The mean time from checkin to completion for a customer serv

The mean time from check-in to completion for a customer served at the Townsburg branch of the DMV is 88 minutes, with a standard deviation of 12 minutes. Assuming a normal distribution, what is the probability that a randomly chosen customer experiences service done between 88 and 112 minutes?

Solution

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    88      
x2 = upper bound =    112      
u = mean =    88      
          
s = standard deviation =    12      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0      
z2 = upper z score = (x2 - u) / s =    2      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.5      
P(z < z2) =    0.977249868      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.477249868   [ANSWER]