Average height of an adult male is normally distributed with
 Average height of an adult male is normally distributed with a mean of 69.5 and standard deviation of 2.4. What percentage of males will have height between 67.1- 73.1 
 
 
 
 0.1570
 0.4525
 0.2140
 0.7745
 0.1451
 0.5922
  Average height of an adult male is normally distributed with a mean of 69.5 and standard deviation of 2.4. What percentage of males will have height between 67.1- 73.1 
 
 
 
 0.1570
 0.4525
 0.2140
 0.7745
 0.1451
 0.5922
 0.1570
 0.4525
 0.2140
 0.7745
 0.1451
 0.5922
 Solution
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
 x1 = lower bound =    67.1      
 x2 = upper bound =    73.1      
 u = mean =    69.5      
           
 s = standard deviation =    2.4      
           
 Thus, the two z scores are          
           
 z1 = lower z score = (x1 - u)/s =    -1      
 z2 = upper z score = (x2 - u) / s =    1.5      
           
 Using table/technology, the left tailed areas between these z scores is          
           
 P(z < z1) =    0.158655254      
 P(z < z2) =    0.933192799      
           
 Thus, the area between them, by subtracting these areas, is          
           
 P(z1 < z < z2) =    0.7745 [ANSWER]      

