The number of accidents on a certain road has a Poisson distribution with mean 0.4 per 50-day period.
- (a)Find the probability that there will be fewer than 3 accidents during a year (365 days).[3]
- (b)The probability that there will be no accidents during a period of n days is greater than 0.95. Find the largest possible value of n.[4]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2020). The mark or mark type is shown in brackets.
- [B1] λ (= 0.4 × 365 ÷ 50) = 2.92
- [M1] e^(-2.92)(1 + 2.92 + (2.92^2)/2); Any λ. Allow one end error
- [A1] = 0.441 (3 sf)
- [M1] e^(-λ) > 0.95; Allow '=' throughout
- [M1] -λ > ln 0.95 or λ < 0.051293 OE; Attempt ln both sides
- [M1] '0.051293' × 50 ÷ 0.4 (= 6.411)
- [A1] Largest n is 6 (3 sf); Allow n = 6 or n ≤ 6 (NOT n < 6 or n ≥ 6 as final answer); SC Trial and Improvement M1 for e^(-λ) > 0.95 SOI; M1 for λ = n × (0.4/50); M1 for use of both; n = 6 giving 0.9531 and n = 7 giving 0.9455; A1 n = 6
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