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0580/41 May/Jun 2024, Question 4: Box-and-Whisker Plots

Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 18 marks

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The question, in short

Use a box-and-whisker plot of car journey times over 195 m and a grouped frequency table for 80 cars' fuel distances to find averages, spread, speeds, histogram bar heights and a probability.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (ai) · 1 mark

Read the median time in seconds from the box-and-whisker plot.

  1. 1Median = line inside the box = 9.3 s

Answer9.3

Part (aii) · 1 mark

Find the interquartile range of the times from the box-and-whisker plot.

  1. 1From the plot: upper quartile = 11.6, lower quartile = 8.2
  2. 2IQR = 11.6 − 8.2 = 3.4 s

Answer3.4

Part (aiii) · 5 marks

Using fastest and slowest times, find the difference in average speeds in kilometres per hour.

  1. 1Fastest car: 6 s, slowest car: 13 s
  2. 2Fastest speed = 195 ÷ 6 = 32.5 m/s = 32.5 × 3600 ÷ 1000 = 117 km/h
  3. 3Slowest speed = 195 ÷ 13 = 15 m/s = 15 × 3600 ÷ 1000 = 54 km/h
  4. 4Difference = 117 − 54 = 63 km/h

Answer63

Part (bi) · 1 mark

Identify which distance class interval contains the median for the 80 cars.

  1. 1Median is the 40th/41st value
  2. 2Cumulative frequencies: 7, 20, 39, 60, 80
  3. 340th and 41st are in 420 < d ≤ 450

Answer420 < d ≤ 450

Part (bii) · 4 marks

Calculate an estimate of the mean distance travelled using the grouped frequencies.

  1. 1Mid-points: 275, 350, 410, 435, 475
  2. 2Σfx = 7×275 + 13×350 + 19×410 + 21×435 + 20×475
  3. 3= 1925 + 4550 + 7790 + 9135 + 9500 = 32 900
  4. 4Mean = 32 900 ÷ 80 = 411.25 km

Answer411.25

Part (biii) · 3 marks

Given the 250 < d ≤ 300 bar is 2.8 cm high, find the heights for the three listed intervals.

  1. 1Frequency density 250−300: 7 ÷ 50 = 0.14 → 2.8 cm, so height = 20 × fd
  2. 2300 < d ≤ 400: 13 ÷ 100 = 0.13 → 0.13 × 20 = 2.6 cm
  3. 3400 < d ≤ 420: 19 ÷ 20 = 0.95 → 0.95 × 20 = 19 cm
  4. 4420 < d ≤ 450: 21 ÷ 30 = 0.7 → 0.7 × 20 = 14 cm

Answer300 < d ≤ 400: 2.6 cm 400 < d ≤ 420: 19 cm 420 < d ≤ 450: 14 cm

Part (biv) · 3 marks

Find the probability that one random car travels over 450 km and the other at most 300 km.

  1. 1More than 450 km: 20 cars; not more than 300 km: 7 cars
  2. 2P = 2 × 20/80 × 7/79 = 280/6320
  3. 3= 7/158

Answer7/158

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Also searched as: 0580/41/M/J/24 · 0580_s24_qp_41 · 0580 s24 41 · question 4.
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