Of the 2500 middle managers in a large company 60 hold MBA d

Of the 2500 middle managers in a large company, 60% hold M.B.A. degrees. You select a random sample of 81 of them. What is the probability that the proportion in the sample with M.B.A. degrees is between 59.5% and 60.5%?

Solution

Normal Distribution
Proportion ( P ) =0.6
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.6*0.4/81)
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.595) = (0.595-0.6)/0.0544
= -0.005/0.0544 = -0.0919
= P ( Z <-0.0919) From Standard Normal Table
= 0.46338
P(X < 0.605) = (0.605-0.6)/0.0544
= 0.005/0.0544 = 0.0919
= P ( Z <0.0919) From Standard Normal Table
= 0.53662
P(0.595 < X < 0.605) = 0.53662-0.46338 = 0.0732