Solution
1) The average customer arrival rate and service rate per hour
Patient arrival time in a day (11 pm to 7 am) = 8 hr/day
Patient arrival time in a year (Total study time) = 8 (hr/day) * 365 = 2920 hrs
For average customer arrival rate, we know that 62 patients are required during that time period.
So average customer arrival rate (λ) = 62/2920 = 0.0212 patients /hr
Service rate (μ) = 60/80.79 = 0.7426 /hr 2) Probabilities
Probability of having Zero (0) patients arriving during night shift
= (1- λμ) = (1- 0.02120.7426) = 0.9714 or 97.14%
Probability of having one patient arriving during night shift
= (1- λμ) ( λμ) = (1- 0.02120.7426) ( 0.02120.7426) = 0.02773 or 2.773%
Probability of arriving two or more patients simultaneously during the night shift = 1-[P(0)+P(1)] = 1-(0.9714+0.02773) = 0.00087 or 0.087% 3)
The criterion is that if the probability of two or more patients simultaneously arriving during the night shift is greater that 1percent, a backup OR team should be employed. Otherwise, then a backup OR team will not be necessary. Based on the calculation for the first two questions, we know that the probability of two or more patients simultaneously arriving during the night shift which also be symbol as px =0.07%, which less than 1%. So, the backup OR team is not necessary. I suggest the community hospital should not put the backup OR team on the hospital every single day. If they have 100 patients, the need for them only 0.07; besides that, for the entire year there were only 62 patients that required OR treatment during the night. Probably, the chef of the resident should make a night on call schedule of their surgeons. Distribute them by quarterly, or monthly, just make sure there must be on duty surgeons. They can get off work normally, go home or do whatever evening stuff