# Mean number of customers in a queue at a car repair workshop

$\textbf{Q:}$ Suppose that a car repair workshop encounters on average $1$ customer per day, where the probability of encountering a customer is proportional to the length of a small time interval, and constant throughout the entire day. Furthermore, service time can be shown by a scaled standard normal distribution with a mean of $10$ hours. Furthermore, assume that there is no queue capacity, and that customers are served in sequence.

(i). Deduce the mean number of customers in the queue.

(ii). Find the variance of time spent by a customer in the queue, from the time he/she arrives and leaves the queue (after servicing their vehicle).

$\textbf{Query:}$ For part (i), I computed the traffic intensity to be $\rho = \frac{\lambda}{\mu}=\frac{10}{24}$, and computed the mean number of customers in the queue (mean queue length in equilibrum of a $M/G/1$ system) to be

$$$$L=\rho + \frac{\lambda^2 \tau^2+\rho^2}{2(1-\rho)}=\frac{10}{24}+\frac{1^2\cdot1^2+(\frac{10}{24})^2}{2(1-\frac{10}{24})}=\frac{239}{168}$$$$

where $\tau^2$ is the variance of the service type distribution, which I took as $1$, since service time is scaled by a standard normal distribution with mean $0$ and variance $1$.

Am I correct for part (i), or is the a conceptual misunderstanding in my working?

Also, for part (ii), am I on the right track if I were to make use of the mean of $10$ hours as given?

Some help and clarification will be deeply appreciated.