Standard deviation of a delivery time after switching to hours

Question

Delivery times measured in minutes, T, have a standard deviation of 10 minutes. You re-express each time in hours using H = \frac{T}{60}. What is the standard deviation of the delivery times in hours?

Accepted Answer

Dividing by 60 is the same as multiplying by the factor 	frac{1}{60}, with no shift. This shrinks the scale, so it shrinks the spread by the same factor. The rule for H = aT is \mathrm{SD}(H) = a\,\mathrm{SD}(T), so the standard deviation is divided by 60 right along with the measurements. Apply it. With a = 	frac{1}{60}: \mathrm{SD}(H) = \frac{1}{60} 	imes 10 = \frac{10}{60} = \frac{1}{6} \approx 0.1667 	ext{ hours}. As a variance check, \mathrm{Var}(H) = \left(	frac{1}{60}ight)^2 	imes 100 = 	frac{100}{3600} = 	frac{1}{36}, and \sqrt{	frac{1}{36}} = 	frac{1}{6}. Takeaway. Rescal…

Standard deviation of a delivery time after switching to hours

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