Poisson Distribution in Customer Service: Estimating Service Times at the Bank Counter

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Exercise

The number of customers served by a bank’s central counter is 2 customers per minute. What is the probability that the counter will:

1. Serve 5 customers in a minute?
2. Serve 4 customers in the next 3 minutes?

Solution

This problem involves a Poisson distribution, as we are dealing with the number of events (customers served) within a fixed period of time.

The Poisson probability mass function is given by:

$$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$$

Where:
– $X$ is the number of customers served,
– $\lambda$ is the average number of customers served in the given time period,
– $k$ is the number of customers we are interested in,
– $e$ is the base of the natural logarithm, approximately 2.71828.

In this problem:
– For Question 1, $\lambda = 2$ (as 2 customers are served per minute),
– For Question 2, $\lambda = 6$ (as 2 customers per minute over 3 minutes gives $2 \times 3 = 6$).

Question 1: Probability of serving 5 customers in one minute.

We are asked to find $P(X = 5)$ for $\lambda = 2$:

$$P(X = 5) = \frac{2^5 e^{-2}}{5!} = \frac{32 \cdot e^{-2}}{120} = \frac{32}{120 \cdot e^2}$$

Calculating this:

$$P(X = 5) \approx \frac{32}{120 \times 7.3891} \approx \frac{32}{886.692} \approx 0.036$$

So, the probability of serving 5 customers in a minute is approximately 0.036, or 3.6%.

Question 2: Probability of serving 4 customers in the next 3 minutes.

We are asked to find $P(X = 4)$ for $\lambda = 6$:

$$P(X = 4) = \frac{6^4 e^{-6}}{4!} = \frac{1296 \cdot e^{-6}}{24}$$

Calculating this:

$$P(X = 4) \approx \frac{1296}{24 \times 403.4288} = \frac{1296}{9682.2912} \approx 0.134$$

So, the probability of serving 4 customers in the next 3 minutes is approximately 0.134, or 13.4%.

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Summary of Answers

1. The probability of serving 5 customers in a minute is approximately 3.6%.
2. The probability of serving 4 customers in the next 3 minutes is approximately 13.4%.