The number of clients who arrive at an information desk has a Poisson distribution with mean 2.2 per 5-minute period.
- (a)Find the probability that, in a randomly chosen 15-minute period, exactly 6 clients arrive at the desk.[3]
- (b)If more than 4 clients arrive during a 5-minute period, they cannot all be served. Find the probability that, during a randomly chosen 5-minute period, not all the clients who arrive at the desk can be served.[2]
- (c)Use a suitable approximating distribution to find the probability that, during a randomly chosen 1-hour period, fewer than 20 clients arrive at the desk.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.
- [B1] λ = 6.6
- [M1] e^-6.6 x 6.6^6 / 6!
- [A1] 0.156 (3 s.f.)
- [M1] 1 – e^-2.2(1 + 2.2 + 2.2^2/2! + 2.2^3/3! + 2.2^4/4!)
- [A1] 0.0725 (3 s.f.)
- [B1] N(26.4, 26.4)
- [M1] (19.5-'26.4') / √'26.4' [= -1.343]
- [M1] Φ('-1.343') = 1 – Φ('1.343')
- [A1] 0.0897 or 0.0896 (3 s.f.)
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