Mathematics Standard • Year 12 • Module 8 • Lesson 5

Box Plots and Outliers — Skill Drill

Build fluency in computing five-number summaries, applying the 1.5 × IQR outlier rule, and drawing and reading box plots.

Build · Skill Drill

1. Quick recall

Answer each in the space provided. 1 mark each

Q1.1 List the five-number summary in order: ____________, ____________, ____________, ____________, ____________.

Q1.2 Write the outlier fence formulas: Lower fence = Q1 − ________ × IQR Upper fence = Q3 + ________ × IQR.

Q1.3 In a box plot, what does the line inside the box represent? ____________. What does the length of the box represent? ____________.

Stuck? Revisit lesson § Key Ideas — Five-number summary, Outlier rule, Box plot components.

2. Worked example — full five-number summary + outlier check

Daily peak temperatures (°C): 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 30, 35. n = 13.

Step 1 — Order the data and find min, max.

Min = 14, Max = 35.

Step 2 — Median (n odd, middle = (n+1)/2 = 7th value).

Median = 7th value = 21.

Step 3 — Q1 (median of lower 6 values: 14, 16, 17, 18, 19, 20).

Q1 = (17 + 18)/2 = 17.5.

Step 4 — Q3 (median of upper 6 values: 22, 23, 24, 25, 30, 35).

Q3 = (24 + 25)/2 = 24.5.

Step 5 — IQR and outlier fences.

IQR = 24.5 − 17.5 = 7. 1.5 × IQR = 10.5.
Lower fence = 17.5 − 10.5 = 7. Upper fence = 24.5 + 10.5 = 35.

Conclusion. Five-number summary: 14, 17.5, 21, 24.5, 35. The value 35 sits exactly on the upper fence — borderline (often not classified as an outlier). The 30 is within fences. No clear outliers. Slight right-skew because 30 and 35 stretch the upper end.

3. Faded example — find the summary, then test for outliers

Data: 12, 14, 15, 16, 18, 20, 21, 22, 25, 28, 45. n = 11. 4 marks

Step 1 — Min: ________. Max: ________.

Step 2 — Median (6th value, n = 11): ________.

Step 3 — Q1 (median of the 5 values BELOW the median: 12, 14, 15, 16, 18) = ________ Q3 (median of the 5 values ABOVE the median: 21, 22, 25, 28, 45) = ________

Step 4 — IQR: ________ − ________ = ________. 1.5 × IQR = ________.

Step 5 — Lower fence: Q1 − 1.5 × IQR = ________. Upper fence: Q3 + 1.5 × IQR = ________.

Step 6 — Outliers: The value ________ is above the upper fence, so it is an outlier.

Stuck? With n odd, the median is its own data point — exclude it when finding Q1 and Q3.

4. Graduated practice — summaries, fences, plots

Foundation — one-step (4 questions)

Q	Problem	Answer
4.1 1	If Q1 = 22 and IQR = 8, what is the lower fence?
4.2 1	If Q3 = 40 and IQR = 6, what is the upper fence?
4.3 1	For Q1 = 12, Q3 = 28, find the IQR.
4.4 1	In a box plot, the median = 50, Q1 = 42, Q3 = 60. What is the width of the box?

Standard — typical HSC difficulty (6 questions)

4.5 For the data 5, 7, 9, 12, 15, 17, 20, 22, 25 (n = 9), find the five-number summary. 2 marks

4.6 For Q1 = 15, Q3 = 27, IQR = 12, decide if 50 is an outlier. 2 marks

4.7 Draw a box plot from the five-number summary: 10, 18, 24, 30, 50. (Use a scale 0-50; assume 50 is an outlier — draw whisker to the largest non-outlier value, then a dot at 50.) 2 marks

4.8 For the data 2, 3, 5, 7, 8, 10, 12, 14, 16, 20 (n = 10), find Q1, median, Q3, IQR. 2 marks

4.9 A box plot shows Min = 5, Q1 = 12, Median = 18, Q3 = 25, Max = 40. The IQR = 13. Is 40 an outlier? 2 marks

4.10 A box plot has a long right whisker and a short left whisker, with the median sitting near Q1. Describe the shape of the distribution. 2 marks

Extension — comparison and interpretation (2 questions)

4.11 Two box plots on the same scale: Class A has 5-num (50, 60, 70, 80, 90); Class B has 5-num (40, 55, 70, 85, 95). State which class has (a) the higher median, (b) the larger IQR, (c) the same range or different ranges. Then state which class performed more consistently. 3 marks

4.12 Data: 4, 5, 6, 7, 8, 9, 10, 11, 12, 60. (a) Find the five-number summary. (b) Test whether 60 is an outlier using the 1.5 × IQR rule. (c) State, in one sentence, whether the median or the mean better represents this data set. 3 marks

Stuck on 4.11? Box plots line up vertically — read median, Q1, Q3 and the whiskers at a glance.

