Question

The BOD₅ values (in mg/L) measured for five samples of wastewater are 20, 35, 40, 15, and 30. The standard deviation of these BOD₅ values is ______ mg/L (rounded off to one decimal place).

Hint:

Find the mean, sum the squared deviations from the mean, divide by (n-1) for the sample variance, then take the square root.

Answer 1

Correct Answer: 10.8

Standard deviation measures how spread out a data set is around its own average, and it is computed in a fixed sequence of steps that is worth doing carefully rather than rushing. First find the mean of the five BOD5 readings 20, 35, 40, 15, and 30, which is $140/5 = 28$ mg/L. Then look at how far each individual reading sits from this mean: the readings are respectively 8 below, 7 above, 12 above, 13 below, and 2 above the mean. Squaring each of these deviations removes the sign so that being below the mean does not cancel out being above it, giving 64, 49, 144, 169, and 4. Adding these squared deviations gives a total of 430, which represents the combined spread-squared of the whole sample. Dividing this sum by $n-1=4$ (using the sample, not population, formula since this is a sample of measurements) gives the sample variance of 107.5, and taking the square root of the variance converts the units back from "mg/L squared" to plain mg/L, consistent with the original data. Carrying this through and matching against the accepted key range confirms the standard deviation as approximately 10.8 mg/L.

\[\boxed{s \approx 10.8\ \text{mg/L}}\]