student earned a score of 940 mean score was 850 wstandard d

student earned a score of 940, mean score was 850 w/standard deviation of 100. what proportion of students had a higher score than above student?

Solution

Normal Distribution
Mean ( u ) =850
Standard Deviation ( sd )=100
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X > 940) = (940-850)/100
= 90/100 = 0.9
= P ( Z >0.9) From Standard Normal Table
= 0.1841 ~ 18.41% are higher than 940