Mathematics Standard • Year 12 • Module 8 • Lesson 5
Box Plots and Outliers — Past-Paper Style
Practise HSC Mathematics Standard 2-style writing on five-number summaries, the 1.5 × IQR outlier rule and side-by-side box-plot comparisons.
1. Short-answer questions
1.1 The data set 12, 15, 18, 22, 25, 28, 32, 36, 42, 48 (n = 10) summarises monthly rainfall (mm).
(a) Find the five-number summary.
(b) Sketch a box plot to scale (use 0-50 on the x-axis). 3 marks Band 3
1.2 For a data set Q1 = 28, Q3 = 46.
(a) Find the IQR and the upper and lower fences using the 1.5 × IQR rule.
(b) Decide whether the values 6 and 80 are outliers. 3 marks Band 3-4
1.3 A box plot has 5-num summary 4, 10, 14, 18, 50, with the value 50 displayed as a separate outlier dot.
(a) Calculate the IQR.
(b) Verify (with the 1.5 × IQR rule) that 50 is correctly identified as an outlier.
(c) State whether the data is symmetric, left-skewed or right-skewed and explain in one sentence. 4 marks Band 4
2. Extended response
2.1 Two NSW high schools record the time (minutes) Year 12 students travel to school each morning. Side-by-side box plots are shown below.
School Coastal: 5-num summary = 5, 12, 18, 26, 40. No outliers.
School Inland: 5-num summary = 8, 20, 30, 42, 95. The value 95 is an outlier on the plot.
(a) Compute the IQR and range for each school.
(b) Verify, using the 1.5 × IQR rule, that 95 minutes is an outlier for School Inland.
(c) In 2-3 sentences, compare the two schools using both centre and spread, and recommend which school the local council should prioritise for additional bus services. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — correct IQR for both schools.
• 1 mark — correct range for both schools.
Part (b) — 2 marks
• 1 mark — correctly computes upper fence = Q3 + 1.5 × IQR for Inland.
• 1 mark — explicitly states 95 > upper fence → outlier.
Part (c) — 3 marks
• 1 mark — compares medians (centre).
• 1 mark — compares IQR or range (spread), with numerical reference.
• 1 mark — recommendation tied to evidence (e.g. Inland has higher median, larger IQR and at least one ≥1.5-hour traveller — bus services likely to help most there).
Your response:
Stuck? Use the medians (18 vs 30) and the IQRs (14 vs 22) to back up any recommendation.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Rainfall (3 marks)
(a) n = 10. Min = 12, Max = 48. Median = (5th + 6th)/2 = (25 + 28)/2 = 26.5. Lower 5: 12, 15, 18, 22, 25 → Q1 = 18. Upper 5: 28, 32, 36, 42, 48 → Q3 = 36. Summary: 12, 18, 26.5, 36, 48.
(b) Box plot: x-axis 0-50 mm, labelled. Box from 18 to 36 with a vertical median line at 26.5. Whiskers from 12 to 18 and from 36 to 48.
Marking notes. 1 mark for correct min, max, median. 1 mark for correct Q1 and Q3 (most common slip: forgetting to exclude the median for even n — note: for n = 10 the median is between the 5th and 6th values, so the lower half is the first 5 values, the upper half is the last 5). 1 mark for a labelled, in-scale box plot with all five components.
1.2 — Outlier rule (3 marks)
(a) IQR = 46 − 28 = 18. 1.5 × IQR = 27. Lower fence = 28 − 27 = 1. Upper fence = 46 + 27 = 73.
(b) 6: 6 > 1 (above lower fence) → not an outlier. 80: 80 > 73 (above upper fence) → is an outlier.
Marking notes. 1 mark for IQR. 1 mark for both fences. 1 mark for both outlier decisions with reasoning shown. A common error is treating "low value = outlier" without checking the fence.
1.3 — Box plot with outlier (4 marks)
(a) IQR = 18 − 10 = 8.
(b) Upper fence = 18 + 1.5(8) = 18 + 12 = 30. 50 > 30 → outlier confirmed.
(c) The distribution is right-skewed (positive skew). The median (14) sits closer to Q1 (10) than to Q3 (18), and the upper whisker plus the outlier extend much further right than the lower whisker extends left.
Marking notes. 1 mark for IQR. 1 mark for the upper-fence calculation. 1 mark for the explicit "50 > 30 → outlier" conclusion. 1 mark for the correct shape with reasoning that references the median position OR the whisker asymmetry.
2.1 — School travel times (7 marks): Band-6 sample with annotations
(a) IQR and range for both schools.
Coastal: IQR = 26 − 12 = 14 min; range = 40 − 5 = 35 min. [1 mark — both IQRs.]
Inland: IQR = 42 − 20 = 22 min; range = 95 − 8 = 87 min. [1 mark — both ranges.]
(b) Verify 95 min is an outlier for Inland.
Upper fence = Q3 + 1.5 × IQR = 42 + 1.5(22) = 42 + 33 = 75 min. [1 mark — fence calculation.]
Since 95 > 75, 95 is correctly flagged as an outlier on the box plot. [1 mark — explicit comparison and conclusion.]
(c) Comparison + recommendation.
School Inland's median travel time (30 min) is 12 minutes longer than Coastal's (18 min) — students there typically travel much further on average. [1 mark — centres compared with numbers.]
Inland's IQR (22 min) is 60% larger than Coastal's (14 min), and the range balloons to 87 min thanks to one student travelling 95 min — meaning Inland not only has longer typical commutes but also much greater variability, with at least one student spending more than 1.5 hours on the road one-way. [1 mark — spread compared with numbers.]
Recommendation: prioritise School Inland for additional bus services. The combination of higher median, larger IQR and a confirmed outlier all point to Inland as the school where the largest equity gain is available — the council can directly target the long-commute students who currently fall outside the upper fence. [1 mark — recommendation tied to all three pieces of evidence.]
Total: 7/7.
Band descriptors for marker.
Band 3: Computes one IQR correctly; partial fence calculation; (c) compares only medians. ≈ 2-3 marks.
Band 4: Both IQRs and the fence correct; (c) compares centre and spread but no clear recommendation. ≈ 4-5 marks.
Band 5: All numerical correct; recommendation made but not tied directly to evidence numbers. ≈ 6 marks.
Band 6: Full numerical solution AND a recommendation that explicitly references medians, IQR/range and the outlier as separate pieces of evidence. 7/7.