Assume the average weight of the adult male is normally dist

Solution

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    66.63      
x2 = upper bound =    72.86      
u = mean =    69.5      
          
s = standard deviation =    2.4      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.195833333      
z2 = upper z score = (x2 - u) / s =    1.4      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.115880803      
P(z < z2) =    0.919243341      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.8041 [ANSWER, OPTION D]