4024/21 Oct/Nov 2020, Question 2: Finding the Median and Quartiles
Cambridge O Level Mathematics 4024 · Paper 21 (Calculator) · Calculator · 13 marks
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The question, in short
Use a cumulative frequency graph for 60 variety A potatoes, compare spread with variety B, estimate a mean from a grouped table for variety C, and find an upper bound for a box of 10 bags.
Summarised in our own words. Use the official paper for the exact wording and the diagram.
Part (aia) · 1 mark
Read off the mass at cumulative frequency 30 to estimate the median.
- 1Median at cumulative frequency 60 ÷ 2 = 30
- 2Read across at 30 to the curve and down: 188 g
Answer188 to 189
Part (aib) · 2 marks
Use the graph to find the lower and upper quartiles, then work out the interquartile range.
- 1Lower quartile at 15: about 170 g
- 2Upper quartile at 45: about 208 g
- 3IQR = 208 − 170 = 38 g
Answer37 to 39
Part (aii) · 3 marks
Find how many potatoes exceed 220 g and express this as a percentage of 60.
- 1At 220 g the cumulative frequency is 52
- 2Number heavier than 220 g = 60 − 52 = 8
- 3Percentage = 8/60 × 100 = 13.3%
Answer13.3
Part (aiii) · 1 mark
Compare the interquartile ranges of A and B to decide which variety has masses closer together.
- 1IQR for A ≈ 38 g, IQR for B = 30 g
- 2B has the smaller interquartile range, so its masses are more consistent: variety B
AnswerB, IQR is smaller
Part (b) · 3 marks
Using midpoints of the mass classes and their frequencies, estimate the mean mass of the 120 potatoes.
- 1Midpoints: 90, 110, 125, 135, 170
- 2Σfx = 900 + 1650 + 5250 + 4860 + 2890 = 15 550
- 3Mean = 15 550 ÷ 120 = 129.6 g
Answer130 or 129.5 to 129.6
Part (c) · 3 marks
With 2.5 kg to the nearest 100 g and box 600 g to the nearest 10 g, find the upper bound for 10 bags plus the box, in kg.
- 1Upper bound of one bag = 2.55 kg
- 2Upper bound of box = 605 g = 0.605 kg
- 3Total = 10 × 2.55 + 0.605 = 26.105 kg
Answer26.105
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