Chance average handle time runs long

Question

A call center's handle time has a true mean of $5$ minutes with standard deviation $\sigma = 1.5$ minutes. Calls are independent, so by the central limit theorem the average of many calls is approximately normal.

Over $n = 25$ calls, what is the probability the average handle time exceeds $5.75$ minutes?

Accepted Answer

The average of 25 independent calls is approximately normal, centered at 5 minutes, with spread equal to the standard error: 	ext{SE} = \frac{\sigma}{\sqrt{n}} = \frac{1.5}{\sqrt{25}} = \frac{1.5}{5} = 0.3 	ext{ minutes}. Turn the 5.75-minute threshold into standard-error units (the z-score): z = \frac{5.75 - 5}{0.3} = \frac{0.75}{0.3} = 2.5. From the normal table: P(Z 2.5) \approx 0.0062, about 0.6\%. An individual call easily runs 45 seconds long, but the average of 25 calls doing so is quite rare. Takeaway. The sample mean's spread is \sigma/\sqrt{n}: a 0.75-minute overshoot on a 25-call…

Chance average handle time runs long

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