Chance average pick time blows past target

Question

A warehouse's item-pick time has a true mean of $40$ seconds with standard deviation $\sigma = 9$ seconds. Picks are independent, so by the central limit theorem the average of many picks is approximately normal.

Over $n = 81$ picks, what is the probability the average pick time exceeds $43$ seconds?

Accepted Answer

The average of 81 independent picks is approximately normal, centered at 40 seconds, with spread equal to the standard error: 	ext{SE} = \frac{\sigma}{\sqrt{n}} = \frac{9}{\sqrt{81}} = \frac{9}{9} = 1 	ext{ second}. Convert the 43-second threshold to standard-error units (the z-score): z = \frac{43 - 40}{1} = 3. From the normal table: P(Z 3) \approx 0.00135, about 0.135\%, roughly one chance in 740. A 3-second overshoot on an 81-pick average is a genuinely rare event because averaging over 81 picks shrinks the spread to just 1 second. Takeaway. With n = 81 the standard error is \sigma/9, so…

Chance average pick time blows past target

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