Mathematics • Year 10 • Unit 4 • Lesson 7
Median and Mode — Skill Drill
Build fluency with Lesson 7's two measures of centre: median (middle value when ordered; for even n, the average of the two middle values) and mode (most frequent value; can be none, one, or many). Practise ordering data first, finding the median position, and identifying bimodal sets.
1. I do — fully worked example
Read every step. Each one has a short reason on the right.
Problem. Find the median and mode of: 7, 3, 8, 5, 3, 9, 6, 4, 3.
Step 1 — Order the data from smallest to largest.
3, 3, 3, 4, 5, 6, 7, 8, 9
Reason: the median is the middle of the ORDERED list. Skipping this step is the #1 cause of wrong medians.
Step 2 — Find the median position.
n = 9 (odd). Median position = (n + 1) ÷ 2 = (9 + 1) ÷ 2 = 5th value.
Reason: for odd n there is exactly one middle value; for even n we need the two middle values (Lesson 7 misconception).
Step 3 — Read off the 5th value.
Counting: 3 (1st), 3 (2nd), 3 (3rd), 4 (4th), 5 (5th). Median = 5.
Step 4 — Find the mode (most frequent value).
3 appears 3 times. Every other value appears once. Mode = 3.
Reason: Lesson 7 Key Terms — "Mode: the value that occurs most frequently".
Answer: median = 5, mode = 3.
2. We do — fill in the missing steps (even n)
This one has even n, so step 3 changes. Fill in each blank. 4 marks
Problem. Find the median and mode of: 12, 7, 9, 12, 14, 7, 18, 12.
Step 1 — Order the data.
_____, _____, _____, _____, _____, _____, _____, _____
Step 2 — Find the middle positions.
n = 8 (even). The two middle positions are the ____th and ____th values.
Step 3 — Average the two middle values.
Median = (____ + ____) ÷ 2 = ________
Notice: the median may NOT be a value in the data set — this is the Lesson 7 misconception.
Step 4 — Find the mode.
The value 12 appears ____ times; the value 7 appears ____ times. Mode = ________.
3. You do — independent practice
Eight graduated questions. Show ordering and middle-position working. Foundation (small odd-n sets), Standard (even n, ties), Extension (bimodal, no-mode, choose best measure).
Foundation — find the median (odd n)
3.1 Find the median of 8, 2, 5, 9, 4. 1 mark
3.2 Find the median of 14, 11, 18, 13, 16, 12, 15. 1 mark
3.3 Find the mode of 2, 5, 4, 2, 7, 2, 9, 4. 1 mark
Standard — even n and mixed measures
3.4 Find the median of 3, 5, 7, 9, 11, 13. (Hint: n = 6, so average the two middle values.) 2 marks
3.5 A group of 10 students recorded their shoe size: 7, 8, 9, 8, 10, 9, 8, 11, 9, 8. Find the median AND the mode. 2 marks
3.6 The data set 4, 6, 8, 10, 12, 14, 16, 18 has no value repeating. State whether it has a mode, and (using Lesson 7) explain in one sentence what to say in this case. 2 marks
Extension — bimodal, no-mode, and choosing the best measure
3.7 The favourite colour survey of 30 students shows: red 9, blue 9, green 6, yellow 4, purple 2. State the mode(s). What is the Lesson 7 Key Term for a data set with two modes? 2 marks
3.8 A real-estate report shows 9 house prices (millions of $): 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 7.5. Find the mean AND the median. Which measure better describes the "typical" price, and why? (Reference Lesson 7's "outlier-resistant" key term.) 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (median and mode of 12, 7, 9, 12, 14, 7, 18, 12)
Step 1 ordered: 7, 7, 9, 12, 12, 12, 14, 18.
Step 2: middle positions are the 4th and 5th values.
Step 3: median = (12 + 12) ÷ 2 = 12.
Step 4: 12 appears 3 times, 7 appears 2 times. Mode = 12.
3.1 — Median of 8, 2, 5, 9, 4
Ordered: 2, 4, 5, 8, 9. n = 5, middle = 3rd value. Median = 5.
3.2 — Median of 14, 11, 18, 13, 16, 12, 15
Ordered: 11, 12, 13, 14, 15, 16, 18. n = 7, middle = 4th value. Median = 14.
3.3 — Mode of 2, 5, 4, 2, 7, 2, 9, 4
2 appears 3 times, 4 appears twice, others once. Mode = 2.
3.4 — Median of 3, 5, 7, 9, 11, 13
n = 6, middle = 3rd and 4th values. Median = (7 + 9) ÷ 2 = 8. Notice 8 is not in the data set — exactly the Lesson 7 misconception.
3.5 — Shoe sizes
Ordered: 7, 8, 8, 8, 8, 9, 9, 9, 10, 11. n = 10, middle = 5th and 6th values = (8 + 9) ÷ 2 = 8.5. 8 appears 4 times, so mode = 8.
3.6 — No-mode case
Every value appears exactly once, so there is no mode. Lesson 7 misconception: "every data set has exactly one mode" is wrong — a data set can have no mode at all.
3.7 — Bimodal colour survey
Red and blue both appear 9 times. Modes = red and blue. Lesson 7 Key Term: a data set with two modes is bimodal.
3.8 — Median vs mean for house prices
Σx = 1.1 + 1.2 + 1.3 + 1.4 + 1.5 + 1.6 + 1.7 + 1.8 + 7.5 = 19.1. Mean = 19.1 ÷ 9 ≈ $2.12 m.
Ordered: 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 7.5. n = 9, middle = 5th value. Median = $1.5 m.
The median ($1.5 m) better describes the typical price. The $7.5 m mansion is an outlier; the mean is pulled up to $2.12 m, well above 8 of the 9 houses. The median is outlier-resistant (Lesson 7) so it barely moves.