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A-LevelMathematicsProbability and statistics9709/63 · May/June 2024 · Q713 Marks

The independent random variables X and Y have the distributions Po(1.9) and Po(2.2) respectively.

  1. (a)Find P(X+Y < 4).[3]
  2. (b)Find the probability that X = 2 given that X+Y < 4.[4]
  3. (c)A sample of 60 randomly chosen pairs of values of X and Y is taken, and the value of X+Y is calculated for each pair. The sample mean of these 60 values is found. Find the probability that the sample mean of X+Y is less than 4.0 .[6]

Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

  1. [B1] λ = 1.9 + 2.2 [= 4.1]
  2. [M1] e⁻⁴·¹ (1 + 4.1 + 4.1²/2 + 4.1³/6) or e⁻⁴·¹ (1 + 4.1 + 8.405 + 11.487) or 0.01657 + 0.06795 + 0.13929 + 0.19037
  3. [A1] = 0.414 (3sf)
  4. [M1] P(X + Y < 4 and X = 2) = P(X=2, Y=0 or X=2, Y=1)
  5. [M1] e⁻¹·⁹ * (1.9)²/2 * (e⁻²·² + e⁻²·² * 2.2)
  6. [M1] P(X=2 | X + Y < 4) = '0.0957' / '0.414'
  7. [A1] 0.231 (3sf)
  8. [B1] E(X + Y) = 4.1 Var(X + Y) = 4.1
  9. [M1] Normal and var = 4.1/60
  10. [M1] (4.0 - 4.1) / √(4.1/60) or continuity correction: (4.0-0.5-4.1) / √(4.1÷60) or (239.5-246) / √246
  11. [A1] = -0.383 (3sf)
  12. [M1] Φ('-0.383') = 1 – Φ('0.383')
  13. [A1] = 0.351 (3sf)

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About this question

Topic
Probability and statistics, A-Level Mathematics (syllabus 9709). Structured question, 13 marks.
Source
A-Level Mathematics, May/June 2024 Paper 6 Variant 3, question 7.

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