Quoting the sample standard deviation $s_{n-1}$ instead of $\sigma_n$ . The HSC convention is the population value $\sigma_n$ (or $\sigma x$ ). The two keys sit side by side on the calculator and $s_{n-1}$ is always a little larger, so read carefully.

Mishandling the median when splitting the data: With an odd number of values, leave the middle value out of both halves before finding $Q_1$ and $Q_3$ ; with an even number, split straight down the middle. Mixing these up shifts both quartiles.
Confusing the IQR with the quartiles: The IQR is the single number $Q_3 - Q_1$ , not the pair of quartiles. Subtract.
Comparing two data sets on centre alone: "Which is more consistent?" is a spread question, not an average question. Quote the range, IQR or standard deviation, not just the mean.
Not clearing the calculator's old data: Leftover values from a previous question quietly corrupt the mean and standard deviation. Clear the list every time.

Exam technique

Always sort the data first, then write the median and quartiles down explicitly - those lines carry the method marks even if a later subtraction slips. State the method you use for the quartiles and stay consistent with it across every part. When a question just says "standard deviation", read $\sigma_n$ from the calculator and round exactly as asked (usually $2$ decimal places), and sanity-check that it is positive and smaller than the range. For a "compare" question, address both centre and spread, name the actual measure you used, and quote the figures, since markers reward a numerical comparison (" $\sigma_A = 5.15 < \sigma_B = 13.17$ ") over a vague "A is less spread out". If extreme values are present, prefer the IQR and say why.

Exam-style practice questions

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

2021 HSC-style3 marksThe ages of seven members of a sports club are

14

16

19

23

28

35

41

. (a) Find the median age. (b) Find the interquartile range. (c) Explain why the interquartile range may describe the spread of these ages better than the range.

Show worked answer →

Median: with $7$ ordered values the median is the $4$ th value, $23$ .

IQR: leaving out the median, the lower half is $14, 16, 19$ (median $16$ ) and the upper half is $28, 35, 41$ (median $35$ ), so $Q_1 = 16$ , $Q_3 = 35$ and IQR $= 35 - 16 = 19$ .

Explanation: the range ( $41 - 14 = 27$ ) depends only on the two extreme ages and is dragged up by the oldest member, whereas the IQR describes the spread of the middle half of the ages, so it is more resistant to a single unusually old or young member.

Markers award one mark for the median, one for a correctly worked IQR, and one for a clear statement that the IQR is less affected by extreme values than the range.

2022 HSC-style4 marksTwo baristas record the time (in seconds) to make a coffee over six orders. Barista X:

38

40

41

43

44

46

. Barista Y:

30

35

41

43

50

53

. Both have a mean of about

42

seconds. (a) Find the range for each barista. (b) State, with a reason, which barista is more consistent. (c) Describe what a larger standard deviation would tell you about a barista's times.

Show worked answer →

Ranges: Barista X has $46 - 38 = 8$ seconds; Barista Y has $53 - 30 = 23$ seconds.

More consistent: Barista X, because with the same mean ( $\approx 42$ s) X's times are far less spread out (range $8$ s versus $23$ s), so X's service times vary less from order to order.

Larger standard deviation: it would mean that barista's times are spread more widely around the mean, so the service is less predictable (some coffees much faster and some much slower than average).

Markers award marks for both correct ranges, for naming Barista X with a spread-based reason (not a mean-based one), and for linking a larger standard deviation to greater variability about the mean.

2023 HSC-style3 marksA data set of

10

values, in order, is

5

8

9

11

12

14

15

17

19

26

. (a) Find

Q_1

, the median and

Q_3

. (b) An outlier is any value above

Q_3 + 1.5 \times \text{IQR}

or below

Q_1 - 1.5 \times \text{IQR}

. Determine whether

26

is an outlier.

Show worked answer →

Quartiles: with $n = 10$ the median is the average of the $5$ th and $6$ th values, $(12 + 14) \div 2 = 13$ . The lower half $5, 8, 9, 11, 12$ gives $Q_1 = 9$ and the upper half $14, 15, 17, 19, 26$ gives $Q_3 = 17$ .

