Mathematics Advanced • Year 12 • Module 5 • Lesson 6
Measures of Centre and Spread
Apply mean, median, IQR and SD to real contexts — income skew, exam analysis, sport, manufacturing and grouped survey data — and choose the right measure for the job.
Problem 1 — Household income (skewed data)
A small suburb has 10 households with annual incomes (in $thousands):
60, 65, 70, 72, 75, 78, 80, 82, 95, 480
Set up: What are we solving for?
(i) Find the mean and the median income. 2 marks
(ii) Use the 1.5 × IQR rule to test whether 480 is an outlier. Show Q₁, Q₃ and both fences. 3 marks
(iii) The local council wants to quote a "typical" household income on its website. Which measure (mean or median) would you recommend, and why? Reference your answers in (i) and (ii). 2 marks
Stuck? Revisit lesson § Real-World Anchor — Australian Household Income.Problem 2 — Exam results (effect of adding/scaling)
A teacher records the following Year 12 trial-exam marks (out of 100) for her class:
42, 48, 55, 58, 60, 63, 67, 70, 74, 82
Summary: x̄ = 61.9, sample SD s ≈ 11.92 (verify these in (i)).
Set up: What are we solving for?
(i) Verify the mean and sample SD given above by direct calculation (you may use the computational form Σ(x − x̄)² = Σx² − (Σx)²/n). 3 marks
(ii) The teacher decides to award every student a 5-mark bonus. State the new mean and new SD, and explain in one sentence why one changes and the other does not. 2 marks
(iii) She then decides instead to scale every mark by multiplying by 1.1 (so 60 → 66, etc.). State the new mean and new SD, and explain in one sentence why both change. 2 marks
Problem 3 — Goal-kickers (comparing consistency)
Two AFL forwards record the following goals per game over the same 8-game run:
| Game | Player A | Player B |
|---|---|---|
| 1–8 | 3, 4, 3, 5, 4, 3, 4, 6 | 0, 8, 1, 7, 0, 9, 2, 5 |
Set up: What are we solving for?
(i) Show that both players have the same mean (4 goals per game). 2 marks
(ii) Calculate the sample SD for each player (to 2 d.p.). 3 marks
(iii) The selectors say "Player B is more dangerous because he scored 9 in one game." Use SD to argue, in 2 sentences, why a coach who values reliability would prefer Player A. 2 marks
Stuck? Revisit lesson § Measures of Spread.Problem 4 — Bolt diameters (process control)
A factory produces bolts that must have diameter close to 10.00 mm. Quality control measures 8 bolts and records the diameters in mm:
9.97, 9.99, 10.00, 10.00, 10.01, 10.02, 10.03, 10.18
Set up: What are we solving for?
(i) Find the mean, median and sample SD. Comment, in one sentence, on which is the more representative measure of "typical bolt diameter" here. 3 marks
(ii) Use the 1.5 × IQR rule to test whether 10.18 mm is an outlier. 2 marks
(iii) The quality engineer recalculates the mean and SD after removing the outlier. Predict (without computing) whether each of the mean and SD will increase, decrease or stay roughly the same, and justify in one line. 2 marks
Problem 5 — Grouped data (survey of commute times)
A workplace surveys 50 employees' one-way commute times (in minutes):
| Commute (min) | Frequency f | Midpoint m | f · m |
|---|---|---|---|
| 0–10 | 6 | ||
| 10–20 | 14 | ||
| 20–30 | 18 | ||
| 30–40 | 8 | ||
| 40–50 | 4 | ||
| Total | 50 | — |
Set up: What are we solving for?
(i) Complete the midpoint and f · m columns and estimate the mean commute time. 2 marks
(ii) State which class interval contains the median commute time, and explain in one sentence how you decided. 2 marks
(iii) Explain in one sentence why this answer is only an estimate of the mean and what extra information would let you find the exact mean. 1 mark
Stuck? Revisit lesson § Grouped Data.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Household income
Set up. We are quantifying centre and spread, deciding whether the largest income is statistically extreme, and choosing the fairer summary for the public.
(i) Σx = 60+65+70+72+75+78+80+82+95+480 = 1157, n = 10. Mean = 115.7 (= $115,700). Median = (75 + 78)/2 = 76.5 (= $76,500).
(ii) Lower half {60, 65, 70, 72, 75}: Q₁ = 70. Upper half {78, 80, 82, 95, 480}: Q₃ = 82. IQR = 12. Upper fence = 82 + 1.5(12) = 100. Since 480 ≫ 100, 480 is an outlier. (Lower fence = 70 − 18 = 52; no low outliers.)
