Snyder Sates does much of its soliciting by telephone Each c

Snyder Sates does much of its soliciting by telephone. Each call lasts a mean of 8 minutes (standard deviation of the population = 3.3). If during one promotional blitz, many operators are each given 75 calls to make, what percent would spend a mean time per call of less than 7 minutes?

Solution

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    7      
u = mean =    8      
n = sample size =    75      
s = standard deviation =    3.3      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -2.624319405      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2.624319405   ) =    0.004341116 or 0.43% [ANSWER, OPTION 4]