Mathematics Advanced • Year 12 • Module 5 • Lesson 6
Measures of Centre and Spread
Practise HSC-style writing on summary statistics — including a structured extended response on choosing the right measure of centre in skewed data.
1. Short-answer questions
1.1 The 8 scores 18, 22, 25, 27, 28, 30, 32, 100 are recorded. (a) Calculate the sample mean and median. (b) Using the 1.5 × IQR rule, show that 100 is an outlier. 3 marks Band 3-4
1.2 A data set of 20 scores has mean 60 and sample standard deviation 8. A constant of 4 is added to every score, then every result is multiplied by 1.5. Find the new mean and the new sample SD, showing reasoning. 2 marks Band 3
1.3 The table below shows grouped commute times in minutes for 40 workers:
| Time (min) | 0–10 | 10–20 | 20–30 | 30–40 |
|---|---|---|---|---|
| Frequency | 6 | 14 | 12 | 8 |
(a) Estimate the mean using mid-interval values. (b) State which interval contains the median, with a one-line justification. 4 marks Band 4
Stuck on 1.3(b)? Build a cumulative-frequency column and find where the count first reaches n/2.2. Extended response
2.1 A real-estate analyst examines weekly rents (in $) for 12 apartments in a Sydney suburb:
520, 545, 560, 575, 580, 595, 600, 615, 625, 640, 660, 1500
(a) Find the mean, median, IQR and sample standard deviation. (b) Use the 1.5 × IQR rule to identify any outliers. (c) The analyst writes a one-paragraph "rental snapshot" for new tenants and must choose either the mean or the median as the headline figure. Recommend which to use and justify your recommendation using the summary statistics you computed, the outlier result in (b) and the concept of robustness. Reference the lesson principle "mean is sensitive to outliers; median is robust" explicitly.
7 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — correct mean (≈ $668) and median ($597.50) with working.
• 1 mark — correct Q₁ (≈ $567.50), Q₃ (≈ $632.50) and IQR (≈ $65).
• 1 mark — correct sample SD (≈ $264, accept 260–270 from rounding).
Part (b) — 1 mark
• 1 mark — computes upper fence = Q₃ + 1.5 × IQR ≈ $730 and concludes $1500 is an outlier (lower fence ≈ $470, no low outliers).
Part (c) — 3 marks
• 1 mark — recommendation: recommends median and contrasts it with the inflated mean.
• 1 mark — quantitative justification: notes the mean ($668) exceeds 10 of the 12 rents while the median sits in the cluster, and connects this to the $1500 outlier identified in (b).
• 1 mark — conceptual link: states the principle "the mean is sensitive to outliers; the median is robust" (or equivalent) and explains why in terms of how each statistic uses the data (mean averages every value; median depends only on rank).
Your response:
Stuck on (c)? Compare how many of the 12 rents lie above the mean vs above the median, and link that to "robustness".How did this worksheet feel?
What I'll revisit before next class:
1.1 — Five-number summary & outlier (3 marks)
Sample response. (a) Σx = 18 + 22 + 25 + 27 + 28 + 30 + 32 + 100 = 282. Mean = 282 / 8 = 35.25. Ordered data, n = 8: median = (4th + 5th)/2 = (27 + 28)/2 = 27.5.
(b) Lower half {18, 22, 25, 27}: Q₁ = (22 + 25)/2 = 23.5. Upper half {28, 30, 32, 100}: Q₃ = (30 + 32)/2 = 31. IQR = 31 − 23.5 = 7.5. Upper fence = 31 + 1.5(7.5) = 42.25. Since 100 > 42.25, 100 is an outlier.
Marking notes. (a) 1 mark — mean and median both correct with the working visible. (b) 1 mark — Q₁ and Q₃ correct. 1 mark — computes the upper fence and explicitly compares 100 with it. Stating "100 is an outlier" without producing the fence value scores 0.5.
1.2 — Linear transformation of every score (2 marks)
Sample response. If x has mean 60 and SD 8, then x + 4 has mean 60 + 4 = 64 and SD 8 (adding a constant does not change spread). Multiplying by 1.5 then gives new mean = 1.5 × 64 = 96 and new SD = 1.5 × 8 = 12.
Marking notes. 1 mark — applies "add c shifts mean, leaves SD unchanged" correctly. 1 mark — applies "multiply by k scales both centre and spread by k" correctly to reach 96 and 12. Common error: forgetting that the SD is unchanged by the +4 step.
