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4024/22 May/Jun 2023, Question 7: Histograms (Unequal Class Intervals)

Cambridge O Level Mathematics 4024 · Paper 22 (Calculator) · Calculator · 9 marks

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The question, in short

Use a grouped frequency table and partial histogram of Monday customer spending to find p, finish the histogram, and find a probability; then estimate Tuesday's mean and judge a claim about adding $41.

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

Part (ai) · 1 mark

Use the histogram bar for 0 < m ≤ 30 with frequency density 0.5 to find p.

  1. 1Frequency = frequency density × class width
  2. 2p = 0.5 × 30 = 15

Answer15

Part (aii) · 3 marks

Draw the missing histogram bars for the remaining class intervals using frequency density.

  1. 140 < m ≤ 50: 24 ÷ 10 = 2.4
  2. 250 < m ≤ 60: 19 ÷ 10 = 1.9
  3. 360 < m ≤ 90: 24 ÷ 30 = 0.8
  4. 4Draw bars of these heights over each class

AnswerCompleted histogram

Part (aiii) · 1 mark

Find the probability a randomly chosen customer spent more than $50.

  1. 1Total = 15 + 18 + 24 + 19 + 24 = 100
  2. 2More than $50: 19 + 24 = 43
  3. 3P = 43/100

Answer43/100 or 0.43 or 43%

Part (bi) · 3 marks

Estimate Tuesday's mean spending using class midpoints and frequencies.

  1. 1Midpoints: 15, 35, 45, 55, 75
  2. 2Σfx = 22 × 15 + 16 × 35 + 24 × 45 + 19 × 55 + 14 × 75 = 4065
  3. 3Σf = 22 + 16 + 24 + 19 + 14 = 95
  4. 4Mean = 4065 ÷ 95 = 42.78… ≈ $42.79

Answer42.80 or 42.78 to 42.79

Part (bii) · 1 mark

Explain without calculating why including a $41 sale raises the estimated mean.

  1. 1$41 is in the class 40 < m ≤ 50, so it would be counted at the mid-value $45
  2. 245 is higher than the estimated mean of 42.78…, so adding it makes the mean higher

Answer41 is in 40 < m ≤ 50 so mid-value of 45 would have been used to estimate the mean which is higher than the estimated mean

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Also searched as: 4024/22/M/J/23 · 4024_s23_qp_22 · 4024 s23 22 · question 7.
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