Boxplots and the five-number summary: VCE General Mathematics Unit 3 Data analysis
“Summary of the distributions of numerical variables: the five-number summary and boxplots (including the use of the lower fence (Q1 - 1.5 x IQR) and upper fence (Q3 + 1.5 x IQR) to identify and display possible outliers), and the use of parallel boxplots to compare distributions”
The five-number summary is the minimum, , median, and maximum; the box runs from to with a line at the median, so it holds the middle of the data. Possible outliers are values below or above ; they are plotted as separate points, and each whisker stops at the most extreme value that is not an outlier. Read shape from the median's position and the whisker lengths, and compare parallel boxplots by their medians (centre) and IQRs (spread), always quoting the values.
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What this dot point is asking
In "Investigating data distributions" the Study Design lists "summary of the distributions of numerical variables; the five-number summary and boxplots (including the use of the lower fence () and upper fence () to identify and display possible outliers)", and in "Investigating association between two variables" it lists parallel boxplots "and their use in identifying and describing associations between a numerical variable and a categorical variable". You need to:
- find the five-number summary (minimum, , median, , maximum) from raw data, a dot plot, a stem plot or CAS;
- calculate the IQR and the fences, and identify possible outliers;
- draw a boxplot correctly, with whiskers ending at the last value inside the fences;
- read a boxplot: quartiles, range, IQR, the percentage of data in each section, and the shape;
- compare distributions with parallel boxplots, quoting medians and IQRs.
Boxplot questions appear in both examinations every year. In 2025 Examination 1 Questions 3 and 6 asked students to compare two boxplots () and count lower outliers (); in 2025 Examination 2 Question 2 asked for the range () and upper fence (); in 2023 Examination 1 Question 2 the shape of a distribution depended on the upper fence ().
The answer
The median and quartiles
Put the data in order. For values:
- the median is in position (the middle value if is odd, the mean of the two middle values if is even);
- the lower half is the values below the median position and the upper half the values above it. When is odd, the median itself is left out of both halves;
- is the median of the lower half and is the median of the upper half.
| median position | lower half | |||
|---|---|---|---|---|
| 11 | 6th | 1st to 5th | 3rd | 9th |
| 12 | mean of 6th and 7th | 1st to 6th | mean of 3rd and 4th | mean of 9th and 10th |
| 15 | 8th | 1st to 7th | 4th | 12th |
| 40 | mean of 20th and 21st | 1st to 20th | mean of 10th and 11th | mean of 30th and 31st |
This is the method used by CAS one-variable statistics, so your hand answers and your calculator should agree. With a dot plot or frequency table, count cumulatively to the positions you need.
The five-number summary is the minimum, , the median , and the maximum. The range is maximum minimum and the interquartile range is , the spread of the middle of the data.
Fences and possible outliers
A data value less than the lower fence or greater than the upper fence is a possible outlier. A value exactly on a fence is not an outlier. The fences are a calculation to decide which points to plot separately; they are not drawn on a finished boxplot.
Drawing the boxplot
- Draw a scale and label it with the variable and units.
- Draw a box from to with a vertical line at the median.
- Plot each possible outlier as a separate point (a dot or a cross).
- Draw each whisker from the box to the most extreme data value that is not an outlier. Without outliers, the whiskers run to the minimum and maximum.
The runners' times from 2023 Examination 1 (40 runners: three at 134 s, eight at 135 s, eleven at 136 s, six at 137 s, four at 138 s, two each at 139 s and 140 s, and one each at 141, 142, 143 and 146 s) give this boxplot.
Reading a boxplot
Each of the four sections (lower whisker, lower half of the box, upper half of the box, upper whisker) holds about 25% of the data. So about of values lie in the box, about lie above , and about lie above . For a sample of 200, about 50 values lie in each section.
The shape follows from the positions of the median and the whiskers:
- positively skewed: the median is closer to and the upper whisker (or the upper tail including outliers) is longer;
- negatively skewed: the median is closer to and the lower whisker is longer;
- approximately symmetric: the median is near the centre of the box and the whiskers are of similar length.
Describe a distribution with shape (including any outliers), centre (median) and spread (IQR and range).
Comparing parallel boxplots
Parallel boxplots share one scale, one boxplot per group, so they show whether a numerical variable is associated with a categorical one. Compare:
- centre: the medians;
- spread: the IQRs (and the ranges);
- shape and outliers, if they differ.
