A teacher models the numbers of girls and boys who arrive late for her class on any day by the independent random variables G ~ Po(0.10) and B ~ Po(0.15) respectively.

Mark scheme

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

1. [B1] [e^(-0.2)] = 0.819 (3 sf) Accept e^(-0.2) as final answer.
2. [B1] λ = 1.25
3. [M1] e^(-1.25) (1 + 1.25 + 1.25²/2) or e^(-1.25) (1 + 1.25 + 0.78125) or 0.2865 + 0.3581 + 0.2238 Any λ Allow one end error. Must see expression (in any form). Accept correct ∑ notation.
4. [A1] = 0.868 (3 sf) SC Answer with no working seen scores B1 (could be implied).
5. [M1] Correct calculations for at least one of P(B=0)P(G=1), P(B=0)P(G=2), P(B=1)P(G=2) OR P(B=0)P(G>0), P(B=1)P(G>1).
6. [M1] Correct combination of terms for the probability. (Method 1: Sum of P(B=0)P(G=1) + P(B=0)P(G=2) + P(B=1)P(G=2). Method 2: Sum of P(B=0)P(G>0) + P(B=1)P(G>1))
7. [A1] = 0.0824 (3 sf)
8. [B1] Ho: λ = 1.25 or 0.25[per day], H₁: λ > 1.25 or 0.25[per day]. Or μ or 'population mean'.
9. [M1] Calculation of P(X >= 4) using Poisson distribution with correct λ. (1 - e^(-1.25) (1 + 1.25 + 1.25²/2 + 1.25³/3!) or 1 – e^(-1.25)(1 + 1.25 + 0.7813 + 0.3255) or 1 - (0.2865 + 0.3581 + 0.2238 + 0.09326))
10. [A1] = 0.0383 SC 0.0383 with no working scores B1.
11. [M1] Valid comparison of calculated probability (0.0383) with significance level (0.05).
12. [A1] [Reject Ho] Conclusion in context, not definite, and consistent with their comparison. E.g., 'Hence there is sufficient evidence to suggest that the teacher's claim is true' or 'There is sufficient evidence to suggest that more students are late on average'. Ft their 0.0383.