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 2 or 3 customers will arrive in a 15-minute period?
A) 0.0099
B) 0.4703
C) 0.0427
D) 0.0053

Question

Quizplus · Accepted Answer

Given that the mean number of customers entering the bank per hour is 10, for a 15-minute period (which is 1/4 of an hour), the mean ($\lambda$) becomes $10 \times \frac{1}{4} = 2.5$. The probability of $k$ events in a Poisson distribution is given by $P(k) = \frac{e^{-\lambda} \lambda^k}{k!}$. For 2 or 3 customers, we calculate $P(2) + P(3)$. $P(2) = \frac{e^{-2.5} 2.5^2}{2!}$ and $P(3) = \frac{e^{-2.5} 2.5^3}{3!}$. Calculating these and adding them gives approximately 0.4703, indicating the probability of 2 or 3 customers arriving in a 15-minute period.

The Number of Customers Who Enter a Bank Is Thought