The independent random variables X and Y have the distributions Po(1.9) and Po(2.2) respectively.
- (a)Find P(X+Y < 4).[3]
- (b)Find the probability that X = 2 given that X+Y < 4.[4]
- (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.
- [B1] λ = 1.9 + 2.2 [= 4.1]
- [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
- [A1] = 0.414 (3sf)
- [M1] P(X + Y < 4 and X = 2) = P(X=2, Y=0 or X=2, Y=1)
- [M1] e⁻¹·⁹ * (1.9)²/2 * (e⁻²·² + e⁻²·² * 2.2)
- [M1] P(X=2 | X + Y < 4) = '0.0957' / '0.414'
- [A1] 0.231 (3sf)
- [B1] E(X + Y) = 4.1 Var(X + Y) = 4.1
- [M1] Normal and var = 4.1/60
- [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
- [A1] = -0.383 (3sf)
- [M1] Φ('-0.383') = 1 – Φ('0.383')
- [A1] = 0.351 (3sf)
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