Mathematics • Year 8 • Unit 4 • Lesson 9
Median and Mode — Mixed Challenge
Pull together everything from Lesson 9: finding median (odd/even n), identifying mode (none/one/bimodal), comparing measures of centre, and choosing the right one for the context.
1. Mixed problems — choose the right move
Each question uses a different idea from Lesson 9. Show your working. 3 marks each
1.1 Find the median, mode, and range of: 22, 18, 25, 22, 19, 28, 22, 24, 20.
1.2 Identify the type of mode for each (no mode / unimodal / bimodal):
(a) 1, 2, 3, 4, 5, 6.
(b) 2, 4, 4, 6, 8, 10.
(c) 1, 1, 2, 3, 3, 5, 5, 5.
(d) 3, 3, 5, 5, 7.
1.3 A data set of 6 values has mean 12, median 11.5, mode 8. (a) What might this tell you about the data's shape? (b) Invent a possible data set of 6 values matching these statistics (use whole numbers).
1.4 For each scenario, decide which measure of centre is best (mean / median / mode) and justify in one sentence:
(a) The "average" sale price in a suburb with several luxury mansions.
(b) Choosing the most popular T-shirt size to manufacture more of.
(c) Calculating a student's term grade from 5 consistent test scores.
1.5 Find the median of an EVEN-sized data set: 12, 8, 14, 6, 10, 18, 15, 9. Show your ordering.
1.6 A 6-value data set has mean 10 and median 9.5. A 7th value of 100 is added. (a) What is the new mean? (b) Without knowing the data, can you say what happens to the median (rises / falls / barely changes)? Justify.
2. Find the mistake
Another student attempted this median problem. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Problem: Find the median of: 17, 23, 11, 19, 25, 15.
Line 1: The data has 6 values, so n = 6.
Line 2: For an even n, median = average of the 2 middle values.
Line 3: Middle values (in original order) are the 3rd and 4th: 11 and 19.
Line 4: Median = (11 + 19) ÷ 2 = 15.
Line 5: Answer: median = 15.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working and the correct median.
Stuck? You MUST order the data first before identifying the middle values.3. Open-ended challenge — design a data set with target statistics
This question has many valid answers. 4 marks
3.1 Your job: invent a data set of EXACTLY 9 whole-number values that satisfies ALL of the following at once:
• mean = 20
• median = 18
• mode = 15 (single mode, appearing at least 3 times)
• range = 30
Write up your answer with:
(i) List your 9 values (ordered smallest to largest).
(ii) Verify the mean by showing the sum and division.
(iii) Verify the median by identifying the middle value.
(iv) Verify the mode by counting frequencies.
(v) Verify the range (max − min).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Median, mode, range
Ordered: 18, 19, 20, 22, 22, 22, 24, 25, 28. n = 9. Median = 5th value = 22. Mode = 22 (3 times). Range = 28 − 18 = 10.
1.2 — Type of mode
(a) No mode (each value appears once). (b) Unimodal (4 appears twice; one mode = 4). (c) Unimodal (5 appears 3 times — only one most-frequent value). (d) Bimodal (3 and 5 both appear twice).
1.3 — Interpreting summary stats
(a) Mean (12) > median (11.5) > mode (8) suggests skewed right — a few high values pull the mean up.
(b) Sample data set: 8, 8, 11, 12, 13, 20 (mean = (8+8+11+12+13+20)/6 = 72/6 = 12 ✓; median = (11+12)/2 = 11.5 ✓; mode = 8 ✓).
1.4 — Choose the measure
(a) Median — luxury mansions are outliers that inflate the mean.
(b) Mode — the most popular size sells the best.
(c) Mean — consistent scores with no outliers means the mean fairly represents the term average.
1.5 — Even-sized median
Ordered: 6, 8, 9, 10, 12, 14, 15, 18. n = 8. Median = (4th + 5th) ÷ 2 = (10 + 12) ÷ 2 = 11.
1.6 — Adding an outlier
(a) Old total = 60. New total = 160. New mean = 160 ÷ 7 ≈ 22.86 — increased by ~13.
(b) The median barely changes — adding one value to a set of 6 makes the median the 4th value of a 7-value set; the new value sits at the top so the 4th position only shifts slightly (resistant to outliers).
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The student took the 3rd and 4th values from the data in its ORIGINAL order. The median requires the data to be ORDERED first; otherwise the "middle" values are arbitrary.
(c) Correct: order the data → 11, 15, 17, 19, 23, 25. 3rd and 4th values = 17 and 19. Median = (17 + 19) ÷ 2 = 18.
3 — Open-ended (sample solution)
(i) Sample data set: 5, 12, 15, 15, 18, 21, 24, 35, 35... wait, that gives two modes. Try: 5, 15, 15, 15, 18, 19, 25, 32, 36. Check: range = 36 − 5 = 31 (not 30). Adjust to: 5, 15, 15, 15, 18, 20, 25, 32, 35. Range = 30 ✓.
(ii) Mean check: Σx = 5+15+15+15+18+20+25+32+35 = 180. Mean = 180 ÷ 9 = 20 ✓.
(iii) Median check: n = 9, position 5 = 18 ✓.
(iv) Mode check: 15 appears 3 times — most frequent → mode = 15 ✓.
(v) Range check: 35 − 5 = 30 ✓.
Marking: 1 mark for any valid 9-value data set; 1 mark for correct mean verification; 1 mark for correct median + mode verification; 1 mark for correct range verification. Note: students can use trial-and-error to find any valid set.