Chance total daily revenue lands in a band

Question

A warehouse ships n = 400 orders in a day. Each order's revenue averages \50 with a standard deviation of \30. Revenues are independent, so by the central limit theorem the day's total revenue is approximately normal. What is the probability the total revenue lands between \19{,}400 and \20{,}600?

Accepted Answer

This is a two-sided question about the total of 400 orders. Mean and variance both add, so the total is approximately normal with 	ext{mean} = n\mu = 400 	imes 50 = \20{,}000, 	ext{SD} = \sigma\sqrt{n} = 30 	imes \sqrt{400} = 30 	imes 20 = \600. The band \19{,}400 to \20{,}600 is exactly \pm\600 around the mean, which is \pm 1 standard error: z = \frac{600}{600} = 1. So we want the probability of landing within \pm 1 standard error: P(-1 < Z < 1) \approx 0.6827, about 68\%. This is the familiar "1 sigma covers about 68\%" rule, applied to the day's total. Takeaway. For the total of a sum…

Chance total daily revenue lands in a band

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