The time a customer has to hold for a customer service repre

The time a customer has to hold for a customer service representative during a phone call to the customer service department of a cable company is normally distributed with a mean of 45 minutes and standard deviation of 10 minutes. What is the probability that a customer will hold for less than 1 hour?

Solution

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    60      
u = mean =    45      
          
s = standard deviation =    10      
          
Thus,          
          
z = (x - u) / s =    1.5      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   1.5   ) =    0.933192799 [ANSWER, D]