A training program designed to upgrade the supervisory skill

A training program designed to upgrade the supervisory skills of production-line supervisors has been offered for the past five years at a Fortune 500 company. Because the program is self-administered, supervisors require different numbers of hours to complete the program. A study of past participants indicates that the mean length of time spent on the program is 600 hours and that this normally distributed random variable has a standard deviation of 100 hours. Suppose the training-program director wants to know the probability that a participant chosen at random would require between 650 and 750 hours to complete the required work. Determine that probability.

Solution

Normal Distribution
Mean ( u ) =600
Standard Deviation ( sd )=100
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 650) = (650-600)/100
= 50/100 = 0.5
= P ( Z <0.5) From Standard Normal Table
= 0.69146
P(X < 750) = (750-600)/100
= 150/100 = 1.5
= P ( Z <1.5) From Standard Normal Table
= 0.93319
P(650 < X < 750) = 0.93319-0.69146 = 0.2417