If the mean of a normal distribution is 40 and the standard

If the mean of a normal distribution is 40 and the standard deviation is 4, find P(38 < x < 46).

Solution

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    38      
x2 = upper bound =    46      
u = mean =    40      
          
s = standard deviation =    4      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.5      
z2 = upper z score = (x2 - u) / s =    1.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.308537539      
P(z < z2) =    0.933192799      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.62465526