Question 285981: in Rivendell memorial hospital the time to complete surgery in a routine tubal ligation without complication is normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes. The next procedure has been scheduled in the same operating room 60 minutes after the beginning of a tubal ligation procedure.
Allowing 20 minutes to vacate and prepare the operating room between procedures, what is the probability that the next procedure will have to be delayed?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! in Rivendell memorial hospital the time to complete surgery in a routine tubal ligation without complication is normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes. The next procedure has been scheduled in the same operating room 60 minutes after the beginning of a tubal ligation procedure.
Allowing 20 minutes to vacate and prepare the operating room between procedures, what is the probability that the next procedure will have to be delayed?
--------------------
If tubal ligation takes > 40 min (procedure delayed)
------------------------------------
z(40) = (40-30)/8 = 5/4
------
P(procedure delayed) = P(x>40) = P(z> 5/4) = normalcdf(5/4,100) = 0.1056
=================================
Cheers,
Stan H.
|
|
|
