Mathematics Standard • Year 12 • Module 8 • Lesson 2
Measures of Central Tendency — Problem Set
Apply mean, median and mode to realistic Australian data sets — pay, housing, marks, customer wait times and grouped data.
Problem 1 — Suburb house prices (mean vs median)
An estate agent reports the following 11 house sale prices ($,000) for a Sydney suburb in one month: 760, 820, 845, 880, 910, 930, 950, 975, 1,020, 1,080, 4,250 (the last is a renovated waterfront).
Set up: What are we solving for?
(i) Calculate the mean sale price for the month (to the nearest $1,000). 1 mark
(ii) Calculate the median sale price. 1 mark
(iii) A real-estate advertisement headlines "Average price in this suburb: $1.31 million". Explain in 2-3 sentences why the median would give a more honest summary, and what number it would give. 3 marks
Stuck? Revisit lesson § When the Mean Misleads — outliers and skewed distributions.Problem 2 — Test marks and the missing student
A class of 25 students sat a maths test. The mean for the whole class was 68 marks. A 26th student joined late and took the test, scoring 94. The teacher needs to update the class mean.
Set up: What are we solving for?
(i) Calculate the sum of marks for the original 25 students. 1 mark
(ii) Calculate the new class mean after the 26th student is included. 2 marks
(iii) The teacher claims "Adding one good student barely changed the mean". State whether this is supported by your calculation and quantify the change in one sentence. 2 marks
Stuck? Use Σx = mean × n to recover the total, then add the new mark and divide by 26.Problem 3 — Cafe wait times (frequency table)
A cafe records customer wait times to the nearest minute over a busy hour:
| Wait time (min) | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|
| Frequency | 4 | 9 | 12 | 15 | 8 | 2 | 1 |
Set up: What are we solving for?
(i) Calculate the mean wait time (1 d.p.). 2 marks
(ii) State the modal wait time. 1 mark
(iii) Find the median wait time (show the cumulative-frequency reasoning). 2 marks
Stuck on (iii)? n = Σf. The median is the value at position (n+1)/2 in the ordered list.Problem 4 — Wages at a small business
A landscaping business pays 10 staff per week:
$960, $960, $980, $980, $1,000, $1,020, $1,020, $1,040, $1,060, $3,500 (owner).
Set up: What are we solving for?
(i) Calculate the mean, median and modal weekly pay. 2 marks
(ii) An employee asks "what's the typical wage?". Which of the three measures should you quote, and why? 2 marks
(iii) Recalculate the mean if the owner's $3,500 is excluded. Comment on the size of the change in one sentence. 2 marks
Stuck? Revisit lesson § Worked Example — outlier inflates the mean.Problem 5 — Grouped data (estimated mean)
A retail manager records the time (minutes) customers spend in store, grouped:
| Class (min) | Midpoint x | Frequency f |
|---|---|---|
| 0 – 9 | 4.5 | 8 |
| 10 – 19 | 14.5 | 22 |
| 20 – 29 | 24.5 | 30 |
| 30 – 39 | 34.5 | 15 |
| 40 – 49 | 44.5 | 5 |
Set up: What are we solving for?
(i) Calculate the estimated mean time spent in store (1 d.p.). 2 marks
(ii) Identify the modal class and state what it tells the manager. 1 mark
(iii) Explain in one sentence why the calculated value in part (i) is an estimate rather than the exact mean. 2 marks
Stuck? Use x̄ = Σ(f × midpoint) / Σf for grouped data.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Suburb house prices
(i) Σ = 760+820+845+880+910+930+950+975+1,020+1,080+4,250 = 13,420 ($,000). Mean = 13,420 / 11 ≈ $1,220,000.
(ii) n = 11, ordered (already ordered). Median = 6th value = $930,000.
(iii) The single $4.25 m waterfront sale pulls the mean above every other house in the data set (10 of the 11 houses sold below $1.1 m). The median ($930,000) is closer to a "typical" sale and is the figure normally quoted by official sources like CoreLogic. Advertising "average $1.31 m" is technically correct but misleads buyers about what they should expect to pay.
Problem 2 — Test marks update
(i) Σx (25 students) = 68 × 25 = 1,700.
(ii) New total = 1,700 + 94 = 1,794. New mean = 1,794 / 26 = 69.0 (to 1 d.p.).
(iii) Mean rose from 68 to 69, a change of just +1 mark. The teacher is right: one good student moves the average by only 1 mark because it averages with the other 25.
Problem 3 — Cafe wait times
(i) n = 4+9+12+15+8+2+1 = 51. Σ(f×x) = 4(2)+9(3)+12(4)+15(5)+8(6)+2(7)+1(8) = 8+27+48+75+48+14+8 = 228. Mean = 228 / 51 = 4.5 min (1 d.p.).
(ii) Modal wait time = 5 minutes (highest frequency, f = 15).
(iii) Median position = (51+1)/2 = 26th. Cumulative frequencies: 4, 13, 25, 40, 48, 50, 51. The 26th value falls in the "5 min" class (positions 26-40). Median = 5 minutes.
Problem 4 — Landscaping wages
(i) Σ = $12,520. Mean = $12,520 / 10 = $1,252. Median = (5th + 6th) / 2 = ($1,000 + $1,020) / 2 = $1,010. Mode = both $960 and $980 (and $1,020) appear twice — strictly the data is multi-modal; quoting "$960 and $980" is acceptable.
(ii) Quote the median ($1,010). The mean is distorted by the owner's $3,500; the median sits squarely in the worker pay range.
(iii) New Σ = $12,520 − $3,500 = $9,020. New mean = $9,020 / 9 = $1,002.22. Removing the owner dropped the mean by about $250 — confirming the mean was being inflated by the outlier.
Problem 5 — Grouped retail times
(i) Σf = 8+22+30+15+5 = 80. Σ(f×x) = 8(4.5)+22(14.5)+30(24.5)+15(34.5)+5(44.5) = 36 + 319 + 735 + 517.5 + 222.5 = 1,830. Estimated mean = 1,830 / 80 = 22.9 min.
(ii) Modal class = 20 - 29 min (f = 30). Tells the manager that the most common length of stay is about 20-29 min — useful for staffing the floor during the heaviest browsing window.
(iii) It is an estimate because we assumed every customer in a class spent exactly the midpoint time — in reality their times are spread across each class, and the true individual values are not known from the table alone.