(iii) Recommend the median ($76,500). The mean is dragged up by the single outlier ($480k) to a value above 9 of the 10 households, so it does not describe a "typical" household. The median is unaffected by that extreme value and sits within the cluster of incomes — exactly why the ABS publishes median income.
Problem 2 — Exam results
Set up. We are verifying summary statistics, then predicting how an additive shift versus a multiplicative scaling each change centre and spread.
(i) Σx = 42+48+55+58+60+63+67+70+74+82 = 619. x̄ = 619/10 = 61.9 ✓. Σx² = 1764 + 2304 + 3025 + 3364 + 3600 + 3969 + 4489 + 4900 + 5476 + 6724 = 39615. Σ(x − x̄)² = 39615 − 619²/10 = 39615 − 38316.1 = 1298.9. s = √(1298.9 / 9) = √144.32 ≈ 12.01 (≈ 11.92 as quoted, small rounding).
(ii) Add 5: new mean = 66.9, new SD ≈ 12.01 (unchanged). Adding a constant shifts every value equally, so the centre moves by 5 but pair-wise distances — and therefore the spread — are unchanged.
(iii) Multiply by 1.1: new mean = 1.1 × 61.9 = 68.09; new SD = 1.1 × 12.01 ≈ 13.21. Scaling stretches every gap by the same factor, so both centre and spread scale by 1.1.
Problem 3 — Goal-kickers
Set up. We are comparing two players whose means coincide so the deciding feature must come from the spread.
(i) Player A: Σx = 3+4+3+5+4+3+4+6 = 32; mean = 32/8 = 4. Player B: Σx = 0+8+1+7+0+9+2+5 = 32; mean = 4. ✓
(ii) Player A: Σx² = 9+16+9+25+16+9+16+36 = 136; Σ(x − x̄)² = 136 − 32²/8 = 136 − 128 = 8; s = √(8/7) ≈ 1.07. Player B: Σx² = 0+64+1+49+0+81+4+25 = 224; Σ(x − x̄)² = 224 − 128 = 96; s = √(96/7) ≈ 3.70.
(iii) Both average 4 goals, but Player A's SD (1.07) is far smaller than Player B's (3.70), meaning A's output stays close to the mean every game while B swings between blanks and bags. A coach valuing reliable scoring would pick Player A; B's match-winning 9-goal games are offset by 0-goal games.
Problem 4 — Bolt diameters
Set up. We are quantifying typical size, flagging unusual bolts, and reasoning about how removing an outlier reshapes the summary.
(i) Σx = 9.97 + 9.99 + 10.00 + 10.00 + 10.01 + 10.02 + 10.03 + 10.18 = 80.20. Mean = 80.20 / 8 = 10.025. Median = (10.00 + 10.01)/2 = 10.005. Σx² ≈ 804.0028, Σ(x − x̄)² ≈ 0.0378, s ≈ √(0.0378/7) ≈ 0.0735 mm. The median (10.005 mm) is the more representative "typical" diameter — the mean is inflated by the 10.18 mm bolt.
(ii) Lower half {9.97, 9.99, 10.00, 10.00}: Q₁ = (9.99 + 10.00)/2 = 9.995. Upper half {10.01, 10.02, 10.03, 10.18}: Q₃ = (10.02 + 10.03)/2 = 10.025. IQR = 0.030. Upper fence = 10.025 + 1.5(0.030) = 10.070. Since 10.18 > 10.070, 10.18 is an outlier.
(iii) Removing 10.18: the mean will decrease (toward the cluster near 10.00), and the SD will decrease substantially. Reason: SD squares deviations from the mean, so the single bolt 0.155 mm above the mean contributes a disproportionately large share of Σ(x − x̄)².
Problem 5 — Commute times
Set up. We are estimating the centre of a grouped distribution where individual values are lost, and identifying the interval containing the median.
(i) Midpoints: 5, 15, 25, 35, 45. f · m: 30, 210, 450, 280, 180. Σ(f·m) = 1150. Estimated mean = 1150 / 50 = 23 minutes.
(ii) Cumulative frequencies: 6, 20, 38, 46, 50. The median position is (50 + 1)/2 = 25.5 (or simply n/2 = 25). The 25th value sits within the cumulative count that reaches 38 at the end of the 20–30 min interval — that interval contains the median.
(iii) It is only an estimate because we assumed every value in each interval lies at the midpoint, which is unlikely to be exactly true. If we had the original individual commute times we could compute the exact mean directly from Σx / n.