5. Self-check the easy 3

For 4.1 I subtracted 1.5 × 8 = 12 from Q1 = 22, getting 10.

For 4.2 I added 1.5 × 6 = 9 to Q3 = 40, getting 49.

For 4.3 I did Q3 − Q1 (not Q3 + Q1), getting 16.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Five-number summary

Min, Q1, Median, Q3, Max.

Q1.2 — Outlier fences

Lower fence = Q1 − 1.5 × IQR. Upper fence = Q3 + 1.5 × IQR.

Q1.3 — Box plot anatomy

Line inside the box = median. Length of the box = IQR (Q3 − Q1, i.e. the middle 50% of the data).

Q3 — Faded example

Min = 12, Max = 45. Median (6th value) = 20. Lower 5: 12, 14, 15, 16, 18 → Q1 = 15. Upper 5: 21, 22, 25, 28, 45 → Q3 = 25. IQR = 25 − 15 = 10. 1.5 × IQR = 15. Lower fence = 15 − 15 = 0. Upper fence = 25 + 15 = 40. The value 45 > 40 → 45 is an outlier.

Q4.1 – Q4.4 — Foundation

4.1: lower fence = 22 − 1.5(8) = 22 − 12 = 10. 4.2: upper fence = 40 + 1.5(6) = 40 + 9 = 49. 4.3: IQR = 28 − 12 = 16. 4.4: box width = Q3 − Q1 = 60 − 42 = 18.

Q4.5 — Five-number summary (n = 9)

Min = 5; Max = 25. Median = 5th value = 15. Lower 4: 5, 7, 9, 12 → Q1 = (7+9)/2 = 8. Upper 4: 17, 20, 22, 25 → Q3 = (20+22)/2 = 21. Summary: 5, 8, 15, 21, 25.

Q4.6 — Is 50 an outlier?

Upper fence = 27 + 1.5(12) = 27 + 18 = 45. 50 > 45 → Yes, 50 is an outlier.

Q4.7 — Box plot

Box from 18 to 30 with median line at 24; left whisker from 10 to 18; right whisker from 30 to the largest non-outlier value (around 35 if we assume 50 is the only outlier); plot a single dot at 50 with no whisker. Scale 0-50 with x-axis labelled.

Q4.8 — n = 10

Median = (8 + 10)/2 = 9. Lower 5: 2, 3, 5, 7, 8 → Q1 = 5. Upper 5: 10, 12, 14, 16, 20 → Q3 = 14. IQR = 14 − 5 = 9. (Summary: 2, 5, 9, 14, 20.)

Q4.9 — Is 40 an outlier?

Upper fence = 25 + 1.5(13) = 25 + 19.5 = 44.5. Since 40 ≤ 44.5, 40 is NOT an outlier.

Q4.10 — Box plot shape

A long right whisker and a short left whisker, with median close to Q1 → right-skewed (positive skew). The upper 50% of the data is more spread out than the lower 50%.

Q4.11 — Compare two classes

(a) Same median = 70. (b) Class A IQR = 80 − 60 = 20. Class B IQR = 85 − 55 = 30. Class B has larger IQR. (c) Range A = 90 − 50 = 40; Range B = 95 − 40 = 55 → different. Class A performed more consistently — smaller IQR and smaller range mean its students' scores cluster more tightly.

Q4.12 — Outlier check with skewed data

(a) Min = 4; Max = 60. Median (n = 10) = (5th + 6th)/2 = (8 + 9)/2 = 8.5. Lower 5: 4, 5, 6, 7, 8 → Q1 = 6. Upper 5: 10, 11, 12, 60 wait — Upper 5: 10, 11, 12, 60 has only 4 — recount: after removing the lower 5 we have 9, 10, 11, 12, 60 → Q3 = 11. Summary: 4, 6, 8.5, 11, 60.
(b) IQR = 11 − 6 = 5. Upper fence = 11 + 1.5(5) = 18.5. 60 > 18.5 → 60 is an outlier.
(c) The median (8.5) better represents this data set than the mean, because the single 60 outlier pulls the mean far away from where most values actually lie.