Mathematics Standard • Year 12 • Module 8 • Lesson 2
Measures of Central Tendency — Skill Drill
Build fluency in calculating mean, median and mode for raw data and frequency tables, and judging which centre best fits a data set.
1. Quick recall
Answer each in the space provided. 1 mark each
Q1.1 Write each formula. Mean: x̄ = ____________________ Mean from a frequency table: x̄ = ____________________
Q1.2 Define each measure in one phrase.
Median = ____________________________________________________
Mode = ______________________________________________________
Q1.3 Which measure of centre is most affected by an outlier? ____________. Which is least affected? ____________.
2. Worked example — mean, median, mode with an outlier
Daily ice-cream van sales ($): 450, 520, 480, 600, 500, 2000, 550, 510, 490, 530. Find mean, median and mode, then comment on which one best represents a typical day.
Step 1 — Order the data.
450, 480, 490, 500, 510, 520, 530, 550, 600, 2000 (n = 10)
Step 2 — Mean.
Σx = 450 + 480 + 490 + 500 + 510 + 520 + 530 + 550 + 600 + 2000 = 6,630
x̄ = 6,630 ÷ 10 = $663
Step 3 — Median (n even → average the middle two).
Median = (510 + 520) ÷ 2 = $515
Step 4 — Mode.
All values appear once → no mode.
Conclusion. The $2,000 day pulls the mean up to $663, far above 9 of the 10 actual days. The median ($515) better represents a typical day.
3. Faded example — mean from a frequency table
Number of pets per household (n = 30): 4 marks
| Pets (x) | Frequency (f) | f × x |
|---|---|---|
| 0 | 6 | 0 |
| 1 | 12 | ____________ |
| 2 | 8 | ____________ |
| 3 | 3 | ____________ |
| 4 | 1 | ____________ |
| Σ | ____________ | ____________ |
Step — Mean. x̄ = Σ(f×x) ÷ Σf = ________ ÷ ________ = ____________ pets per household.
Median position. n = 30, so median is the (30+1)/2 = 15.5th value → the mean of the 15th and 16th data points → Median = ____________.
Mode. Most frequent x-value = ____________.
4. Graduated practice — find the centre
Foundation — one-step (4 questions)
| Q | Problem | Answer |
|---|---|---|
| 4.1 1 | Find the mean of: 6, 8, 10, 12, 14. | |
| 4.2 1 | Find the median of: 12, 7, 15, 9, 10, 8, 11. | |
| 4.3 1 | Find the mode of: 3, 5, 5, 7, 8, 5, 9, 10. | |
| 4.4 1 | Find the median of: 14, 18, 20, 22, 25, 30 (n = 6). |
Standard — typical HSC difficulty (6 questions)
4.5 Class test scores: 62, 68, 71, 71, 74, 78, 82, 85, 88, 95. Find the mean, median and mode. 2 marks
4.6 A frequency table shows shoe sizes for 25 students: size 6 (3), size 7 (6), size 8 (8), size 9 (5), size 10 (3). Find the mean shoe size (1 d.p.). 2 marks
4.7 A small business pays 9 staff weekly: $880, $880, $880, $920, $940, $940, $960, $980, $4,200 (CEO). Find the mean and median. Which is more typical of a worker's pay? 2 marks
4.8 The mean of five numbers is 14. Four of them are 10, 12, 15 and 17. Find the fifth. 2 marks
4.9 A frequency table: x = 1 (f=4), x = 2 (f=7), x = 3 (f=5), x = 4 (f=3), x = 5 (f=1). Find the median. 2 marks
4.10 Find the mean of: 25, 28, 30, 32, 35, 36, 38, 40, 42 (n = 9). 2 marks
Extension — work backwards (2 questions)
4.11 A student has the test marks 72, 68, 75, 80. What mark on the fifth test will raise the mean to 76? 3 marks
4.12 Daily customer counts at a cafe: 60, 65, 70, 72, 75, 80, 82, 85, 90, 600 (where 600 is a sports-day spike). (a) Calculate the mean with and without the spike. (b) Calculate the median with and without the spike. (c) State which measure is more robust to the outlier, with a one-sentence justification. 3 marks
5. Self-check the easy 3
Tick once you've checked your reasoning works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Formulas
x̄ = Σx ÷ n. From frequency table: x̄ = Σ(f×x) ÷ Σf.
Q1.2 — Definitions
Median = middle value of the ordered data set (average of the two middle values if n is even). Mode = the most frequently occurring value.
Q1.3 — Outlier effect
Most affected: mean. Least affected: median.
Q3 — Faded example (pets)
f×x column: 0, 12, 16, 9, 4. Σf = 30, Σ(f×x) = 41.
Mean = 41 ÷ 30 = 1.37 pets per household (2 d.p.).
Median: the 15th and 16th values are both x = 1 (because 6 households have 0 pets and the next 12 have 1 pet, covering positions 7 to 18). Median = 1 pet.
Mode = 1 pet (highest frequency f = 12).
Q4.1–4.4 — Foundation
4.1: mean = (6+8+10+12+14)/5 = 50/5 = 10.
4.2: ordered: 7, 8, 9, 10, 11, 12, 15 → median = 10 (4th value, n = 7).
4.3: mode = 5 (appears 3 times).
4.4: n = 6, middle two are 20 and 22 → median = (20+22)/2 = 21.
Q4.5 — Test scores
Σx = 774. Mean = 774/10 = 77.4. Median = (74+78)/2 = 76. Mode = 71 (appears twice).
Q4.6 — Shoe sizes from frequency table
Σf = 25. Σ(f×x) = 3(6) + 6(7) + 8(8) + 5(9) + 3(10) = 18 + 42 + 64 + 45 + 30 = 199.
Mean = 199/25 = 7.96 ≈ 8.0.
Q4.7 — Pay with CEO outlier
Σ = $11,580. Mean = $11,580 / 9 = $1,286.67. Median = 5th value (ordered) = $940.
The median ($940) is more typical of a worker's pay — the CEO's $4,200 inflates the mean far above what any worker actually earns.
Q4.8 — Missing value to give mean = 14
Σx = 14 × 5 = 70. Sum of known four = 10 + 12 + 15 + 17 = 54. Fifth = 70 − 54 = 16.
Q4.9 — Median from frequency table
Σf = 20, so median is the average of the 10th and 11th values. Cumulative: x=1 covers positions 1-4; x=2 covers 5-11; x=3 covers 12-16. The 10th and 11th values are both 2 → median = 2.
Q4.10 — Mean of 9 values
Σx = 25+28+30+32+35+36+38+40+42 = 306. Mean = 306/9 = 34.
Q4.11 — Fifth test for mean 76
Required Σx = 76 × 5 = 380. Current sum = 72 + 68 + 75 + 80 = 295. Fifth test = 380 − 295 = 85.
Q4.12 — Cafe customer counts with sports-day spike
(a) With spike: Σ = 1,279; mean = 1,279/10 = 127.9. Without spike: Σ = 679; mean = 679/9 = 75.4. Outlier inflated the mean by ~$52.5.
(b) Ordered (with spike): 60, 65, 70, 72, 75, 80, 82, 85, 90, 600. Median = (75+80)/2 = 77.5. Without spike: middle of 9 → median = 75. Median barely moved.
(c) The median is more robust: removing the outlier changed the mean by 52.5 but the median by only 2.5, because the median only depends on the position of the middle value, not the size of any extreme value.