1. **Problem Statement:** Customers arrive at a shop following a Poisson distribution with a mean arrival rate $\lambda = 5$ per hour. The service time is exponentially distributed with an average service time of 10 minutes, so the service rate $\mu$ is the reciprocal of the average service time. We need to find: i) The average number of customers in the shop. ii) The percentage of time a customer can walk in and get served immediately. iii) The percentage of customers who have to wait to get served. 2. **Formulas and Important Rules:** This is a classic M/M/1 queue problem where: - $\lambda$ = arrival rate (customers per hour) - $\mu$ = service rate (customers per hour) The service rate $\mu$ is calculated as: $$\mu = \frac{60}{\text{average service time in minutes}} = \frac{60}{10} = 6 \text{ customers per hour}$$ Key formulas for M/M/1 queue: - Traffic intensity $\rho = \frac{\lambda}{\mu}$ - Average number of customers in the system (in queue + being served): $$L = \frac{\rho}{1 - \rho}$$ - Probability that the server is idle (no customers in system): $$P_0 = 1 - \rho$$ - Percentage of customers who have to wait is the probability that the server is busy: $$P_{wait} = \rho$$ 3. **Calculations:** - Calculate $\rho$: $$\rho = \frac{5}{6} = 0.8333$$ - Average number of customers in the shop: $$L = \frac{0.8333}{1 - 0.8333} = \frac{0.8333}{0.1667} = 5$$ - Percentage of time a customer can walk in and get served immediately (server idle): $$P_0 = 1 - 0.8333 = 0.1667 = 16.67\%$$ - Percentage of customers who have to wait (server busy): $$P_{wait} = 0.8333 = 83.33\%$$ 4. **Interpretation:** - On average, there are 5 customers in the shop (either waiting or being served). - About 16.67% of the time, a customer can walk in and get served immediately without waiting. - About 83.33% of customers will have to wait before being served. **Final answers:** - Average number of customers in the shop: 5 - Percentage of time a customer can get served immediately: 16.67% - Percentage of customers who have to wait: 83.33%