Mathematics • Year 8 • Unit 4 • Lesson 8
Mean — Mixed Challenge
Pull together everything from Lesson 8: mean from raw data, frequency tables, and grouped data; effect of outliers; finding a missing value to achieve a target mean.
1. Mixed problems — choose the right move
Each question uses a different idea from Lesson 8. Show your working. 3 marks each
1.1 Calculate the mean of: 14, 9, 7, 12, 11, 8, 6, 10, 9, 14.
1.2 A frequency table: x = 2 (f=5), x = 4 (f=8), x = 6 (f=4), x = 8 (f=3). Find the mean using mean = Σ(f × x) ÷ Σf.
1.3 Estimate the mean from this grouped table (n = 30): 0–<5 (f=6), 5–<10 (f=10), 10–<15 (f=8), 15–<20 (f=4), 20–<25 (f=2). Show midpoint and f × midpoint columns.
1.4 A data set of 6 values has a mean of 12. Five of the values are 8, 10, 11, 13, 14. Find the missing 6th value using the mean formula.
1.5 A student's first 4 test marks are 65, 72, 78, 81. They want a mean of 75 across 5 tests. What must they score on the 5th test? Show working.
1.6 A test class has 8 students with a mean score of 70. Two new students join, scoring 90 and 80. What is the new mean across all 10 students?
2. Find the mistake
Another student attempted this frequency-table mean 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: A frequency table: x = 1 (f=3), x = 2 (f=5), x = 3 (f=4). Find the mean.
Line 1: f × x: 1×3 = 3, 2×5 = 10, 3×4 = 12.
Line 2: Σ(f × x) = 3 + 10 + 12 = 25.
Line 3: Number of rows = 3.
Line 4: Mean = 25 ÷ 3 ≈ 8.33.
Line 5: Sanity check: mean (8.33) should be between min (1) and max (3) — it is NOT, so something is wrong.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write the corrected working and the correct mean.
Stuck? Mean from a frequency table divides by total frequency (Σf), not by the number of rows.3. Open-ended challenge — design a data set
This question has many valid answers. 4 marks
3.1 Your job: invent TWO data sets, each with EXACTLY 8 values, where:
• DATA SET A: the mean is a fair representation of the typical value (data is fairly symmetric, no outliers).
• DATA SET B: the mean is misleading because of a single outlier.
Write up your answer with:
(i) The 8 values for each data set (use a real-world context for both).
(ii) The mean of each set, with working shown.
(iii) For Set B: identify the outlier, calculate the mean WITHOUT the outlier, and explain in one sentence by how much the outlier shifted the mean.
(iv) Recommend which measure (mean or median) you would report to a journalist for each set.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Mean of 10 values
Σx = 14+9+7+12+11+8+6+10+9+14 = 100. Mean = 100 ÷ 10 = 10.
1.2 — Mean from frequency table
f×x: 2×5=10, 4×8=32, 6×4=24, 8×3=24. Σ(f×x) = 90. Σf = 20. Mean = 90 ÷ 20 = 4.5.
1.3 — Estimated mean from grouped data
Midpoints: 2.5, 7.5, 12.5, 17.5, 22.5. f × mid: 6×2.5=15, 10×7.5=75, 8×12.5=100, 4×17.5=70, 2×22.5=45. Σ = 305. Mean ≈ 305 ÷ 30 ≈ 10.17.
1.4 — Find missing value
Σ of all 6 = 12 × 6 = 72. Sum of known 5 = 8+10+11+13+14 = 56. Missing value = 72 − 56 = 16.
1.5 — Target mean
Sum needed for 5 tests = 75 × 5 = 375. Sum so far = 65+72+78+81 = 296. Required 5th score = 375 − 296 = 79.
1.6 — New mean after adding students
Old total = 70 × 8 = 560. New total = 560 + 90 + 80 = 730. New mean = 730 ÷ 10 = 73.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) For a frequency table, you must divide by the total frequency (Σf), not the number of rows. Σf = 3 + 5 + 4 = 12, not 3.
(c) Correct: Σf = 12. Mean = 25 ÷ 12 ≈ 2.08. The sanity check now passes — 2.08 sits between min (1) and max (3).
3 — Open-ended (sample solution)
(i) Set A — Heights of 8 students (cm): 158, 160, 161, 163, 164, 165, 167, 169.
(i) Set B — Annual incomes ($000s) of 8 employees: 45, 48, 50, 52, 55, 58, 60, 320.
(ii) Set A mean: Σx = 1307. Mean = 1307 ÷ 8 = 163.4 cm. Set B mean: Σx = 688. Mean = 688 ÷ 8 = $86 000.
(iii) Set B outlier: $320 000. Without outlier: Σx = 368. n = 7. Mean = 368 ÷ 7 ≈ $52 600. The outlier shifted the mean upward by about $33 400 — a huge misleading effect.
(iv) Recommendation: Set A → report the mean (no outliers, fair summary). Set B → report the median (≈ $53 500), because the mean misrepresents the typical employee's income.
Marking: 1 mark for two plausible 8-value data sets matching the brief; 1 mark for both means correct; 1 mark for Set B outlier handling; 1 mark for sensible reporting recommendation with reason.