Question 740770
The mean time to complete a task for a group of employees is 5.3 seconds with a standard deviation of 1.6 seconds. 
If a random sample of 60 employees is selected, what is the probability that the mean time for these employees is less than 5 seconds?
------------
z(5) = (5-5.3)/[1.6/sqrt(60)] = 1.4524
----
P(x < 5) = P(z < 1.4524) = normalcdf(-100,1.4524) = 0.9268
================
Cheers,
Stan H.
================