A comparison earns marks only when it states the direction and quotes values: "The median life expectancy for Sample T (about 76.5 years) is higher than for Sample H (about 67.2 years)." To claim an association, say that the value of the numerical variable tends to differ by group (for example, life expectancy tends to be higher for T countries), supported by the medians.
Useful facts for multiple choice: "all values in T exceed " is a statement about T's minimum; "at least of T exceed H's median" is true whenever T's is above H's median; and quartiles can be compared across groups directly ( of H against of T).
. Range max min. Lower fence ; upper fence . Outlier: less than the lower fence or greater than the upper fence. Each of the four sections of a boxplot holds about of the data.
From a dot plot to a boxplot (2023 Examination 1 Questions 1 and 2)
- Step 1: cumulative counts
- 134: 3; 135: 11; 136: 22; 137: 28; 138: 32; 139: 34; 140: 36; 141: 37; 142: 38; 143: 39; 146: 40.
- Step 2: median
- , so the median is the mean of the 20th and 21st values. Both are 136 (positions 12 to 22), so s. (Question 1: B, .)
- Step 3: quartiles
- is the mean of the 10th and 11th values, both 135. is the mean of the 30th and 31st, both 138. .
- Step 4: fences
- Lower: (no values below). Upper: , so 143 and 146 are possible outliers.
- Step 5: shape
- Most times are at 135 to 137 and the tail stretches towards larger times: positively skewed with possible outliers. (Question 2: A, .)
An odd number of values
Fifteen commuting times (minutes) are 21, 24, 25, 27, 28, 28, 30, 31, 33, 34, 35, 37, 38, 41, 58.
- Step 1: median
- Position : .
- Step 2: quartiles
- Leave the 8th value out. The lower half is the first seven values, whose middle (the 4th value) is . The upper half is the 9th to 15th values, whose middle (the 12th value) is .
- Step 3: fences
- , so the fences are and . The time 58 is a possible outlier.
- Step 4: the boxplot
- Box from 27 to 37 with the median at 31, lower whisker to 21, upper whisker to 41 (the largest value below 52), and 58 plotted separately.
Counting outliers from partial information (2025 Examination 1 Question 6)
- Step 1
- .
- Step 2
- lower fence .
- Step 3
- compare each listed value with 69.5: 68.5, 68.6 and 69.0 are below it. Three outliers (C). The upper quartile is only needed for the IQR; the maximum is not needed at all.
Range and upper fence from a boxplot (2025 Examination 2 Question 2)
- Step 1: read the five values
- Minimum $400 000, $600 000, median $750 000, $900 000, maximum $1 300 000.
- Step 2: range
- dollars.
- Step 3: upper fence
- , so the upper fence is dollars. The maximum is below this, which is why no outlier appears.
Comparing two boxplots (2025 Examination 1 Question 3)
Step 1: read both five-number summaries (approximately). H: 56, 63, 67.2, 70.8, 81. T: 68, 73.6, 76.5, 79.2, 83.
- Step 2: compare centre
- T's median (76.5) is about 9.3 years higher than H's (67.2).
- Step 3: compare spread
- T's IQR (5.6) is smaller than H's (7.8), and T's range (15) is smaller than H's (25).
- Step 4: test the claim in D
- T's minimum (68) exceeds H's median (67.2), so every T country exceeds H's median. Answer: D.
A full written comparison: "Life expectancy tends to be higher in Sample T than Sample H, with a median of about 76.5 years compared with 67.2 years. Life expectancy is less variable in Sample T, with an IQR of about 5.6 years compared with 7.8 years."
- Adding the IQR instead of
- For 2025 Examination 2 Question 2 this gives $1 200 000 instead of $1 350 000.
- Drawing the whisker to the fence
- A whisker ends at a data value, the most extreme one that is not an outlier (142 s for the runners, not 142.5).
- Including the median in both halves when is odd
- For 15 values, the halves are the first seven and the last seven.
- Taking the median of the unordered list
- Always sort first (or count along a dot plot). In 2023 Examination 1 Question 14 the common wrong answer was the middle of the unordered time series.
- Calling a value on the fence an outlier
- It must be strictly beyond the fence.
- Comparing without numbers
- "T has a higher median" earns less than "T has a higher median (76.5 years compared with 67.2 years)".
- Using the range as the main spread measure
- For boxplots compare IQRs first; the range is sensitive to extreme values.
Write the five-number summary as a list before you do anything else, then the IQR and both fences. For a multiple-choice comparison of boxplots, jot the five values of each plot beside it and test each option; most distractors swap a quartile for a median or compare the wrong ends of the boxes. In a written comparison, one sentence for centre and one for spread, each with the group names, the direction and two values.
