NSWMaths Standard 2Syllabus dot point

How is a five-number summary turned into a box plot, and how do parallel box plots let you compare two groups by centre, spread, skew and outliers?

Construct and interpret box-and-whisker plots and use them, including parallel (side-by-side) box plots, to compare data sets in terms of centre, spread, skewness and outliers

A focused answer to the HSC Maths Standard 2 dot point on box-and-whisker plots. Building a box plot from the five-number summary, flagging an outlier with the 1.5 times IQR rule, drawing parallel box plots, and comparing two groups by centre, spread, skew and outliers, with worked Australian examples.

Generated by Claude Opus 4.815 min answerUpdated 2026-06-21

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

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What this dot point is asking

NESA wants you to turn a five-number summary into a box-and-whisker plot, and to read one the other way, off the page. You need to draw the box from the quartiles, mark the median, run the whiskers out to the extreme values, and flag any outlier as a separate dot using the $1.5 \times \text{IQR}$ rule. Then you need to put two box plots on the same scale (parallel box plots) and compare the groups by centre, spread, skew and outliers. The skill being tested is reading a distribution at a glance: a box plot is a five-number summary made visual, and comparing two of them is the quickest fair way to say which group is higher, which is more spread out, and which is lopsided.

The answer

A box-and-whisker plot is a picture of the five-number summary drawn against a number line. The box spans the middle half of the data, from the lower quartile $Q_1$ to the upper quartile $Q_3$ , with a line inside it at the median. From each end of the box a whisker reaches out to the smallest and largest values. Because the box covers $Q_1$ to $Q_3$ , its width is the interquartile range, and the whole picture lets you read centre, spread and shape without seeing a single raw number twice.

What the five pieces mean

Read the plot from left to right and each piece has a fixed meaning:

the left whisker end is the minimum, the smallest value,
the left edge of the box is the lower quartile $Q_1$ , one quarter of the way through the sorted data,
the line inside the box is the median, the middle value,
the right edge of the box is the upper quartile $Q_3$ , three quarters of the way through,
the right whisker end is the maximum, the largest value.

Each section of the plot holds about a quarter of the data: from the minimum to $Q_1$ , from $Q_1$ to the median, from the median to $Q_3$ , and from $Q_3$ to the maximum. A short section means the data is bunched there; a long section means it is spread out there. That is why an uneven plot tells you the shape at a glance.

Constructing a box plot from a five-number summary

The five-number summary comes first; the picture is just a careful plot of it. The routine is always the same:

Draw and scale a number line that comfortably covers the minimum and maximum, with even gaps.
Draw the box from $Q_1$ to $Q_3$ above the line.
Draw the median line inside the box, at the median value.
Draw the whiskers from each end of the box out to the minimum and the maximum (unless there is an outlier; see below).

The single thing that earns or loses marks is alignment: every mark must sit at its correct value on the scale. If the scale runs $0$ to $30$ and the box should reach $20$ , its right edge must line up exactly with $20$ on the axis, not "about there".

Flagging an outlier

When a data set has a value that sits far from the rest, you do not let the whisker stretch all the way out to it. Instead you test it with the $1.5 \times \text{IQR}$ rule and, if it fails, draw it as a separate dot.

The two fences are

\text{lower fence} = Q_1 - 1.5 \times \text{IQR}, \qquad \text{upper fence} = Q_3 + 1.5 \times \text{IQR}.

Any value below the lower fence or above the upper fence is an outlier. On the plot you mark each outlier as a dot at its value, and the whisker on that side stops at the most extreme value that is inside the fence. For example, with $Q_1 = 23$ , $Q_3 = 29$ and $\text{IQR} = 6$ , the upper fence is $29 + 1.5 \times 6 = 38$ , so a value of $48$ is an outlier: it becomes a dot at $48$ and the right whisker stops at $30$ (the largest value not over the fence).

Parallel box plots: comparing two groups

A parallel box plot (also called a side-by-side box plot) draws two or more box plots one above the other against the same number line. Sharing one scale is what makes the comparison fair and instant: you can see at a glance which group is centred higher and which is more spread out, because the boxes are measured against identical axis marks.

