find the probability P026
find the probability P(0.26<z<1.33) using the standard normal deviation<1.33)>
 find the probability P(0.26<z<1.33) using the standard normal deviation<1.33)>
 find the probability P(0.26<z<1.33) using the standard normal deviation<1.33)>
Solution
z1 = lower z score =    0.26      
 z2 = upper z score =     1.33      
           
 Using table/technology, the left tailed areas between these z scores is          
           
 P(z < z1) =    0.602568113      
 P(z < z2) =    0.908240864      
           
 Thus, the area between them, by subtracting these areas, is          
           
 P(z1 < z < z2) =    0.305672751   [ANSWER]  

