Assume the random variable X is normally distributed with me

Assume the random variable X is normally distributed with mean as 50 and the standard deviation as 7. Find the 67th percentile.

Solution

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.67      
          
Then, using table or technology,          
          
z =    0.439913166      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    50      
z = the critical z score =    0.439913166      
s = standard deviation =    7      
          
Then          
          
x = critical value =    53.07939216   [ANSWER]