Mathematics Standard • Year 12 • Module 8 • Lesson 2
Measures of Central Tendency — Past-Paper Style
Practise HSC Mathematics Standard 2-style writing on mean, median, mode, frequency-table means and outlier interpretation.
1. Short-answer questions
1.1 The marks (out of 20) for a class quiz were: 12, 15, 15, 16, 17, 18, 18, 18, 19, 20. Calculate the mean, median and mode. 3 marks Band 3
1.2 A frequency table records the number of cars per household in a small street:
| Cars (x) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Frequency (f) | 2 | 8 | 14 | 5 | 1 |
(a) Calculate the mean number of cars per household (2 d.p.).
(b) State the modal number of cars. 3 marks Band 3-4
1.3 A small accountancy firm publishes the "average annual salary" of its 8 staff as $115,000. The salaries are: $62,000, $64,000, $65,000, $68,000, $70,000, $72,000, $74,000, $445,000 (managing partner).
(a) Calculate the mean and median.
(b) Explain in one sentence why publishing the mean as the "average salary" is misleading in this situation, and state which measure should be published instead. 4 marks Band 4
2. Extended response
2.1 A NSW high school records its Year 12 mathematics class results on a 0-100 scale:
52, 54, 58, 60, 62, 65, 65, 67, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 95.
(a) Calculate the mean and the median of the 20 marks.
(b) The teacher wants to add another 5 marks to lift the class mean to exactly 75. What must the sum of those 5 new marks be?
(c) A parent says "the mean mark is higher than the median, so the class did really well". Evaluate this statement in 2-3 sentences, referring to what the relationship between mean and median tells you about the shape of the data. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — correct mean using Σx / n with working.
• 1 mark — correct median (average of 10th and 11th values).
Part (b) — 3 marks
• 1 mark — sets up required new total Σxnew = 75 × 25.
• 1 mark — subtracts the current Σx (20 marks) from the required total.
• 1 mark — correct value of the required sum of the 5 new marks.
Part (c) — 2 marks
• 1 mark — correctly states whether the data is left-skew, right-skew or symmetric, given the mean-vs-median relationship.
• 1 mark — challenges the parent's reasoning (a higher mean than median does not by itself mean "did well" — it means a few high marks pull the mean up).
Your response:
Stuck on (b)? New total − current total = sum of the 5 new marks.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Quiz marks (3 marks)
Sample response. Σx = 168. Mean = 168 / 10 = 16.8. Median = (17 + 18) / 2 = 17.5. Mode = 18 (appears three times).
Marking notes. 1 mark each for mean, median, mode. A common error is to take the median as the 5th OR 6th value rather than averaging them for an even n.
1.2 — Cars per household (3 marks)
(a) Sample response. Σf = 30. Σ(f×x) = 0(2) + 1(8) + 2(14) + 3(5) + 4(1) = 0 + 8 + 28 + 15 + 4 = 55. Mean = 55 / 30 = 1.83 cars.
(b) Sample response. Mode = 2 cars (highest frequency, f = 14).
Marking notes. 1 mark for correct Σf and Σ(f×x). 1 mark for the mean. 1 mark for the mode. Quoting "14" as the mode (the frequency, not the x-value) loses the mode mark — a very common error.
1.3 — Accountancy salaries (4 marks)
(a) Sample response. Σ = $920,000. Mean = $920,000 / 8 = $115,000. n = 8, so median = average of 4th and 5th values = ($68,000 + $70,000) / 2 = $69,000.
(b) Sample response. The mean is pulled massively up by the managing partner's $445,000, so the headline "average $115,000" overstates what 7 of the 8 staff actually earn. The median ($69,000) is a more honest summary of typical pay.
Marking notes. 1 mark for the mean. 1 mark for the median. 1 mark for naming the outlier as the cause of the inflated mean. 1 mark for naming the median as the better summary. A response that says "median is better" with no reason scores ½ at most.
2.1 — Year 12 maths class results (7 marks): Band-6 sample with annotations
(a) Mean and median of the 20 marks.
Σx = 52+54+58+60+62+65+65+67+70+72+74+76+78+80+82+84+86+88+90+95 = 1,456.
Mean = 1,456 / 20 = 72.8. [1 mark — mean with working.]
Median (n = 20) = (10th + 11th) / 2 = (72 + 74) / 2 = 73. [1 mark — median with correct positions identified.]
(b) Sum of the 5 new marks needed for class mean = 75.
Required total for 25 marks: Σxnew = 75 × 25 = 1,875. [1 mark — sets up required total.]
Current Σx = 1,456. [1 mark — subtracts current total.]
Required new-marks sum = 1,875 − 1,456 = 419. [1 mark — correct value, units understood as "out of 5 × 100 = 500 max".]
(Sanity check: the 5 new marks must average 419 / 5 ≈ 83.8 — achievable but well above the current 72.8 average.)
(c) Mean > median — what does it mean?
The mean (72.8) is just below the median (73), so the data is approximately symmetric with a very slight left-skew, not right-skewed. [1 mark — correctly characterises the distribution shape.]
The parent's reasoning is wrong on two counts. First, in this data set the mean is actually lower than the median (72.8 < 73), not higher — so the premise of the claim is incorrect. Second, even if the mean had been higher, that would only mean a few high outliers were pulling it up, not that the class as a whole "did really well". A class with most students at 60 and one student at 100 has mean > median, but most students did not do well. [1 mark — challenges the parent's reasoning with a counter-example or principle.]
Total: 7/7.
Band descriptors for marker.
Band 3: Calculates mean correctly; struggles with the even-n median or with part (b). ≈ 2-3 marks.
Band 4: Mean, median and required-total set up correctly; arithmetic slip in (b); (c) gives a vague comment with no challenge to the parent. ≈ 4-5 marks.
Band 5: All numerical parts correct; (c) correctly describes shape but does not directly address the parent's reasoning. ≈ 6 marks.
Band 6: Full numerical solution AND a sharp evaluation of (c) that both notes the actual mean/median ordering and explains why "mean > median" does not by itself imply strong performance. 7/7.