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4024/21 May/Jun 2023, Question 10: Averages from Frequency Tables

Cambridge O Level Mathematics 4024 · Paper 21 (Calculator) · Calculator · 10 marks

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The question, in short

Using frequency tables of sweets per bag in boxes, find an unknown frequency from a given mean, then use a two-bag probability to form and solve a quadratic for another unknown.

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

Part (ai) · 1 mark

Explain why the total number of bags cannot be 87, using the given frequencies.

  1. 115 + 26 + 38 + 9 = 88
  2. 288 > 87, so p would be 87 − 88 = −1
  3. 3p cannot be negative, so the total cannot be 87

Answerp cannot be negative

Part (aii) · 3 marks

Given the mean is 12.8 sweets per bag, find the value of p.

  1. 1Total sweets = 11 × 15 + 12 × 26 + 13 × 38 + 14p + 15 × 9 = 1106 + 14p
  2. 2Total bags = 88 + p
  3. 3(1106 + 14p)/(88 + p) = 12.8
  4. 41106 + 14p = 1126.4 + 12.8p
  5. 51.2p = 20.4
  6. 6p = 17

Answer17

Part (bi) · 3 marks

With two bags drawn without replacement, use the probability both contain 15 sweets, 4/539, to show r² + 175r − 2046 = 0.

  1. 1Total bags = 12 + 28 + 39 + r + 9 = 88 + r
  2. 2P(both 15) = 9/(88 + r) × 8/(87 + r) = 4/539
  3. 372 × 539 = 4(88 + r)(87 + r)
  4. 438 808 = 4(r² + 175r + 7656)
  5. 59702 = r² + 175r + 7656
  6. 6r² + 175r − 2046 = 0

AnswerCorrect expansion of brackets leading to r² + 175r − 2046 = 0

Part (bii) · 3 marks

Solve r² + 175r − 2046 = 0, showing working, to find r.

  1. 1(r − 11)(r + 186) = 0
  2. 2r = 11 or r = −186
  3. 3r cannot be negative, so r = 11

Answer11

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