The number of customers who enter a bank is thought to be Poisson distributed with a mean equal to 10 per hour. What are the chances that no customers will arrive in a 15-minute period?

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BlackTom AI · Accepted Answer

We start by identifying the Poisson parameter for the 15-minute interval. The bank sees an average of 10 customers per hour, so for a 0.25-hour period the expected number of arrivals (λ) is 10 × 0.25 = 2.5.

Option 1: 0.0821. This value matches the probability of observing zero arrivals in a Poisson distribution with λ = 2.5, which is P(N=0) = e^{-λ} = e^{-2.5} ≈ 0.082085. Rounding to four decimals gives 0.0821, so this option is numerically correct.

Option 2: Approximately zero. While probabilities for rare events can be small, P(N=0) with λ = 2.5 is not near zero; it is about 8.2%. Describing it as approximately zero would be misleading because there is a definite non-negligible chance.

Option 3: 0.0250. This value does not correspond to P(N=0) for λ = 2.5. A probability of 0.0250 might arise from a different event or a miscalculation, but it does not reflect the zero-arrival probability for this Poisson parameter.

Option 4: 0.0067. This is even smaller and further from the actual P(N=0) for λ = 2.5. Since e^{-2.5} ≈ 0.0821, 0.0067 represents a probability far removed from the correct zero-arrival probability for the given rate and interval.

OSC 2030-009 Fall 2025 Exam 1

The number of customers who enter a bank is thought to be Poisson distributed with a mean equal to 10 per hour. What are the chances that no customers will arrive in a 15-minute period?

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