Outlier test: IQR $= 17 - 9 = 8$ , so the upper fence is $Q_3 + 1.5 \times \text{IQR} = 17 + 1.5 \times 8 = 17 + 12 = 29$ . Since $26 < 29$ , the value $26$ is not an outlier.

Markers award one mark for the median, one for both quartiles, and one for a correctly computed upper fence ( $29$ ) with the conclusion that $26$ is not an outlier.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksThe daily maximum temperatures (in degrees Celsius) for a week are

12

18

7

15

22

9

14

. (a) Find the range. (b) State what the range tells you about the data.

Show worked solution →

Part (a) - range is largest minus smallest. Reading the list, the largest value is $22$ and the smallest is $7$ , so

\text{range} = 22 - 7 = 15 \text{ degrees}

Part (b) - what it tells you. The range measures the total spread: it says the hottest and coldest days that week were $15$ degrees apart. (The range uses only the two extreme values, so a single unusually hot or cold day would change it a lot.)

foundation3 marksA set of nine data values, written in order, is

4

7

8

11

13

15

18

21

24

. Find the five-number summary (minimum,

Q_1

, median,

Q_3

, maximum).

Show worked solution →

Find the minimum, maximum and median: The data is already in order. The minimum is $4$ and the maximum is $24$ . With $9$ values the median is the $5$ th value (the middle one), so the median is $13$ .
Split into a lower and an upper half, excluding the median: Removing the middle value $13$ leaves a lower half $4, 7, 8, 11$ and an upper half $15, 18, 21, 24$ .
$Q_1$ is the median of the lower half: With four values, $Q_1$ is the average of the middle two:

Q_1 = \frac{7 + 8}{2} = 7.5

$Q_3$ is the median of the upper half.

Q_3 = \frac{18 + 21}{2} = 19.5

State the five-number summary. Minimum $4$ , $Q_1 = 7.5$ , median $13$ , $Q_3 = 19.5$ , maximum $24$ .

core3 marksThe number of goals scored by a netball team in eight games was

3

5

8

9

12

14

17

20

. (a) Find

Q_1

and

Q_3

. (b) Hence find the interquartile range.

Show worked solution →

Part (a) - split the ordered data in half. There are $8$ values (an even number), so the data splits cleanly into a lower half $3, 5, 8, 9$ and an upper half $12, 14, 17, 20$ .

$Q_1$ is the median of the lower half:

Q_1 = \frac{5 + 8}{2} = 6.5

$Q_3$ is the median of the upper half:

Q_3 = \frac{14 + 17}{2} = 15.5

Part (b) - the IQR is $Q_3 - Q_1$ .

\text{IQR} = 15.5 - 6.5 = 9

so the interquartile range is $9$ goals. (The IQR is the spread of the middle $50\%$ of the games, ignoring the lowest and highest quarters.)

core3 marksThe reaction times (in hundredths of a second) of eleven people, in order, are

2

4

5

7

8

10

12

13

15

18

20

. Find the interquartile range.

Show worked solution →

Find the median first. With $11$ values the median is the $6$ th value (the middle one):

\text{median} = 10

Split into halves, leaving out the median. Below the median sit $2, 4, 5, 7, 8$ and above it sit $12, 13, 15, 18, 20$ .

Find $Q_1$ and $Q_3$ as the medians of those halves. Each half has $5$ values, so each quartile is the middle one of its half:

Q_1 = 5, \qquad Q_3 = 15

Compute the IQR.

\text{IQR} = Q_3 - Q_1 = 15 - 5 = 10

so the interquartile range is $10$ . (Because the IQR throws away the bottom and top quarters, it is unaffected by the extreme value $20$ , unlike the range.)

core3 marksA small data set is

4

6

8

10

12

. (a) Find the mean. (b) Using your calculator's statistics mode, find the population standard deviation

\sigma_n

, correct to two decimal places.

