The numbers of customers arriving at service desks A and B during a 10-minute period have the independent distributions Po(1.8) and Po(2.1) respectively.
- (a)Find the probability that during a randomly chosen 15-minute period more than 2 customers will arrive at desk A.[2]
- (b)Find the probability that during a randomly chosen 5-minute period the total number of customers arriving at both desks is less than 4.[3]
- (c)An inspector waits at desk B. She wants to wait long enough to be 90% certain of seeing at least one customer arrive at the desk. Find the minimum time for which she should wait, giving your answer correct to the nearest minute.[4]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.
- [B1] λ = 1.8 * (15/10) = 2.7
- [M1] 1 – e⁻².⁷(1 + 2.7 + 2.7²/2!) or 1 – e⁻².⁷(1 + 2.7 +3.645) or 1- (0.06721 +0.1815 +0.2450)
- [A1] = 0.506 (3 sf)
- [B1] λ_A = 1.8 * (5/10) = 0.9, λ_B = 2.1 * (5/10) = 1.05. Total λ = 0.9 + 1.05 = 1.95
- [M1] e⁻¹.⁹⁵(1 + 1.95 + 1.95²/2! + 1.95³/3!) or e⁻¹.⁹⁵(1 + 1.95 + 1.90125 +1.2358)
- [A1] = 0.866
- [M1] 1 – e⁻².¹x ≥ 0.90 or 1 – e⁻ᶫ ≥ 0.90
- [M1] [e⁻².¹x < 0.1] or e⁻ᶫ < 0.1 -2.1x < ln0.1 or -λ < ln0.1 [ λ > 2.3026, 2.3026/2.1 ]
- [*A1] 1.096 or 10.96 accept 1.097 or 10.97
- [A1 dep] She must wait for at least 11 minutes
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