When you compare parallel box plots, always cover the same three things, in this order:

Centre. Compare the medians. The group with the higher median line is centred higher. (Here Class B's median of $75$ is well to the right of Class A's $60$ .)
Spread. Compare the IQRs (box widths) and the ranges (whisker to whisker). A wider box or longer whiskers means a more spread out, less consistent group.
Skew (shape). Look at where the median sits in the box and how the whiskers compare. A median near the left of the box with a long right whisker is positive skew; near the right with a long left whisker is negative skew; central with even whiskers is symmetric.

Then mention any outliers shown as separate dots. The exam wants a genuine comparison ("Class B is centred about $15$ marks higher than Class A"), not two separate descriptions.

Reading shape from a box plot

The position of the median inside the box is the quickest read on shape:

Symmetric: the median sits near the middle of the box and the two whiskers are about equal in length.
Positively skewed: the median sits toward the left ( $Q_1$ ) side, and the right whisker is longer; the long tail is at the high values.
Negatively skewed: the median sits toward the right ( $Q_3$ ) side, and the left whisker is longer; the long tail is at the low values.

The whisker that is longer points the way the data is skewed.

How exam questions ask about box plots

The wording varies, but each version maps to one of the methods above:

"Construct / draw a box plot for this data." Find the five-number summary, scale a number line, draw the box from $Q_1$ to $Q_3$ , mark the median, and run the whiskers to the min and max (stopping at the fence if there is an outlier).
"Show that ... is an outlier" or "Are there any outliers?" Compute the IQR and both fences, compare the suspect value, and state the conclusion. The marks are in the fence calculation, not the yes/no.
"Describe the shape of the distribution." Read the median's position in the box and the whisker lengths: symmetric, positively skewed or negatively skewed.
"Compare the two data sets" or "... using the parallel box plots." Compare centre (medians), spread (IQR and range) and skew, with figures, then note outliers. Compare, do not just describe each.
"What does the box plot tell you about the middle $50\%$ ?" That is the box, from $Q_1$ to $Q_3$ , a span equal to the IQR.

Box plots from five-number summaries

Four tasks, one for each skill: build a box plot from a list, flag an outlier on a box plot, and compare two groups with parallel box plots. Each starts from the five-number summary and aligns every mark to the scale.

Draw a box plot from a list of values

The lengths (in cm) of $11$ fish caught off a wharf are $3, 5, 6, 8, 9, 11, 13, 14, 16, 18, 19$ . Find the five-number summary and describe the box plot on a scale from $0$ to $20$ .

Find the median of the $11$ sorted values. The median is the $6$ th value:

\text{median} = 11 \text{ cm}

Find the quartiles from the halves. The lower half (below the median) is $3, 5, 6, 8, 9$ and the upper half is $13, 14, 16, 18, 19$ , so

Q_1 = 6 \text{ cm}, \qquad Q_3 = 16 \text{ cm}.

Write the five-number summary.

\text{min} = 3, \quad Q_1 = 6, \quad \text{median} = 11, \quad Q_3 = 16, \quad \text{max} = 19.

Describe the plot. On a $0$ to $20$ scale, draw the box from $6$ to $16$ with the median line at $11$ , then a left whisker from $6$ out to $3$ and a right whisker from $16$ out to $19$ . The IQR is $16 - 6 = 10$ cm. The median $11$ sits a little right of centre in the box, and the whiskers are similar in length, so the distribution is roughly symmetric.

Flag an outlier on a box plot

The weekly water use (in kL) of $11$ households is $31, 34, 36, 38, 40, 42, 44, 46, 48, 50, 72$ . Test the largest value and explain how the box plot changes.

Find the median and quartiles. The median (the $6$ th value) is $42$ . The lower half is $31, 34, 36, 38, 40$ giving $Q_1 = 36$ , and the upper half is $44, 46, 48, 50, 72$ giving $Q_3 = 48$ .

Compute the IQR and the upper fence.

\text{IQR} = 48 - 36 = 12, \qquad \text{upper fence} = 48 + 1.5 \times 12 = 48 + 18 = 66.

Test the value $72$ . Since $72 > 66$ , the value $72$ is an outlier.

