percentage of scores falling between zs of 80 and 80 percent

Solution

z1 = lower z score =    -0.8      
z2 = upper z score =     0.8      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.211855399      
P(z < z2) =    0.788144601      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.576289203 = 57.63% [ANSWER]