Why boxplots are built on the median and quartiles
The median and quartiles are resistant to extreme values: moving the largest runner's time from 146 s to 200 s would not change the box at all, whereas the mean and standard deviation would change. That makes the boxplot the natural display for skewed data and data with outliers, and the fence rule gives a consistent, checkable definition of "unusually far from the middle". For symmetric, bell-shaped data the mean and standard deviation (and the 68-95-99.7% rule) are the better summary.
Sort the data and find the middle value (the median), then the middle of the bottom half () and the middle of the top half (). The box goes from to . Work out the gap , multiply by 1.5, and go that far out from each end of the box: anything past that is an outlier and gets its own dot. The whiskers stretch to the furthest values that are not outliers. To compare two boxplots, say which has the higher median and which has the bigger box, and give the numbers.
Exam-style questions
Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.
2025 VCAA Exam 1 Q6Past paper1 markThe following information relating to life expectancy comes from a sample of nations in the Oceania region. The first quartile is 74.9 years. The third quartile is 78.5 years. The lowest five values recorded are 68.5, 68.6, 69.0, 70.1 and 74.8 years. How many outliers would be displayed at the lower end of a boxplot showing this sample of Oceania data? A. 1 B. 2 C. 3 D. 4
Show worked answer →
A value is a possible outlier if it is less than 69.5. That is true of 68.5, 68.6 and 69.0, but not of 70.1 or 74.8. So 3 outliers are displayed, option C ( correct).
The whisker would then stop at 70.1, the smallest value that is not an outlier.
Source: VCAA 2025 General Mathematics Examination 1, Question 6, and the 2025 examination report.
2025 VCAA Exam 2 Q2Past paper2 marksA boxplot for the sale prices of a sample of 203 homes is shown. [The boxplot, on a scale from $300 000 to $1 500 000, has minimum $400 000, first quartile $600 000, median $750 000, third quartile $900 000 and maximum $1 300 000, with no outliers shown.] a. Calculate the range of the sale price data in the boxplot. (1 mark) b. Calculate the upper fence for the sale price data in the boxplot. (1 mark)
Show worked answer →
a. Range maximum minimum:
The range is $900 000 ( correct).
b. , so
The upper fence is $1 350 000 ( correct). This is consistent with the plot: the maximum, $1 300 000, is below the fence, so it is not an outlier and the whisker runs all the way to it.
Source: VCAA 2025 General Mathematics Examination 2, Question 2, and the 2025 examination report.
2025 VCAA Exam 1 Q3Past paper1 markThe boxplots below show the life expectancy, in years, for a sample of countries from two different continents (Sample H and Sample T). [Read from the plot: Sample H has minimum about 56, Q1 about 63, median about 67.2, Q3 about 70.8, maximum about 81. Sample T has minimum about 68, Q1 about 73.6, median about 76.5, Q3 about 79.2, maximum about 83.] Which one of the following statements is correct? A. The interquartile range for Sample T is greater than the interquartile range for Sample H. B. The median for Sample T is more than 10 years greater than the median for Sample H. C. The third quartile for Sample H is greater than the first quartile for Sample T. D. Life expectancy in all Sample T countries exceeds the median life expectancy in Sample H.
Show worked answer →
Test each statement against values read from the plots.
- A. IQR for T ; IQR for H . T's is smaller. False.
- B. Medians and : a difference of about 9.3 years, not more than 10. False.
- C. for H () is less than for T (). False.
- D. The minimum for T () is above the median for H (), so every T country exceeds H's median. True.
The answer is D ( correct). The key idea: "all values exceed" is a statement about the minimum, the left end of the whisker.
Source: VCAA 2025 General Mathematics Examination 1, Question 3, and the 2025 examination report.
2023 VCAA Exam 1 Q2Past paper1 markThe dot plot below shows the times, in seconds, of 40 runners in the qualifying heats of their 800 m club championship. [Dot plot counts: 134 s: 3, 135 s: 8, 136 s: 11, 137 s: 6, 138 s: 4, 139 s: 2, 140 s: 2, 141 s: 1, 142 s: 1, 143 s: 1, 146 s: 1.] The shape of this distribution is best described as A. positively skewed with one or more possible outliers. B. positively skewed with no outliers. C. approximately symmetric with one or more possible outliers. D. approximately symmetric with no outliers. E. negatively skewed with one or more possible outliers.
Show worked answer →
Shape. The times cluster at 135 to 137 s with a long tail towards the slower (larger) times, so the distribution is positively skewed.