Describe the plot. Draw the box from $36$ to $48$ with the median at $42$ . The left whisker runs to the minimum $31$ . On the right, instead of a whisker to $72$ , draw a separate dot at $72$ , and stop the right whisker at $50$ (the largest value not above the fence). Labelling the dot with its value, $72$ , makes the outlier unmistakable.

Compare two classes with parallel box plots

Two classes sit the same trial. Class A's marks are $40, 46, 50, 54, 58, 60, 63, 66, 70, 76, 84$ and Class B's are $55, 58, 65, 70, 73, 75, 80, 82, 85, 90, 95$ . Compare them.

Find both five-number summaries. Each list has $11$ values, so the median is the $6$ th value.

\text{Class A: } 40, \; 50, \; 60, \; 70, \; 84.

\text{Class B: } 55, \; 65, \; 75, \; 85, \; 95.

Compare centre: Class B's median ( $75$ ) is $15$ marks above Class A's ( $60$ ), so a typical Class B student scored about $15$ marks higher.
Compare spread: Both have the same IQR: Class A is $70 - 50 = 20$ and Class B is $85 - 65 = 20$ . The ranges are close too ( $84 - 40 = 44$ for A, $95 - 55 = 40$ for B). So the two classes are equally spread; they differ in level, not in consistency.
Compare skew: In both plots the median sits in the middle of the box with fairly even whiskers, so both distributions are roughly symmetric.
Answer: Class B is centred about $15$ marks higher than Class A, with the same spread (IQR $= 20$ ) and a similar, roughly symmetric shape. Class B performed better overall, and just as consistently.

Read a five-number summary off a box plot

A box plot of bus waiting times (in minutes), drawn on a scale from $0$ to $30$ , has its left whisker at $2$ , a box from $7$ to $17$ with the median line at $9$ , and its right whisker at $26$ . Read off the summary and describe the shape.

Read each feature. The whisker ends give min $= 2$ and max $= 26$ ; the box edges give $Q_1 = 7$ and $Q_3 = 17$ ; the line inside is the median $= 9$ .

State the summary and spreads.

\text{min} = 2, \; Q_1 = 7, \; \text{median} = 9, \; Q_3 = 17, \; \text{max} = 26, \quad \text{IQR} = 17 - 7 = 10.

Describe the shape. The median $9$ sits close to the left edge of the box ( $Q_1 = 7$ ) and the right whisker (to $26$ ) is much longer than the left (to $2$ ). The long tail is on the high side, so the distribution is positively skewed: most waits are short, but a few are much longer.

Common traps

Stretching a whisker out to an outlier: Once a value fails the $1.5 \times \text{IQR}$ test it is drawn as a separate dot, and the whisker stops at the last value inside the fence. A whisker that reaches a far-flung point is the tell-tale sign the outlier was not tested.
Saying "yes it is an outlier" without showing the fence: The marks live in the fence calculation $Q_3 + 1.5 \times \text{IQR}$ (or the lower fence) and the comparison, not the conclusion.
Reading skew backwards: The data is skewed toward the longer whisker. A long right whisker is positive skew (high tail); a long left whisker is negative skew (low tail).
Comparing parallel box plots on different scales: Parallel box plots must share one axis. If you sketch them against different scales the visual comparison is meaningless.
Describing two groups instead of comparing them: "Compare" wants sentences that put the two side by side ("B's median is $15$ higher"), with figures, not two separate descriptions.
Mis-aligning a mark with the scale: Every value must sit exactly at its number on the axis. A box edge that should reach $20$ but is drawn near $18$ loses the accuracy mark.

Exam technique

Always find and write down the full five-number summary before you draw, then plot each of the five marks against the scale, checking the box edges and median line land on their exact values. State the IQR explicitly: it is the box width and you need it for the outlier fences. When a value looks extreme, do the fence test in full ( $Q_1 - 1.5 \times \text{IQR}$ and $Q_3 + 1.5 \times \text{IQR}$ ) and only then decide on the dot; show that line of working. For parallel box plots, compare in the fixed order centre, spread, skew, then outliers, and quote numbers from the plots. Finish a "compare" answer with one sentence of plain-English judgement (which group is higher and which is more consistent), because that is where the top mark sits.

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.

