Mathematics • Year 8 • Unit 4 • Lesson 8
Mean (Average)
Build fluency calculating the mean from raw data, frequency tables, and grouped data using midpoints — and judging when outliers make the mean misleading.
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. Calculate the mean of the eight test scores: 62, 71, 68, 74, 70, 69, 72, 20. Then calculate the mean without the outlier and compare.
Step 1 — Apply the formula mean = Σx ÷ n.
Σx = 62 + 71 + 68 + 74 + 70 + 69 + 72 + 20 = 506. n = 8.
Reason: sum all values first, then divide by the count.
Step 2 — Divide.
Mean = 506 ÷ 8 = 63.25
Step 3 — Identify the outlier and recalculate without it.
Outlier = 20 (well below the cluster 62–74). Without 20: Σx = 506 − 20 = 486. n = 7. Mean = 486 ÷ 7 ≈ 69.43.
Reason: the outlier dragged the mean down by over 6 marks. The cleaned mean better represents the typical student.
Answer: Mean with outlier = 63.25; mean without = ≈ 69.4. The outlier (20) drags the mean noticeably below the typical performance.
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. Calculate the mean of the six values: 4, 7, 9, 6, 14, 8.
Step 1 — Sum the values:
Σx = 4 + 7 + 9 + 6 + 14 + 8 = ______
Step 2 — Count the values:
n = ______
Step 3 — Divide:
Mean = ______ ÷ ______ = ______
Step 4 — Sanity check:
Mean should be between min (______) and max (______). It is ______, so ______ (check).
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 — direct mean
3.1 Calculate the mean of: 3, 5, 7, 9, 11. 1 mark
3.2 Calculate the mean of: 12, 15, 18, 21, 24, 30. 1 mark
3.3 A data set has a mean of 10. A new data value of 100 is added. Briefly, what happens to the mean? 1 mark
3.4 A frequency table: value 1 with freq 3, value 2 with freq 5, value 3 with freq 4. Find the mean using mean = Σ(f × x) ÷ Σf. 1 mark
Standard — frequency tables and grouped data
3.5 Calculate the mean from this frequency table — show the f × x column.
Values and frequencies: (3, 4), (5, 6), (7, 8), (9, 5), (11, 2). 2 marks
3.6 Estimate the mean from this grouped frequency table (n = 40). Show midpoint and f × midpoint columns.
0–<10 (f=6), 10–<20 (f=12), 20–<30 (f=14), 30–<40 (f=8). 2 marks
Extension — outliers and reasoning
3.7 Five test scores: 72, 75, 78, 80, 20. (a) Calculate the mean. (b) Calculate the mean without the outlier. (c) Which mean is more representative of the four other students, and by how much was the original mean dragged down? 2 marks
3.8 A class of 5 students has a mean test score of 76. A 6th student joins with a score of 88. (a) What is the total score of the original 5 students? (b) What is the new total for all 6? (c) What is the new mean? 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (mean of 6 values)
Step 1: Σx = 48. Step 2: n = 6. Step 3: 48 ÷ 6 = 8. Step 4: min = 4, max = 14; mean 8 lies between them ✓.
3.1 — Direct mean
Σx = 3 + 5 + 7 + 9 + 11 = 35. Mean = 35 ÷ 5 = 7.
3.2 — Direct mean
Σx = 12 + 15 + 18 + 21 + 24 + 30 = 120. Mean = 120 ÷ 6 = 20.
3.3 — Effect of large new value
The mean will increase significantly — the new value is much higher than the current mean, so it pulls the mean upward.
3.4 — Mean from frequency table (small)
f × x: 1×3 = 3, 2×5 = 10, 3×4 = 12. Σ(f×x) = 25. Σf = 12. Mean = 25 ÷ 12 ≈ 2.08.
3.5 — Mean from frequency table
f × x: 3×4 = 12, 5×6 = 30, 7×8 = 56, 9×5 = 45, 11×2 = 22. Σ(f×x) = 165. Σf = 25. Mean = 165 ÷ 25 = 6.6.
3.6 — Estimated mean from grouped data
Midpoints: 5, 15, 25, 35. f × mid: 6×5 = 30, 12×15 = 180, 14×25 = 350, 8×35 = 280. Σ = 840. Mean ≈ 840 ÷ 40 = 21 (estimate from grouped data).
3.7 — Outlier effect
(a) Σx = 72+75+78+80+20 = 325. Mean = 325 ÷ 5 = 65.
(b) Without 20: Σx = 305. n = 4. Mean = 305 ÷ 4 = 76.25.
(c) The mean WITHOUT the outlier (76.25) better represents the four other students. The outlier dragged the original mean down by about 11 marks.
3.8 — Adding a new value
(a) Total of original 5 = 76 × 5 = 380.
(b) New total = 380 + 88 = 468.
(c) New mean = 468 ÷ 6 = 78.