Given that z is a standard normal random variable P 12 z 15

Given that z is a standard normal random variable, P ( -1.2 z 1.5) is:

A   0.8181
B   0.3849
C   0.4772
D   0.5228

Solution

z1 = lower z score =    -1.2      
z2 = upper z score =     1.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.11506967      
P(z < z2) =    0.933192799      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.818123129= 0.8181 [ANSWER, OPTION A]