Outliers. With 40 values, the median is the mean of the 20th and 21st (both 136). is the mean of the 10th and 11th values (both 135) and the mean of the 30th and 31st (both 138). So and
The times 143 s and 146 s are above 142.5, so there are possible outliers. The answer is A ( correct).
Source: VCAA 2023 General Mathematics Examination 1, Question 2, and the 2023 examination report.
Practice questions
Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.
foundation3 marksThe ordered resting heart rates (beats per minute) of 11 athletes are 48, 52, 55, 55, 58, 60, 62, 64, 66, 70, 88. (a) Find the five-number summary. (b) Show that 88 is a possible outlier. (c) Where does the upper whisker of the boxplot end?
Show worked solution →
(a) , so the median is the 6th value, 60. The lower half (first five values) is 48, 52, 55, 55, 58, so . The upper half is 62, 64, 66, 70, 88, so . Five-number summary: . (1 mark)
(b) . Upper fence , and . (1 mark)
(c) At 70, the largest value that is not an outlier (not at 82.5, the fence). (1 mark)
foundation2 marksA five-number summary is 12, 20, 26, 31, 50. (a) Find the upper and lower fences. (b) Is the maximum a possible outlier?
Show worked solution →
(a) and . Lower fence ; upper fence . (1 mark)
(b) Yes: . On the boxplot, 50 is plotted as a separate point and the upper whisker ends at the largest value that is at most 47.5 (which the summary alone does not tell you). (1 mark)
core3 marksThe scores of 12 students are 3, 14, 15, 17, 18, 18, 20, 21, 23, 24, 26, 29. (a) Find the five-number summary. (b) Identify any possible outliers. (c) Describe where each whisker ends.
Show worked solution →
(a) : median . Lower half 3, 14, 15, 17, 18, 18 gives . Upper half 20, 21, 23, 24, 26, 29 gives . Summary: . (1 mark)
(b) and . Lower fence and upper fence . The score 3 is below 4.75, so it is a possible outlier. (1 mark)
(c) The lower whisker ends at 14 (the smallest non-outlier) and the upper whisker at 29. (1 mark)
core2 marksA boxplot of the reaction times of 200 drivers has minimum 0.31 s, s, median 0.52 s, s and maximum 0.74 s. (a) About how many drivers had a reaction time between 0.45 s and 0.60 s? (b) What percentage of drivers had a reaction time of more than 0.52 s?
Show worked solution →
(a) The box holds the middle : about drivers. (1 mark)
(b) The median splits the data in half: about . (1 mark)
exam2 marksThe upper fence for a data set is 94 and the interquartile range is 16. (a) Find and . (b) Find the lower fence.
Show worked solution →
(a) , so , and . (1 mark)
(b) Lower fence . (1 mark)
exam2 marksParallel boxplots show the daily sales at two cafes over 60 days. Cafe A: 12, 18, 22, 27, 40. Cafe B: 15, 24, 29, 33, 38 (five-number summaries). Neither boxplot shows outliers. Compare the two distributions in terms of centre and spread, quoting appropriate values.
Show worked solution →
Centre: the median daily sales at cafe B (29) are higher than at cafe A (22). (1 mark)
Spread: the IQRs are the same ( for A and for B), but the range is larger for cafe A () than for cafe B (). (1 mark)
Both parts need the comparison and the values.
exam3 marksThe commuting times, in minutes, of 15 employees are 21, 24, 25, 27, 28, 28, 30, 31, 33, 34, 35, 37, 38, 41, 58. (a) Find the five-number summary. (b) Show that one value is a possible outlier. (c) Describe the shape of the distribution.
Show worked solution →
(a) : the median is the 8th value, 31. The lower half is the first seven values, so is the 4th value, 27; the upper half is the 9th to 15th values, so is the 12th value, 37. Summary: . (1 mark)
(b) , upper fence , and . (1 mark)
(c) Positively skewed with a possible outlier: the upper tail (to 58) is much longer than the lower one. Without the outlier the data are close to symmetric (whiskers 21 to 41, median 31). (1 mark)
Practise this
Sources & how we know this
- VCE General Mathematics Examination 1 (2023) and examination report — VCAA (2023)
- VCE General Mathematics Examination 1 (2025) and examination report — VCAA (2025)
- VCE General Mathematics Examination 2 (2025) and examination report — VCAA (2025)
- VCE Mathematics Study Design: General Mathematics (from 2023) — VCAA
- VCE General Mathematics: examination specifications, past examinations and reports — VCAA