Show worked solution →

Part (a) - the mean. Add the values and divide by how many there are:

\bar{x} = \frac{4 + 6 + 8 + 10 + 12}{5} = \frac{40}{5} = 8

Part (b) - the standard deviation from the calculator. Put the calculator into statistics (STAT / SD) mode, clear any old data, type each value pressing the data-entry key after each ( $4$ , $6$ , $8$ , $10$ , $12$ ), then read off the population standard deviation, the key labelled $\sigma_n$ (or $\sigma x$ ):

\sigma_n \approx 2.83

so the population standard deviation is $2.83$ (correct to two decimal places). (Be sure to read $\sigma_n$ , not the sample value $s_{n-1}$ , which is slightly larger; $\sigma_n$ is the HSC convention.)

exam5 marksTwo Year 11 classes sat the same test (marks out of

80

). Class A scored

52

55

58

60

62

65

68

. Class B scored

40

48

55

60

66

73

81

. (a) Find the range of each class. (b) Using a calculator, find the population standard deviation

\sigma_n

of each class, correct to two decimal places. (c) Both classes have a mean of about

60

. Compare the two classes' spread, referring to your figures.

Show worked solution →

Part (a) - ranges. Range is largest minus smallest:

\text{Class A: } 68 - 52 = 16, \qquad \text{Class B: } 81 - 40 = 41

Part (b) - standard deviations from the calculator. Enter each class's marks in statistics mode and read $\sigma_n$ :

\sigma_A \approx 5.15, \qquad \sigma_B \approx 13.17

Part (c) - compare the spread. Both classes centre on a mean of about $60$ , so the comparison is about spread, not centre. Every measure of spread is much larger for Class B: its range ( $41$ ) is more than double Class A's ( $16$ ), and its standard deviation ( $13.17$ ) is more than double Class A's ( $5.15$ ). So Class B's marks are far more spread out, while Class A's marks are tightly bunched near the mean. In plain terms, Class A performed consistently and Class B's results were much more variable. (Because the means are equal, the smaller standard deviation - Class A - identifies the more consistent class.)

exam6 marksThe masses (in kg) of ten parcels are

12

15

15

18

20

22

25

27

30

36

. (a) Find the five-number summary. (b) Find the range and the interquartile range. (c) Using a calculator, find the mean and the population standard deviation

\sigma_n

, correct to two decimal places. (d) The courier flags any parcel heavier than

Q_3 + 1.5 \times \text{IQR}

as oversized. Find this cut-off mass and state whether the

36

kg parcel is flagged.

Show worked solution →

Part (a) - five-number summary. The data is already in order; $n = 10$ . Minimum $12$ , maximum $36$ . The median is the average of the $5$ th and $6$ th values:

\text{median} = \frac{20 + 22}{2} = 21

The lower half is $12, 15, 15, 18, 20$ and the upper half is $22, 25, 27, 30, 36$ , so

Q_1 = 15, \qquad Q_3 = 27

The five-number summary is: minimum $12$ , $Q_1 = 15$ , median $21$ , $Q_3 = 27$ , maximum $36$ .

Part (b) - range and IQR.

\text{range} = 36 - 12 = 24, \qquad \text{IQR} = 27 - 15 = 12

Part (c) - mean and standard deviation. Adding the ten masses gives $220$ , so the mean is

\bar{x} = \frac{220}{10} = 22 \text{ kg}

Entering the data in statistics mode and reading the population standard deviation,

\sigma_n \approx 7.16 \text{ kg}

Part (d) - the oversized cut-off. Apply the upper-fence formula with $Q_3 = 27$ and IQR $= 12$ :

Q_3 + 1.5 \times \text{IQR} = 27 + 1.5 \times 12 = 27 + 18 = 45 \text{ kg}

The $36$ kg parcel is below $45$ kg, so it is not flagged as oversized. (Even the heaviest parcel sits inside the fence, so this data set has no high outlier by the $1.5 \times$ IQR rule.)

What this dot point is asking

The answer

Quartiles and the five-number summary

The range

The interquartile range

The standard deviation (population, σn\sigma_nσn​)

Finding the standard deviation on a calculator

Comparing the spread of two data sets

How exam questions ask about spread

Exam-style practice questions

Practice questions

Related dot points

The standard deviation (population, $\sigma_n$ )