Scores on a test are normally distributed with a mean of 689

Scores on a test are normally distributed with a mean of 68.9 and a standard deviation of 11.6 Find p81, which separates the bottom 81% from the top 19%

Solution

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.81      
          
Then, using table or technology,          
          
z =    0.877896295      
          
As x = u + z * s,          
          
where          
          
u = mean =    68.9      
z = the critical z score =    0.877896295      
s = standard deviation =    11.6      
          
Then          
          
x = critical value =    79.08359702   [ANSWER]