Mathematics • Year 7 • Unit 4 • Lesson 9
The Median — Mixed Challenge
Combine every median skill: odd n, even n, finding a missing value to hit a target median, comparing mean and median, and deciding which to use. Then spot one plausible student error, and design your own median-vs-mean investigation.
1. Mixed problems — apply every skill
Each question uses a different idea from the lesson. Show working. 2 marks each
1.1 Find the median of: 14, 8, 19, 22, 11, 25, 6, 15.
1.2 Find the median of these 5 daily temperatures (°C): 27, 31, 24, 29, 28.
1.3 Six numbers have a median of 15. Five of them are 8, 12, 14, 18, 20. What is the sixth, and where does it sit when sorted?
1.4 For the data {5, 6, 6, 7, 8, 9}: (a) calculate the median, (b) calculate the mean, (c) explain in one sentence why the two values are so close.
1.5 A teacher reports: "Median test score = 71." If 23 students took the test, how many scored AT LEAST 71? Explain how you know without seeing every score.
1.6 Take the dataset {10, 20, 30, 40, 50}. (a) Find the median. (b) Now replace 50 with 5,000. Find the new median. (c) Now find the mean before and after the replacement. Use your numbers to explain in one sentence why the median is called "resistant".
2. Find the mistake
Another Year 7 student answered the prompt: "Find the median of: 8, 14, 22, 11, 17, 25." Their working has exactly one error. Spot it, explain why it's wrong, then write the correct solution. 3 marks
Student's working:
Line 1: The data is: 8, 14, 22, 11, 17, 25.
Line 2: n = 6 (even).
Line 3: Middle positions are 3rd and 4th.
Line 4: 3rd value = 22, 4th value = 11.
Line 5: Median = (22 + 11) ÷ 2 = 16.5. [Answer: 16.5]
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write the correct calculation and the correct median.
Stuck? Revisit lesson § "Spot the Trap" — Wrong: NOT sorting data first.3. Open-ended challenge — design your own median investigation
This question has many correct answers. Show your work clearly. 4 marks
3.1 Design a one-page investigation that demonstrates WHY the median is preferred to the mean in certain situations. You must:
- (i) Choose a real-world variable likely to contain extreme outliers (e.g. house prices, athlete salaries, response times, daily screen time);
- (ii) Invent a small dataset (8–12 values) for this variable, including AT LEAST ONE clear outlier;
- (iii) Calculate BOTH the mean and the median;
- (iv) Compare the two figures in one short sentence;
- (v) Decide which figure you would report to the public, and justify in one or two sentences using the word "resistant".
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Median of {14,8,19,22,11,25,6,15}
Sort: 6, 8, 11, 14, 15, 19, 22, 25. n = 8, middle = 4th and 5th = 14 and 15. Median = (14 + 15) ÷ 2 = 14.5.
1.2 — Median of {27,31,24,29,28}
Sort: 24, 27, 28, 29, 31. n = 5, position = 3rd = 28 °C.
1.3 — Finding the 6th value
Sort the known five: 8, 12, 14, 18, 20. With n = 6 and median = 15, the average of the 3rd and 4th values (sorted) must equal 15.
Trying the unknown = 16: sorted becomes 8, 12, 14, 16, 18, 20 → 3rd = 14, 4th = 16, median = (14+16)÷2 = 15 ✓.
Sixth value = 16, sitting between 14 and 18 (4th in sorted list).
1.4 — {5,6,6,7,8,9}
(a) Sorted already. n = 6, middle = 3rd & 4th = 6 and 7. Median = (6 + 7) ÷ 2 = 6.5.
(b) Mean = (5+6+6+7+8+9) ÷ 6 = 41 ÷ 6 ≈ 6.83.
(c) The values are close together with no outliers, so the mean and median agree closely — both are valid summaries of this symmetric-ish dataset.
1.5 — How many scored ≥ 71?
n = 23 (odd), median position = (23+1)÷2 = 12th. By definition, the median (71) AND the 11 values above it all sit at or above 71. So at least 12 students scored 71 or higher. (Could be more if any students below position 12 also scored 71, but we know at least 12.)
1.6 — Median is resistant
(a) Sorted {10, 20, 30, 40, 50}. n = 5, median = 3rd = 30.
(b) Replace 50 with 5000. Sorted: 10, 20, 30, 40, 5000. Median = 3rd = 30 (unchanged!).
(c) Original mean = (10+20+30+40+50) ÷ 5 = 30. New mean = (10+20+30+40+5000) ÷ 5 = 5100 ÷ 5 = 1020. The mean jumped from 30 to 1020 because one extreme value was added to the SUM. The median didn't move because it depends only on the POSITION of the middle value in the sorted list — that's what "resistant" means.
2 — Find the mistake
(a) Line 1 (or "missing sort step") — the student went straight to finding middle positions without sorting first.
(b) The median is the middle of the SORTED list, not the original order. The "3rd value = 22" is the 3rd value of the unsorted list, not the actual 3rd-smallest.
(c) Sort first: 8, 11, 14, 17, 22, 25. Middle positions 3rd and 4th = 14 and 17. Median = (14 + 17) ÷ 2 = 15.5.
3 — Median investigation (sample answer)
(i) Variable: daily screen time (minutes) for 10 friends.
(ii) Values: 60, 75, 80, 85, 90, 95, 100, 110, 120, 480. The 480-minute value (8 h) is a clear outlier — possibly a sick day spent on the phone.
(iii) Sorted as above. n = 10, median = (5th + 6th) ÷ 2 = (90 + 95) ÷ 2 = 92.5 min. Mean = (60+75+80+85+90+95+100+110+120+480) ÷ 10 = 1295 ÷ 10 = 129.5 min.
(iv) The mean (~130 min) is 37 min above the median (~92 min); the outlier has pulled it up.
(v) I would report the median (92.5 min) because the median is RESISTANT to the 480-min outlier — it depends only on position, not magnitude, so it reflects what a typical friend actually uses each day. The mean would mislead readers into thinking everyone spends ~130 min daily, when really 9 of 10 are between 60 and 120.
Marking: 1 mark each for (i)+(ii), (iii) calculations, (iv) comparison, (v) "resistant" justification.