Mathematics Advanced • Year 12 • Module 5 • Lesson 7
Representing Data
Practise HSC-style writing on data representation, including a structured extended response on choosing and critiquing displays.
1. Short-answer questions
1.1 A small data set has five-number summary 5, 12, 18, 23, 38. (a) Compute the IQR. (b) Use the 1.5 × IQR rule to determine whether 38 is an outlier. 2 marks Band 3
1.2 A histogram of weekly sales (in $) uses class intervals 0–100 (frequency 12), 100–200 (frequency 24), 200–500 (frequency 30). Calculate the frequency density of each class and state which class produces the tallest bar in a density histogram. 3 marks Band 3-4
1.3 The cumulative-frequency table below summarises the IQ scores of 200 students:
| Upper boundary | 80 | 90 | 100 | 110 | 120 | 130 |
|---|---|---|---|---|---|---|
| Cumulative | 10 | 40 | 90 | 150 | 185 | 200 |
(a) Estimate the median by linear interpolation. (b) Estimate the 90th percentile. 4 marks Band 4
Stuck on 1.3? Find which class interval contains the target cumulative count, then interpolate linearly within it.2. Extended response
2.1 A school principal wants to publish a one-page report on student performance in the Year 12 trial exam. He has the following 25 marks (out of 100):
38, 42, 45, 48, 50, 52, 55, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 85, 88, 90, 92, 99
(a) Find the five-number summary and the IQR. (b) Test whether either extreme value (38 or 99) is an outlier under the 1.5 × IQR rule. (c) The principal is choosing between three single-display options for the front page: a histogram (5 equal classes of width 12 starting at 36), a box plot, or a stem-and-leaf plot. For each display, state one feature of these 25 marks it shows clearly and one feature it hides. Conclude with a recommendation and a single-sentence justification using the lesson principle "No single representation tells the whole story".
8 marks Band 5-6
Explicit marking criteria
Part (a), 2 marks
• 1 mark correct min/max (38, 99) and median (= 13th value = 68).
• 1 mark correct Q₁ (= 7th value = 55) and Q₃ (= 19th value = 80), so IQR = 25.
Part (b), 2 marks
• 1 mark computes lower fence = 55 − 1.5(25) = 17.5 and upper fence = 80 + 1.5(25) = 117.5.
• 1 mark explicitly concludes neither 38 nor 99 is an outlier (38 ≥ 17.5 and 99 ≤ 117.5).
Part (c), 4 marks
• 1 mark, histogram: identifies a strength (e.g. shows distribution shape / symmetry) and a weakness (loses individual values; class boundaries can hide or create features).
• 1 mark, box plot: identifies a strength (compact, shows five-number summary and outliers, here, none, instantly) and a weakness (cannot reveal multiple peaks or the gap structure of the data).
• 1 mark, stem-and-leaf: identifies a strength (preserves every raw value; both shape and exact data visible) and a weakness (clumsy for large data sets; less compact for one-page report).
• 1 mark, recommendation: states a clear choice and justifies it using the principle "no single representation tells the whole story", e.g. recommends a box plot + histogram pair, or stem-and-leaf for this small data set.
Your response:
Stuck on (c)? For each display, ask: "what does the principal's audience get to see?" and "what is lost?".How did this worksheet feel?
What I'll revisit before next class:
1.1, IQR & outlier test (2 marks)
Sample response. (a) IQR = Q₃ − Q₁ = 23 − 12 = 11. (b) Upper fence = 23 + 1.5(11) = 39.5. Since 38 ≤ 39.5, 38 is not an outlier.
Marking notes. 1 mark, IQR correctly computed from the five-number summary. 1 mark, upper fence quoted and explicitly compared with 38. Just writing "no, it's not an outlier" without computing the fence scores 0.5.
1.2, Frequency density (3 marks)
Sample response. Class widths 100, 100, 300; densities 12/100 = 0.12, 24/100 = 0.24, 30/300 = 0.10. Tallest density bar = 100–200 class (density 0.24).
Marking notes. 1 mark, each density correct (cumulative 1 mark across the three). 1 mark, identifies the 100–200 class as the tallest. 1 mark, recognises that the 200–500 class has the highest frequency but the lowest density because of its width.
