Assume the hold time of callers to a cable company is normally distributed with a mean of 3.53.5 minutes and a standard deviation of 0.20.2 minute. Determine the percent of callers who are on hold between 3.43.4 minutesminutes and 3.83.8 minutes.
Accepted Solution
A:
We calculate z-scores associated with x = 3.4 and x = 3.8. For x = 3.4, z = (3.4 - 3.5) / (0.2) = -0.5 For x = 3.8, z = (3.8 - 3.5) / (0.2) = 1.5 Then we calculate the probability that -0.5 < z < 1.5 from z-tables. P(-0.5 < z < 1.5) = 0.6247 = 62.47% of callers on hold between 3.4-3.8 minutes.