percentage of scores falling between zs of 53 and 84 percent

Solution

z1 = lower z score =    -0.53      
z2 = upper z score =     0.84      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.298055965      
P(z < z2) =    0.799545807      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.501489841 = 50.15% [ANSWER]