1.3 — Grouped commute times (4 marks)
Sample response. (a) Midpoints 5, 15, 25, 35. f·m: 30, 210, 300, 280. Σ(f·m) = 820, Σf = 40. Estimated mean = 820 / 40 = 20.5 min.
(b) Cumulative frequencies: 6, 20, 32, 40. The median position is n/2 = 20, which is reached at the end of the 10–20 min interval. The 21st value (the median for n = 40 is the average of the 20th and 21st) falls in the next interval, so the median lies in the 20–30 min interval (it equals the upper boundary 20 of the previous group, so strictly the 20–30 interval contains the median).
Marking notes. (a) 1 mark — correct midpoints; 1 mark — correct Σ(f·m) and mean. (b) 1 mark — cumulative-frequency table; 1 mark — correctly identifies the interval containing the median, with a one-line justification that mentions n/2 (or the 20th/21st value). Accept "10–20 min" only if the student justifies it from the 20th value lying at the boundary.
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Summary statistics. Σx = 520 + 545 + 560 + 575 + 580 + 595 + 600 + 615 + 625 + 640 + 660 + 1500 = 8015. Mean = 8015 / 12 ≈ $668. Ordered data, n = 12: median = (6th + 7th)/2 = (595 + 600)/2 = $597.50. [1 mark — mean and median.]
Lower 6 values {520, 545, 560, 575, 580, 595}: Q₁ = (560 + 575)/2 = $567.50. Upper 6 values {600, 615, 625, 640, 660, 1500}: Q₃ = (625 + 640)/2 = $632.50. IQR = 632.50 − 567.50 = $65. [1 mark — Q₁, Q₃, IQR.]
Σx² ≈ 6,118,775 (key contribution: 1500² = 2,250,000). Σ(x − x̄)² = Σx² − (Σx)²/n ≈ 6,118,775 − 8015²/12 ≈ 6,118,775 − 5,353,352.08 ≈ 765,423. Sample SD s = √(765,423 / 11) ≈ $264. [1 mark — sample SD.]
(b) Outlier test. Lower fence = 567.50 − 1.5(65) = $470. Upper fence = 632.50 + 1.5(65) = $730. The value $1500 is well above $730, so $1500 is an outlier; no rents fall below the lower fence. [1 mark — outlier identified with fence calculation.]
(c) Recommendation. I recommend the analyst publish the median ($597.50) as the headline figure, not the mean ($668). [1 mark — clear recommendation contrasting the two figures.]
Quantitatively, 10 of the 12 apartments rent for less than $660 — i.e. less than even the second-highest rent — yet the mean ($668) is higher than 10 of the 12 actual rents. The median ($597.50) sits squarely within the cluster of rents from $560 to $640, so it describes what a typical tenant would actually pay. The whole gap between mean and median is driven by the single $1500 rent identified as an outlier in (b). [1 mark — quantitative justification connecting the mean/median gap to the outlier in (b).]
Conceptually, the mean is calculated as Σx / n and so includes every data point linearly: one extreme value moves it directly. The median depends only on the rank order of the data — once the $1500 rent is placed at the top of the ordered list, the median (centred at positions 6 and 7) does not feel its magnitude at all. This is exactly the lesson principle "the mean is sensitive to outliers; the median is robust", and it is the reason the ABS reports median household income rather than mean income for the same kind of skewed distribution. [1 mark — conceptual link to robustness, explained from how each statistic uses the data.]
Total: 7/7.
Band descriptors for marker.
Band 3: Computes mean and median correctly but mis-handles Q₁/Q₃, SD or the outlier test; recommends median by intuition without quantitative or conceptual link. ≈ 3-4 marks.
Band 4: Full (a) summary statistics and (b) outlier test; recommendation present but justification only quantitative or only conceptual, not both. ≈ 5 marks.
Band 5: All of (a), (b) correct; recommendation justified both with a quantitative comparison (mean above 10 of 12 rents) and a conceptual statement of robustness. ≈ 6 marks.
Band 6: Complete (a) and (b), recommendation justified quantitatively, conceptually, and linked explicitly to the lesson principle "mean is sensitive to outliers; median is robust" with reference to how each statistic uses the data (or to the ABS / household-income analogue). 7/7.