2022 HSC-style3 marksThe heights of

11

plants (in cm) are

12, 15, 16, 18, 19, 20, 21, 22, 23, 24, 40

. (a) Find the five-number summary. (b) Show that

40

is an outlier using the

1.5 \times \text{IQR}

rule.

Show worked answer →

Five-number summary: median is the $6$ th value $= 20$ ; lower half $12, 15, 16, 18, 19$ gives $Q_1 = 16$ ; upper half $21, 22, 23, 24, 40$ gives $Q_3 = 23$ . So min $= 12$ , $Q_1 = 16$ , median $= 20$ , $Q_3 = 23$ , max $= 40$ .

Outlier test: IQR $= 23 - 16 = 7$ ; upper fence $= 23 + 1.5 \times 7 = 23 + 10.5 = 33.5$ . Since $40 > 33.5$ , the value $40$ is an outlier.

Markers award one mark for a correct median and quartiles, one for the full five-number summary, and one for the fence calculation with the comparison and conclusion. A bare "yes it is an outlier" with no fence shown does not earn the outlier mark.

2021 HSC-style4 marksParallel box plots show the daily maximum temperature (in degrees Celsius) at two towns over summer. Town A: min

14

Q_1 = 22

, median

= 26

Q_3 = 30

, max

= 38

. Town B: min

20

Q_1 = 24

, median

= 25

Q_3 = 27

, max

= 32

. Compare the temperatures at the two towns, referring to centre, spread and skew.

Show worked answer →

Centre: the medians are close ( $26$ for Town A, $25$ for Town B), so a typical day is about the same temperature in both towns.

Spread: Town A has range $38 - 14 = 24$ and IQR $30 - 22 = 8$ ; Town B has range $32 - 20 = 12$ and IQR $27 - 24 = 3$ . Town A's temperatures are much more spread out, so Town B is far more consistent.

Skew: Town A is roughly symmetric (median near the middle of the box). Town B is positively skewed (the median $25$ sits close to $Q_1$ , with a longer upper whisker to $32$ ).

Markers reward a comparison of medians, a comparison of spread using range or IQR with numbers, and a comment on shape. The strongest answers compare (not just describe each town separately) and quote figures from the plots.

2023 HSC-style3 marksA box plot of finishing times for a charity fun run has its median line very close to the left edge of the box, with a short left whisker and a long right whisker. Describe the shape of the distribution and explain, in context, what this tells you about the finishing times.

Show worked answer →

Shape: the distribution is positively skewed, because the long tail (whisker) is on the higher (right) side and the median sits near the lower quartile.

Interpretation: most runners finished in a fairly tight band of faster times (the compact left side and box), but a smaller number took much longer to finish, stretching the upper tail. So the typical (median) time is on the faster side, while a few slow finishers pull the maximum well to the right.

Markers award one mark for naming positive skew, one for linking it to the long upper whisker / median near $Q_1$ , and one for a sensible context interpretation (a cluster of faster times with a few slow finishers).

Practice questions

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

foundation2 marksA data set has the five-number summary minimum

= 4

Q_1 = 9

, median

= 15

Q_3 = 22

, maximum

= 28

. Describe, in order, the five marks you would draw on a box plot for these values.

Show worked solution →

Recall what a box plot shows. A box-and-whisker plot marks the five-number summary against a number-line scale: the two whisker ends, the two ends of the box, and the median line inside the box.

List the five marks in order from left to right.

left whisker end at the minimum, $4$ ,
left edge of the box at the lower quartile $Q_1$ , $9$ ,
the median line inside the box, $15$ ,
right edge of the box at the upper quartile $Q_3$ , $22$ ,
right whisker end at the maximum, $28$ .

The box runs from $9$ to $22$ and holds the middle half of the data, with the median line drawn at $15$ . The left whisker runs from $4$ to $9$ and the right whisker from $22$ to $28$ . (Check: the five numbers increase from left to right, $4 < 9 < 15 < 22 < 28$ , as they must on a box plot.)

foundation3 marksSeven players scored these goals over a season:

5, 8, 9, 12, 15, 18, 21

. (a) Find the five-number summary. (b) Find the interquartile range.

