Hint
Trials must still be independent with a common p for the initial binomial model. Multiply the trial count by the defect probability.
Worked solution
(a)
\(np=2\)
Answer: \(2\).
Answer: \(2\)
(b)
\(e^{-np}=0.135335283236613\)
Answer: \(0.135335283236613\).
Answer: \(\approx 0.135335283236613\)
(c)
\(1-P(Y\le 1)=0.593994150290162\)
Answer: \(0.593994150290162\).
Answer: \(\approx 0.593994150290162\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit valid alternative reasoning and consequential working; do not award a reasoning mark for an unexplained final answer.
(a) · 2 marks
- 1 mark: Multiply the trial count by the defect probability.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Do not use p as the Poisson mean.
(b) · 2 marks
- 1 mark: Use the Poisson zero-count probability.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Approximation does not remove the need to translate integer endpoints.
(c) · 2 marks
- 1 mark: Evaluate the Poisson survival probability at k−1.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Do not use p as the Poisson mean.
Total: 6 marks.