Answer:
A) Single-server single-phase model (M/M/1).
![\lambda=2.5 \,customers/hour\\\\\mu=6\,customers/hour](https://img.qammunity.org/2021/formulas/business/college/zdcvdad8air411ms8y405s2bkx92cjt7ac.png)
B) The goal is not met, as the average time waiting for service is 5.56 minutes.
C) The new mean service rate is 7.5 customers/hour.
In this case, the average time waiting for service is 4 minutes, so the goal is met.
Step-by-step explanation:
A) This situation can be modeled as a single-server single-phase model (M/M/1).
The mean arrival rate is 2.5 customers per hour.
![\lambda=2.5 \,customer/h](https://img.qammunity.org/2021/formulas/business/college/fabfm2shlsuombed9m7kpzw5rfavpszsok.png)
The mean service rate is 6 customers per hour, calculated as:
![\mu=(60\, min/h)/(10 \,min/customer)=6\, customer/h](https://img.qammunity.org/2021/formulas/business/college/2noqzi4sy5asiur894847nxkmjjbj1knup.png)
B) The average waiting time for a customer can be expressed as:
![W_q=(\lambda)/(\mu)(1)/(\mu-\lambda) =(2.5)/(6)(1)/(6-2.5) =0.417*0.222=0.093\,hours\\\\W_q=0.093\,hours*(60min/h)=5.56 \,min](https://img.qammunity.org/2021/formulas/business/college/kcqo4gd73cmpw5yn00pg7kmn057yl8cw9f.png)
The average waiting time is 5.56 minutes, so it is more than the goal of 5 minutes.
C) If the average time spent per customer to 8 minutes, the mean service rate becomes
![\mu=(60\, min/h)/(8 \,min/customer)=7.5\, customer/h](https://img.qammunity.org/2021/formulas/business/college/5e6a2rt6l614j3igwymtib6jykhmq4h1lh.png)
An the average waiting time for the service now becomes:
![W_q=(\lambda)/(\mu)(1)/(\mu-\lambda) =(2.5)/(7.5)(1)/(7.5-2.5) =0.333*0.2=0.067\,hours\\\\W_q=0.067\,hours*(60min/h)=4 \,min](https://img.qammunity.org/2021/formulas/business/college/g22ss848olpsg7ks1xmvvvkh3xqr2jrysm.png)
The average time is now 4 minutes, so the goal is achieved.