Show worked solution →

The list is already in order, so read off the median first. With $7$ values the median is the $4$ th value:

\text{median} = 12

Find the quartiles from the two halves. Splitting either side of the median, the lower half is $5, 8, 9$ and the upper half is $15, 18, 21$ . The quartile is the middle of each half:

Q_1 = 8, \qquad Q_3 = 18

Part (a) five-number summary. The smallest value is $5$ and the largest is $21$ , so the summary is

\text{min} = 5, \quad Q_1 = 8, \quad \text{median} = 12, \quad Q_3 = 18, \quad \text{max} = 21.

Part (b) interquartile range. The IQR is the spread of the middle box:

\text{IQR} = Q_3 - Q_1 = 18 - 8 = 10.

(Check: the five numbers increase in order and $Q_1$ and $Q_3$ sit either side of the median, as required.)

foundation2 marksA box plot is drawn on a scale from

0

50

. Its left whisker starts at

10

, the box runs from

18

34

with the median line at

24

, and the right whisker ends at

46

. (a) Write down the five-number summary. (b) State the range and the interquartile range.

Show worked solution →

Read each feature off the plot. The whisker ends give the minimum and maximum, the box edges give the quartiles, and the line inside the box is the median.

Part (a) five-number summary.

\text{min} = 10, \quad Q_1 = 18, \quad \text{median} = 24, \quad Q_3 = 34, \quad \text{max} = 46.

Part (b) range and IQR. The range is the full whisker-to-whisker spread and the IQR is the width of the box:

\text{range} = 46 - 10 = 36, \qquad \text{IQR} = 34 - 18 = 16.

(Check: the box width $16$ is smaller than the full range $36$ , which is always true because the box holds only the middle half.)

core4 marksThe masses (in kg) of

11

parcels are

10, 12, 13, 15, 16, 18, 20, 21, 23, 24, 26

. (a) Find the five-number summary. (b) Find the interquartile range. (c) Describe the steps you would take to draw the box plot on a scale from

0

30

Show worked solution →

The data is already sorted, so find the median. With $11$ values the median is the $6$ th value:

\text{median} = 18 \text{ kg}

Find the quartiles. Excluding the median, the lower half is $10, 12, 13, 15, 16$ and the upper half is $20, 21, 23, 24, 26$ . The middle of each half is

Q_1 = 13 \text{ kg}, \qquad Q_3 = 23 \text{ kg}.

Part (a) five-number summary.

\text{min} = 10, \quad Q_1 = 13, \quad \text{median} = 18, \quad Q_3 = 23, \quad \text{max} = 26.

Part (b) interquartile range.

\text{IQR} = 23 - 13 = 10 \text{ kg}.

Part (c) drawing it. Draw a number line from $0$ to $30$ with even gaps. Above the line, draw a box from $13$ to $23$ and a vertical line inside it at the median $18$ . From the left edge of the box draw a whisker out to $10$ , and from the right edge draw a whisker out to $26$ . Label the scale. (Check: $13 < 18 < 23$ confirms the median sits inside the box, and the whiskers reach the smallest and largest masses.)

core5 marksThe reaction times (in hundredths of a second) of

11

people are

20, 22, 23, 24, 25, 26, 27, 28, 29, 30, 48

. (a) Find the five-number summary. (b) Use the

1.5 \times \text{IQR}

rule to test the largest value for being an outlier. (c) Explain how the outlier is shown on a box plot.

Show worked solution →

Find the median of the $11$ sorted values. The median is the $6$ th value:

\text{median} = 26

Find the quartiles. The lower half (excluding the median) is $20, 22, 23, 24, 25$ and the upper half is $27, 28, 29, 30, 48$ , so

Q_1 = 23, \qquad Q_3 = 29, \qquad \text{IQR} = 29 - 23 = 6.

Part (a) five-number summary.

\text{min} = 20, \quad Q_1 = 23, \quad \text{median} = 26, \quad Q_3 = 29, \quad \text{max} = 48.

Part (b) outlier test. An upper outlier lies above the upper fence $Q_3 + 1.5 \times \text{IQR}$ :

29 + 1.5 \times 6 = 29 + 9 = 38.

Since $48 > 38$ , the value $48$ is an outlier.

