The test scores for the MZZZ admission test are normally dis

The test scores for the MZZZ admission test are normally distributed with a mean of 450 and standard deviation of 100. A certain highly ranked university would not accept anyone scoring below 480 in this test. What percentage of the applicants would be acceptable to the University?

Solution

Mean ( u ) =450
Standard Deviation ( sd )=100
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X > 480) = (480-450)/100
= 30/100 = 0.3
= P ( Z >0.3) From Standard Normal Table
= 0.3821                  
              
38.21% of the applicants would be acceptable to the University