what proportion of hisher specimens will fail to meet the qu

what proportion of his/her specimens will fail to meet the quality specifications?

Solution

Mean ( u ) =28
Standard Deviation ( sd )=1.6
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X > 30) = (30-28)/1.6
= 2/1.6 = 1.25
= P ( Z >1.25) From Standard Normal Table
= 0.1056                  
10.56% proportion of his/her specimens will fail to meet the quality specifications