Part (c) showing it. On the box plot the outlier $48$ is drawn as a separate dot beyond the right whisker. The right whisker then stops at the largest value that is not an outlier, which is $30$ , rather than reaching $48$ . (Check: the lower fence is $23 - 9 = 14$ and the smallest value $20$ is above it, so there is no lower outlier.)

exam5 marksParallel box plots compare the trial marks of two classes, North and South. North has five-number summary

42, 52, 60, 70, 80

. South has five-number summary

50, 60, 70, 81, 92

. (a) Compare the two classes by median. (b) Compare them by spread (range and IQR). (c) State, with a reason, which class performed better overall.

Show worked solution →

Compare the centres first. The median is the fair measure of a typical mark.

\text{North median} = 60, \qquad \text{South median} = 70.

South's median is $10$ marks higher, so a typical South student scored higher than a typical North student.

Part (a) median comparison. South has the higher median ( $70$ versus $60$ ), so South is centred about $10$ marks above North.

Part (b) spread comparison. Compare both the full range and the middle-half IQR.

\text{North range} = 80 - 42 = 38, \qquad \text{South range} = 92 - 50 = 42.

\text{North IQR} = 70 - 52 = 18, \qquad \text{South IQR} = 81 - 60 = 21.

South has a slightly larger range and IQR, so South's marks are a little more spread out (less consistent) than North's.

Part (c) overall judgement. South performed better overall: its median is $10$ marks higher and its whole box sits to the right of North's, so South scored higher across the board. The trade-off is that South's marks were slightly less consistent (larger IQR), but the clear lift in the centre outweighs the small extra spread. (Check: both medians lie inside their own boxes, $52 < 60 < 70$ and $60 < 70 < 81$ , as required.)

exam6 marksA cafe records the number of customers in the hour after opening on

11

weekdays and

11

weekend days. Weekdays:

4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16

. Weekends:

8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 45

. (a) Find the five-number summary for each group. (b) Test the largest weekend value for being an outlier. (c) Compare the two groups by centre, spread and skew, ready to draw as parallel box plots.

Show worked solution →

Weekdays: find the median and quartiles: With $11$ sorted values the median is the $6$ th value, $10$ . The lower half is $4, 6, 7, 8, 9$ and the upper half is $11, 12, 13, 14, 16$ , so $Q_1 = 7$ and $Q_3 = 13$ .
Weekends: find the median and quartiles: The median is the $6$ th value, $18$ . The lower half is $8, 10, 12, 14, 16$ and the upper half is $20, 22, 24, 26, 45$ , so $Q_1 = 12$ and $Q_3 = 24$ .
Part (a) five-number summaries

\text{Weekday: } 4, \; 7, \; 10, \; 13, \; 16.

\text{Weekend: } 8, \; 12, \; 18, \; 24, \; 45.

Part (b) outlier test on the weekend maximum. The weekend IQR is $24 - 12 = 12$ , so the upper fence is

24 + 1.5 \times 12 = 24 + 18 = 42.

Since $45 > 42$ , the value $45$ is an outlier and is drawn as a separate dot; the weekend right whisker then stops at $26$ .

Part (c) compare by centre, spread and skew.

Centre: the weekend median ( $18$ ) is well above the weekday median ( $10$ ), so weekends are busier on a typical day.
Spread: the weekday IQR is $13 - 7 = 6$ and the weekend IQR is $12$ , so weekend counts are roughly twice as spread out; weekends are less predictable.
Skew: the weekday plot is fairly symmetric (the median $10$ sits near the middle of the box, with similar whiskers). The weekend plot is positively skewed: the upper whisker and the outlier $45$ stretch far to the right while the lower part is compact.

(Check: each median lies inside its own box, $7 < 10 < 13$ and $12 < 18 < 24$ , and the weekend lower fence $12 - 18 = -6$ means no value can be a low outlier, consistent with the compact left tail.)

What this dot point is asking

The answer

What the five pieces mean

Constructing a box plot from a five-number summary

Flagging an outlier

Parallel box plots: comparing two groups

Reading shape from a box plot

How exam questions ask about box plots

Exam-style practice questions

Practice questions

Related dot points