Mathematics • Year 8 • Unit 4 • Lesson 7
Histograms — Mixed Challenge
Pull together everything from Lesson 7: identifying histograms, modal class, cumulative frequency, midpoints, estimated mean, distribution shape, and reconstructing tables from a histogram.
1. Mixed problems — choose the right move
Each question uses a different idea from Lesson 7. Show your working. 3 marks each
1.1 Decide whether each data set should be displayed as a histogram or bar chart, and justify in one sentence:
(a) Number of pets owned by 30 students.
(b) Heights (cm) of 30 students.
(c) Weekly hours of homework (rounded to the nearest hour) of 30 students.
1.2 Find the midpoint of each interval: (a) 35–<45, (b) 5–<15, (c) 100–<150.
1.3 A grouped table: 0–<10 (f=4), 10–<20 (f=9), 20–<30 (f=7), 30–<40 (f=3), 40–<50 (f=2). (a) State n and the modal class. (b) Build the cumulative frequency column.
1.4 Using the same table from 1.3, estimate the mean using midpoints. Show all working.
1.5 Describe each distribution shape (symmetric / skewed left / skewed right) for these frequency lists:
(a) 2, 5, 9, 8, 4 (intervals 0–10, 10–20, ...).
(b) 12, 8, 5, 3, 1.
(c) 1, 3, 5, 8, 12.
1.6 A histogram has bars at heights 4, 9, 12, 8, 2 for 5 equal-width intervals. State n and the modal class.
2. Find the mistake
Another student attempted this estimated-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: Grouped data: 10–<20 (f=4), 20–<30 (f=9), 30–<40 (f=7). Estimate the mean.
Line 1: n = 4 + 9 + 7 = 20.
Line 2: Midpoints: 10, 20, 30.
Line 3: Σ(f × midpoint) = 4×10 + 9×20 + 7×30 = 40 + 180 + 210 = 430.
Line 4: Estimated mean = 430 ÷ 20 = 21.5.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the correct midpoints and redo the calculation.
Stuck? Midpoint = (lower + upper) ÷ 2. For 10–<20, midpoint = (10 + 20) ÷ 2 = 15, NOT 10.3. Open-ended challenge — design a histogram analysis
This question has many valid answers. 4 marks
3.1 Your job: pick ONE topic and invent realistic continuous data (n = 30 values total) suitable for a histogram. Possible topics:
• Hand-span (cm) of 30 students.
• Daily time on social media (minutes) for 30 students.
• Birthweight (kg) of 30 newborns.
• 200 m sprint times (seconds) of 30 athletes.
Write up your analysis with the following:
(i) State your topic and choose 5 equal-width, non-overlapping class intervals that span the data.
(ii) Build a grouped frequency table summing to 30. Add midpoint, f × midpoint, and cumulative frequency columns.
(iii) State the modal class.
(iv) Calculate the estimated mean from your table. Show working.
(v) Describe the SHAPE of the distribution in one sentence (symmetric / skewed left / skewed right) and explain why this shape makes sense for your chosen topic.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Choose histogram or bar chart
(a) Bar chart — number of pets is a discrete count (0, 1, 2, …); use gaps to show separate values.
(b) Histogram — height is continuous; group into intervals like 140–<150 with no gaps.
(c) Either, but bar chart is more common because hours rounded to the nearest hour are treated as discrete categories.
1.2 — Midpoints
(a) (35+45)÷2 = 40. (b) (5+15)÷2 = 10. (c) (100+150)÷2 = 125.
1.3 — n, modal class, cumulative
(a) n = 4+9+7+3+2 = 25. Modal class = 10–<20 (f = 9).
(b) Cumulative: 4, 13, 20, 23, 25.
1.4 — Estimated mean
Midpoints 5, 15, 25, 35, 45. Σ(f × mid) = 4×5 + 9×15 + 7×25 + 3×35 + 2×45 = 20 + 135 + 175 + 105 + 90 = 525. Estimated mean = 525 ÷ 25 = 21.
1.5 — Shape descriptors
(a) Roughly symmetric (peak in the middle, tails on both sides). (b) Skewed right (high values on the left, long tail to the right). (c) Skewed left (peak on the right, long tail to the left).
1.6 — Histogram totals
n = 4 + 9 + 12 + 8 + 2 = 35. Modal class = the 3rd interval (f = 12).
2 — Find the mistake
(a) The mistake is on Line 2.
(b) The "midpoints" listed (10, 20, 30) are actually the lower boundaries of each class. The midpoint must be (lower + upper) ÷ 2.
(c) Correct midpoints: (10+20)÷2 = 15; (20+30)÷2 = 25; (30+40)÷2 = 35. Recompute Σ(f×mid) = 4×15 + 9×25 + 7×35 = 60 + 225 + 245 = 530. Estimated mean = 530 ÷ 20 = 26.5.
3 — Open-ended (sample solution — social-media time)
(i) Topic: Daily time on social media (minutes) for 30 Year 8 students. Intervals: 0–<60, 60–<120, 120–<180, 180–<240, 240–<300.
(ii) Table: 0–<60: f=4, mid=30, f×m=120, cum=4. 60–<120: f=8, mid=90, f×m=720, cum=12. 120–<180: f=10, mid=150, f×m=1500, cum=22. 180–<240: f=6, mid=210, f×m=1260, cum=28. 240–<300: f=2, mid=270, f×m=540, cum=30. Σ(f×m) = 4140.
(iii) Modal class: 120–<180 minutes (f = 10).
(iv) Estimated mean: 4140 ÷ 30 = 138 minutes (about 2 h 18 min).
(v) Shape: Slightly skewed right — most students use 60–180 min, but a few heavy users (240–300 min) pull the tail right. This is realistic because there's a hard lower bound (0 min) but no real upper limit on screen time.
Marking: 1 mark for plausible data and equal-width intervals summing to 30; 1 mark for correct midpoint and cumulative columns; 1 mark for accurate estimated mean; 1 mark for correct shape description with a sensible context-based reason.