Question

It is known that the waiting times, for those who make calls to a local cable television company, are normally distributed with a standard deviation of 1.3 minutes. Find the average time that the caller expects, if the company claims that no more than 8% of callers wait more than 6.7 minutes

Answer 1

No more than 6.7 minutes Means (X 6.7) and this is equal to 8% = 0.08

Therefore P(X < 6.7) = 0.08

The corresponding Z score for p = 0.08 is -1.405

Therefore -1.405 = (6.7 - ) / 1.3

Solving for , we get = 6.7 - (-1.405 * 1.3) = 8.5265 minutes or

(8.5 minutes, rounded to 1 decimal place)

(8.53 minutes, rounded to 2 decimal places)

(8.527 minutes, rounded to 3 decimal places)