Question

Consider an espresso stand with a single barista. Customers arrive at the stand at the rate of 28 per hour according to a Poisson distribution. Service times are exponentially distributed with a service rate of 35 customers per hour. The probability that the server is busy is:

Answer 1

Answer:

The probability that the server is busy is P=0.56.

Step-by-step explanation:

We have a quieing theory problem with M/M/1

In queing theory, the probability of the server being busy can be expressed as:

$P=\frac{\lambda}{\lambda+\mu}$

being μ: the time between services and λ: the time between customers arrival.

Then we can calculate:

$P=\frac{\lambda}{\lambda+\mu}=\frac{\frac{1}{28} }{\frac{1}{28}+\frac{1}{35} } =\frac{0.036}{0.036+0.029}=0.56$

The probability that the server is busy is P=0.56.