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0580/41 May/Jun 2023, Question 3: Estimating the Mean for Grouped Data

Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 13 marks

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The question, in short

Use a grouped frequency table for 1000 eggs to estimate the mean and a probability, round a profit, and work with bounds for hens and egg totals.

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

Part (ai) · 4 marks

Estimate the mean mass of the 1000 eggs using the grouped frequency table with midpoints of each class.

  1. 1Midpoints: 45, 53, 60, 67
  2. 2Σfm = 126 × 45 + 520 × 53 + 154 × 60 + 200 × 67
  3. 3= 5670 + 27 560 + 9240 + 13 400 = 55 870
  4. 4Mean = 55 870 ÷ 1000 = 55.87 g

Answer55.87

Part (aii) · 2 marks

Find, as a simplified fraction, the probability a randomly chosen egg has mass greater than 56 g.

  1. 1Eggs over 56 g = 154 + 200 = 354
  2. 2P = 354/1000 = 177/500

Answer177/500

Part (bi) · 1 mark

Write the profit $24 730 correct to 2 significant figures.

  1. 124 730 ≈ 25 000 (2 s.f.)

Answer25000

Part (bii) · 1 mark

Write the profit $24 730 in standard form.

  1. 124 730 = 2.473 × 10⁴

Answer2.473 × 10⁴

Part (ci) · 3 marks

Using 500 hens (nearest 10) and mean 320 eggs per hen (nearest 20), find the upper bound of total eggs.

  1. 1Upper bound of hens = 500 + 5 = 505
  2. 2Upper bound of mean eggs = 320 + 10 = 330
  3. 3Upper bound of total = 505 × 330 = 166 650

Answer166 650 or 165 816

Part (cii) · 2 marks

With 800 hens correct to the nearest 20 and 500 hens correct to the nearest 10, find the lower bound of the difference in hen numbers.

  1. 1Lower bound of 800 hens = 800 − 10 = 790
  2. 2Upper bound of 500 hens = 505
  3. 3Lower bound of difference = 790 − 505 = 285

Answer285 or 286

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Also searched as: 0580/41/M/J/23 · 0580_s23_qp_41 · 0580 s23 41 · question 3.
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