Mens heights are normally distributed with mean 680 inches a

Men’s heights are normally distributed with mean 68.0 inches and standard deviation 3.0 inches. What is the 90^th percentile for man’s heights? (Show work and round the answer to 2 decimal places)

Solution

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.9      
          
Then, using table or technology,          
          
z =    1.281551566      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    68      
z = the critical z score =    1.281551566      
s = standard deviation =    3      
          
Then          
          
x = critical value =    71.8446547 = 71.85 (answer)