Handling Data 3
4 marks · HardThe table gives the frequencies of heights (in cm) for a group of sunflowers.
| Height (cm) | Frequency |
|---|---|
| 40 < h ≤ 60 | 8 |
| 60 < h ≤ 80 | 17 |
| 80 < h ≤ 100 | 24 |
| 100 < h ≤ 120 | 11 |
a) Complete the cumulative frequency table. b) On graph paper, draw a cumulative frequency graph using your table (plot cumulative frequency against the upper boundary of each height interval, scale not required). c) Estimate the median and inter-quartile range for the heights of these sunflowers from your graph.
Complete the cumulative frequency table with required totals.
Draw the cumulative frequency graph. Label axes clearly. (You may sketch with reasonable accuracy if graph paper is not available.)
Estimate the median and inter-quartile range from your graph (show clearly how these are found).
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) — · must show Cumulative frequency column: 8, 25, 49, 60
(b) — · must show Points plotted at (60,8), (80,25), (100,49), (120,60)
(c) — · must show Median estimate: height at 30th sunflower (~85 cm); IQR = height at 45th (~98 cm) – height at 15th (~70 cm) = ~28 cm
Must show: table completed, axes labelled, curve/line plotted, reading of median/Q1/Q3 indicated
Also accept: neat line/curve; clearly marked reading lines
Common mistakes: plotting at midpoints instead of upper boundaries
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Correct cumulative frequency values: 8, 25, 49, 60 written or used.
M2 (b) · 1 mark · Correct points plotted at upper end of intervals and curve/line drawn; axes labelled.
M3 (c) · 1 mark · Reads off median from cumulative frequency graph at 30th sunflower; answer consistent with graph. (follow-through)
M4 (c) · 1 mark · Reads off lower quartile (15th) and upper quartile (45th); subtracts for IQR; answer consistent with graph. Follow-through if points are slightly off. (follow-through)
Worked solution
a) Cumulative frequencies: 8 (to 60 cm), 25 (to 80 cm), 49 (to 100 cm), 60 (to 120 cm).
| Upper boundary (cm) | Cumulative frequency |
|---|---|
| 60 | 8 |
| 80 | 25 |
| 100 | 49 |
| 120 | 60 |
b) Plot (60,8), (80,25), (100,49), (120,60). Join with a smooth curve. Axes: height (cm), cumulative frequency.
c) Median is at 60/2=30th sunflower. From the graph, median ≈ 85 cm. Lower quartile (15th): ≈ 70 cm; upper quartile (45th): ≈ 98 cm. IQR ≈ 98 − 70 = 28 cm.