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.