The number, X, of books received at a charity shop has a constant mean of 5.1 per day.
- (a)State, in context, one condition for X to be modelled by a Poisson distribution.[1]
- (b)Assume now that X can be modelled by a Poisson distribution. Find the probability that exactly 10 books are received in a 3-day period.[2]
- (c)Use a suitable approximating distribution to find the probability that more than 180 books are received in a 30-day period.[4]
- (d)The number of DVDs received at the same shop is modelled by an independent Poisson distribution with mean 2.5 per day. Find the probability that the total number of books and DVDs that are received at the shop in 1 day is more than 3.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.
- [1] Books received independently or singly or randomly. OE. Must be in context. If more than one condition given, ignore extras.
- [1] e⁻¹⁵.³ × 15.3¹⁰ / 10! Allow incorrect λ.
- [1] = 0.0439 (3sf) SC No working shown but correct answer seen scores B1.
- [1] N(153, 153) Seen or implied.
- [1] (180.5-153) / √153 [= 2.223] For standardising with their values (can be implied). Allow with wrong or missing continuity correction.
- [1] 1 – Φ('2.223') For correct probability area consistent with their values.
- [1] = 0.0131 (3sf)
- [1] (λ =) 5.1 + 2.5 [= 7.6] Give at early stage (seen or implied).
- [1] 1 - e⁻⁷.⁶(1 + 7.6 + 7.6²/2 + 7.6³/3!) = 1 - e⁻⁷.⁶(1 + 7.6 + 28.88 + 73.16) = 1 – (0.0005005+0.003803+0.01445+0.03661) Allow incorrect λ. Allow one end error. Must see an expression (accept correct sigma notation).
- [1] = 0.945 (3sf) SC No working, 0.945 B1(could be implied) SC B1.
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