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0580/43 Oct/Nov 2023, Question 5: Histograms (Unequal Class Intervals)

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 14 marks

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The question, in short

Use the partly filled table and histogram of journey times to complete missing frequencies and bars, estimate the mean, assess a claim about the maximum time, and find a probability for three randomly chosen workers over 25 minutes.

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

Part (a) · 5 marks

Use the histogram bar heights and the given frequencies to find the missing frequencies and draw the missing histogram bars.

  1. 1Frequency = frequency density × class width
  2. 20 < t ≤ 10: 2.8 × 10 = 28; 10 < t ≤ 25: 3 × 15 = 45
  3. 3Heights: 57 ÷ 15 = 3.8, 38 ÷ 20 = 1.9, 12 ÷ 20 = 0.6
  4. 4Draw bars of height 3.8 (25−40), 1.9 (40−60) and 0.6 (60−80)

Answer28 and 45 in the table; histogram completed with blocks of height 3.8, 1.9 and 0.6

Part (b) · 4 marks

Calculate an estimate of the mean journey time from the completed grouped frequency table.

  1. 1Midpoints: 5, 17.5, 32.5, 50, 70
  2. 2Σfx = 28 × 5 + 45 × 17.5 + 57 × 32.5 + 38 × 50 + 12 × 70 = 140 + 787.5 + 1852.5 + 1900 + 840 = 5520
  3. 3Σf = 28 + 45 + 57 + 38 + 12 = 180
  4. 4Mean = 5520 ÷ 180 = 30.66… ≈ 30.7 minutes

Answer30.7 or 30.66 to 30.67

Part (c) · 1 mark

Explain why the longest journey time might not be 80 minutes, referring to the grouped data.

  1. 1The data are grouped, so exact values are not known: 80 is only the upper limit of the class 60 < t ≤ 80, and the longest time could be less than 80

AnswerExact values are not known

Part (d) · 4 marks

For three workers chosen at random from those over 25 minutes, find the probability that one is 60 minutes or less and two are over 60 minutes.

  1. 1Workers over 25 min: 57 + 38 + 12 = 107; 95 take 60 min or less, 12 take more than 60
  2. 2P(≤60, >60, >60) = 95/107 × 12/106 × 11/105
  3. 3The ≤60 worker can be 1st, 2nd or 3rd, so multiply by 3
  4. 43 × 95/107 × 12/106 × 11/105 = 37 620/1 190 910 = 1254/39 697

Answer1254/39697

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Also searched as: 0580/43/O/N/23 · 0580_w23_qp_43 · 0580 w23 43 · question 5.
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