Mathematics • Year 8 • Unit 4 • Lesson 3
Frequency Tables — Mixed Challenge
Pull together everything from Lesson 3: tallying, relative frequency, grouped tables, class centres, cumulative frequency, and the estimated mean from grouped data. Six mixed problems, one "find the mistake", and one open-ended challenge.
1. Mixed problems — choose the right move
Each question uses a different idea from Lesson 3. Show your working. 3 marks each
1.1 Convert each of these tallies to a frequency:
(a) ||||̅ ||||̅ ||||̅ ||
(b) ||||̅ ||||̅ ||||
(c) ||||̅ |
1.2 A frequency table has 4 categories with relative frequencies 0.35, 0.25, 0.30, 0.10. Total n = 60. Find the frequency of each category and verify they sum to 60.
1.3 Which set of class intervals is correctly written to avoid overlap? Briefly say why the others are wrong.
(a) 0–10, 10–20, 20–30
(b) 0–<10, 10–<20, 20–<30
(c) 0–9, 10–20, 21–30
1.4 A grouped table has frequencies 3, 5, 9, 7, 4. (a) Build the cumulative frequency column. (b) What is the cumulative frequency after the 3rd row? (c) What is n?
1.5 A class of 28 students sat a test. Grouped frequencies: 0–<40: 4, 40–<60: 8, 60–<80: 12, 80–100: 4. (a) State the modal class. (b) Estimate the mean using class centres.
1.6 Match each statement to the correct term (frequency, relative frequency, cumulative frequency, modal class):
(a) The running total of values up to and including the current row.
(b) The class interval with the highest frequency.
(c) The count of values in a single category.
(d) The proportion of the total represented by one category.
2. Find the mistake
Another student attempted this relative-frequency 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 has 4 categories with frequencies 8, 12, 16, 4. Find the relative frequency of each, as a decimal.
Line 1: Total n = 8 + 12 + 16 + 4 = 40.
Line 2: Category 1: 8 ÷ 40 = 0.20.
Line 3: Category 2: 12 ÷ 40 = 0.30.
Line 4: Category 3: 16 ÷ 40 = 0.40.
Line 5: Category 4: 4 ÷ 40 = 0.20.
Line 6: Check: 0.20 + 0.30 + 0.40 + 0.20 = 1.10 ✓
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected line(s) so the relative frequencies sum to 1.
Stuck? Recalculate each rel freq carefully. They must sum to 1, not 1.10.3. Open-ended challenge — design and analyse your own frequency table
This question has many valid answers. 4 marks
3.1 Choose ONE topic from the list below (or invent your own with similar style):
• Number of hours slept per night by 20 Year 8 students.
• Number of text messages sent in a day by 25 students.
• Daily maximum temperature for 30 days in your suburb.
• Time (minutes) spent travelling to school for 24 students.
Write up your analysis with the following:
(i) State your topic and invent 20–30 realistic data values.
(ii) Construct a grouped frequency table with 4 or 5 equal-width, non-overlapping intervals. Include Tally, Frequency, and Cumulative Frequency columns.
(iii) State the modal class.
(iv) Estimate the mean using class centres. Show working.
(v) Comment in one sentence: does the modal class match where the mean falls? What does this tell you about the shape of the data?
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Convert tallies to frequencies
(a) 3 full groups + 2 = 17. (b) 2 full groups + 4 = 14. (c) 1 full group + 1 = 6.
1.2 — Frequencies from relative frequencies (n = 60)
Cat 1: 0.35 × 60 = 21. Cat 2: 0.25 × 60 = 15. Cat 3: 0.30 × 60 = 18. Cat 4: 0.10 × 60 = 6. Check: 21 + 15 + 18 + 6 = 60 ✓.
1.3 — Class intervals
Correct: (b) 0–<10, 10–<20, 20–<30. (a) is wrong because the value 10 belongs to two classes (overlap). (c) is wrong because there are gaps — values like 9.5 or 20.7 aren't in any class.
1.4 — Cumulative frequency
(a) Cumulative frequencies: 3, 8, 17, 24, 28. (b) After 3rd row = 17. (c) n = last cumulative frequency = 28.
1.5 — Modal class and estimated mean
(a) Modal class = 60–<80 (f = 12 is highest).
(b) Centres: 20, 50, 70, 90. Σ(f × centre) = 4×20 + 8×50 + 12×70 + 4×90 = 80 + 400 + 840 + 360 = 1680. Estimated mean = 1680 ÷ 28 = 60.
1.6 — Match the term
(a) Cumulative frequency. (b) Modal class. (c) Frequency. (d) Relative frequency.
2 — Find the mistake
(a) The mistake is on Line 5.
(b) 4 ÷ 40 = 0.10, not 0.20. The student likely confused 0.10 with the earlier 0.20. This shows up immediately in the check: 0.20 + 0.30 + 0.40 + 0.20 = 1.10, which is impossible because relative frequencies must always sum to exactly 1.
(c) Corrected Line 5: "Category 4: 4 ÷ 40 = 0.10." Corrected Line 6 check: 0.20 + 0.30 + 0.40 + 0.10 = 1.00 ✓.
3 — Open-ended (sample solution — Hours slept per night)
(i) Topic: Hours slept per night by 20 Year 8 students. Sample data: 6, 7, 7, 8, 8, 8, 9, 9, 9, 9, 9, 9, 10, 10, 7, 8, 9, 8, 10, 7.
(ii) Table: 4–<6: f = 0, cum 0. 6–<8: f = 5 (6, 7, 7, 7, 7), cum 5. 8–<10: f = 13 (8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9), cum 18. 10–<12: f = 2 (10, 10, 10 — wait, recount). Recount: 6 ×1, 7 ×4, 8 ×5, 9 ×7, 10 ×3 → total 20. So 6–<8: 5, 8–<10: 12, 10–<12: 3. Cumfreq: 5, 17, 20. (Yes — depends on student's data; this is one valid version.)
(iii) Modal class: 8–<10 (f = 12).
(iv) Estimated mean: centres 5, 7, 9, 11. Σ(f × centre) = 0×5 + 5×7 + 12×9 + 3×11 = 0 + 35 + 108 + 33 = 176. Mean ≈ 176 ÷ 20 = 8.8 hours.
(v) Comment: The mean (8.8 h) sits inside the modal class (8–<10), so the data is roughly symmetric around 9 hours — most students get a healthy amount of sleep with only a few getting less.
Marking: 1 mark for plausible data + complete table; 1 mark for cumulative frequency column; 1 mark for correctly identifying modal class and computing estimated mean; 1 mark for a sensible comment linking the mean and modal class to the shape of the data.