On hold times for callers to a local cable television compan

On hold\" times for callers to a local cable television company are known to be normally distributed with a standard deviation of 1.3 minutes. Find the average caller \"on hold\" time if the company maintains that no more than 14% of callers wait more than 6.9 minutes. (Answer in two decimal questions)

Solution

No more than 14% of callers is at the 86th percentile. (.14 + .86 = 1.00)

The z score at the 86th percentile is 1.08, from the standard normal table.

The x value at the 86th percentile is 6.9 minute.

Z = (x