For a normal distrubtion with u90 and 5 what proportion of t

For a normal distrubtion with u=90 and @5, what proportion of the scores have greater vaules than X=87

Solution

Normal Distribution
Mean ( u ) =90
Standard Deviation ( sd )=5
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X > 87) = (87-90)/5
= -3/5 = -0.6
= P ( Z >-0.6) From Standard Normal Table
= 0.7257