1.3, Median and 90th percentile by interpolation (4 marks)
Sample response. (a) Median position n/2 = 100, inside the 100–110 class (c.f. rises from 90 to 150). Median ≈ 100 + 10 × (100 − 90)/(150 − 90) = 100 + 10(10/60) ≈ 101.7.
(b) P₉₀ position 0.9 × 200 = 180, inside the 110–120 class (c.f. 150 → 185). P₉₀ ≈ 110 + 10 × (180 − 150)/(185 − 150) = 110 + 10(30/35) ≈ 118.6.
Marking notes. (a) 1 mark, identifies median position 100 and the 100–110 class; 1 mark, interpolation arithmetic correct. (b) 1 mark, identifies P₉₀ position 180 and the 110–120 class; 1 mark, interpolation arithmetic correct. Accept answers within ±0.2 due to rounding.
2.1, Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Five-number summary. n = 25; median = 13th value = 68. Lower 12 values (positions 1–12): Q₁ = median of these = (6th + 7th)/2 of lower half, equivalently the 7th value of the full data, actually for n = 25 the standard quartile positions are Q₁ at position (n + 1)/4 = 6.5, taken as the average of the 6th and 7th values = (52 + 55)/2 = 53.5; Q₃ at position 3(n + 1)/4 = 19.5, the average of the 19th and 20th values = (80 + 82)/2 = 81. IQR = 81 − 53.5 = 27.5. Min = 38, Max = 99. [1 mark, min/max/median; 1 mark, Q₁/Q₃/IQR.]
(b) Outlier test. Lower fence = 53.5 − 1.5(27.5) = 12.25. Upper fence = 81 + 1.5(27.5) = 122.25. Since 38 ≥ 12.25 and 99 ≤ 122.25, neither extreme is an outlier. [1 mark, fences computed; 1 mark, explicit conclusion for both 38 and 99.]
(c) Comparing the three displays.
Histogram (5 classes of width 12, starting at 36). Bins 36–48, 48–60, 60–72, 72–84, 84–96, 96–108 have frequencies 4, 5, 6, 5, 4, 1. Shows clearly: the overall shape, roughly symmetric, possibly slightly bimodal with a small upper-end mode. Hides: the exact value of each mark, and any structure finer than 12-mark bins. [1 mark, histogram strength + weakness.]
Box plot. Whiskers at 38 and 99; box from 53.5 to 81 with internal line at 68. Shows clearly: centre (median 68), middle 50% spread (IQR 27.5), the absence of outliers, and roughly symmetric shape (median is close to the centre of the box). Hides: bimodality and individual values, two very different data sets could produce this identical box plot. [1 mark, box plot strength + weakness.]
Stem-and-leaf (key 3 | 8 = 38).
3 | 8
4 | 2 5 8
5 | 0 2 5 8
6 | 0 2 4 6 8
7 | 0 2 4 6 8
8 | 0 2 5 8
9 | 0 2 9
Shows clearly: every individual mark, the central peak in the 60s–70s, and the long-thin upper tail to 99. Hides: nothing critical, but takes up more space than the other two; less compact for a one-page summary. [1 mark, stem-and-leaf strength + weakness.]
Recommendation. Because "no single representation tells the whole story", I recommend the principal print a box plot for the headline at-a-glance summary plus a small stem-and-leaf in the appendix: the box plot conveys centre and spread instantly, while the stem-and-leaf preserves every value for any teacher who wants to drill into individual marks. With only 25 students the stem-and-leaf remains compact, so this combination loses nothing the histogram would have shown. [1 mark, clear recommendation that explicitly invokes the lesson principle and justifies the pairing.]
Total: 8/8.
Band descriptors for marker.
Band 3: Computes most of (a) and (b) but with arithmetic slips; for (c) names displays without distinguishing strengths from weaknesses. ≈ 3-4 marks.
Band 4: (a) and (b) fully correct; in (c) one strength/weakness per display, but the recommendation is generic and does not invoke the lesson principle. ≈ 5-6 marks.
Band 5: All parts correct; (c) recommends a single display with a complete justification but does not pair displays or quote the principle. ≈ 7 marks.
Band 6: Full computational accuracy in (a) and (b); in (c) clearly states one specific strength and one specific weakness for each display, and concludes with a justified recommendation that explicitly references "no single representation tells the whole story" (or equivalent) and pairs / sequences displays to cover both shape and individual values. 8/8.