Mathematics • Year 8 • Unit 4 • Lesson 9
Median and Mode
Build fluency finding the median (odd and even n), identifying the mode (including bimodal sets), and choosing the right measure of centre.
1. I do — fully worked example
Read every line. Each step has a short reason so you can see why we do it, not just what we do.
Problem. Find the median and mode of: 14, 9, 7, 14, 11, 8, 6, 14, 10, 9.
Step 1 — Order the data smallest to largest.
Ordered: 6, 7, 8, 9, 9, 10, 11, 14, 14, 14.
Reason: median needs ordered data; mode is easier to spot when values are grouped.
Step 2 — Find the median position (n = 10, even).
For even n, median = average of (n/2)th and (n/2 + 1)th values = average of 5th and 6th values.
5th value = 9; 6th value = 10. Median = (9 + 10) ÷ 2 = 9.5.
Step 3 — Find the mode (most frequent value).
14 appears 3 times — more than any other value. Mode = 14.
Answer: Median = 9.5, Mode = 14.
2. We do — fill in the missing steps
Same shape as Section 1, but with the working faded. Fill in each blank. 4 marks
Problem. Find the median and mode of: 11, 5, 9, 3, 15, 7, 13.
Step 1 — Order:
Ordered: ______, ______, ______, ______, ______, ______, ______.
Step 2 — Median (n = ____ , odd):
Position = (n+1)/2 = ______. Median = ______ (value at that position).
Step 3 — Mode:
Each value appears how often? ______. Mode = ______.
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation. The middle two are standard. The last two are extension.
Foundation — quick recall
3.1 Find the median of: 14, 4, 10, 6, 12, 8. 1 mark
3.2 Find the mode of: 3, 7, 5, 7, 2, 9, 7, 4, 5. 1 mark
3.3 Which data set is bimodal? (a) 1, 2, 3, 4, 5. (b) 2, 2, 3, 4, 4, 5. (c) 1, 1, 1, 2, 3. (d) 5, 6, 7, 8, 9. 1 mark
3.4 Define "resistant" in the context of measures of centre. 1 mark
Standard — choose the right measure
3.5 Find both the mean AND the median of: 40, 45, 48, 50, 52, 55, 96. Which is more representative of "typical" performance? 2 marks
3.6 A real-estate report says the average house price in a suburb is $1.4 million but most homes sell for around $900 000. (a) Which measure of centre is likely being reported? (b) Which measure would be fairer? Explain in one sentence. 2 marks
Extension — missing-value reasoning
3.7 A data set of 5 values has a mean of 6.2. Four of the values are: 4, 7, 8, 5. (a) Find the missing fifth value using the mean formula. (b) Then find the median of the complete data set. 2 marks
3.8 Why is the median almost unchanged when an extreme outlier is added, while the mean changes significantly? Use the words "all values" and "middle position" in your explanation. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (median & mode of 7 values)
Step 1: Ordered = 3, 5, 7, 9, 11, 13, 15. Step 2: n = 7; position = (7+1)/2 = 4; median = 9. Step 3: Each value appears once → no mode.
3.1 — Median (even n)
Ordered: 4, 6, 8, 10, 12, 14. n = 6. Median = (8 + 10) ÷ 2 = 9.
3.2 — Mode
7 appears 3 times — more than any other value. Mode = 7.
3.3 — Bimodal
(b) 2, 2, 3, 4, 4, 5 — both 2 and 4 appear twice (tied for most frequent).
3.4 — Resistant
A measure is resistant if it is not strongly affected by outliers — its value barely moves when an extreme value is added or removed. The median is resistant; the mean is not.
3.5 — Mean vs median with outlier
Σx = 40+45+48+50+52+55+96 = 386. Mean = 386 ÷ 7 ≈ 55.1. n = 7, position 4 → Median = 50. The median better represents typical performance because the outlier (96) pulled the mean up; 4 of the 7 scores are between 45 and 55.
3.6 — House prices
(a) The reported average is the mean.
(b) The median would be fairer — a few luxury houses pull the mean up well above what most buyers actually pay, while the median is resistant to those extreme high values.
3.7 — Missing value + median
(a) Total = 6.2 × 5 = 31. Known sum = 4+7+8+5 = 24. Missing value = 31 − 24 = 7.
(b) Full data ordered: 4, 5, 7, 7, 8. Median = 3rd value = 7.
3.8 — Why median is resistant
The mean adds all values together and divides by n, so a large outlier pulls the total — and therefore the mean — up or down significantly. The median only depends on the value at the middle position when the data is ordered, so as long as the outlier doesn't replace the middle value, the median stays the same.