Standard Deviation: Don't Stop at the Variance

The question

The number of hours eight Form 4 students spent on volunteer work in one month is shown below.

3, \quad 14, \quad 7, \quad 11, \quad 18, \quad 5, \quad 9, \quad 13

(a) Find the mean number of hours. [1 mark]

(b) Find the standard deviation of the data, correct to 3 significant figures. [3 marks]

The common wrong answer

Most students start well. They find the mean, square each value and use the formula for the variance:

\bar{x} = \frac{80}{8} = 10

\sum x^2 = 9 + 196 + 49 + 121 + 324 + 25 + 81 + 169 = 974

\sigma^2 = \frac{974}{8} - 10^2 = 121.75 - 100 = 21.75

Then they write “standard deviation $= 21.75$ hours” and move on.

The mistake is stopping one step early. The formula gives $\sigma^2$ , the variance. The standard deviation is $\sigma$ , so you still have to take the square root. Here the working is right all the way, and the student still loses the final mark.

A quick check shows the answer can’t be right. The data go from 3 to 18 hours, a range of only 15 hours. A “typical distance from the mean” of 21.75 hours is bigger than the whole spread of the data, which is impossible.

The correct working

(a) Add the eight values and divide by 8:

\bar{x} = \frac{3 + 14 + 7 + 11 + 18 + 5 + 9 + 13}{8} = \frac{80}{8} = 10

The mean is 10 hours.

(b) Square each value. A table keeps it tidy:

$x$	3	14	7	11	18	5	9	13
$x^2$	9	196	49	121	324	25	81	169

\sum x^2 = 974

Find the variance:

\sigma^2 = \frac{\sum x^2}{N} - \bar{x}^2 = \frac{974}{8} - 10^2 = 121.75 - 100 = 21.75

Now take the square root to get the standard deviation:

\sigma = \sqrt{21.75} = 4.6636\ldots

The standard deviation is 4.66 hours (3 s.f.).

You can check it with the other formula, $\sigma^2 = \dfrac{\sum (x - \bar{x})^2}{N}$ . The deviations from 10 are $-7, 4, -3, 1, 8, -5, -1, 3$ , and their squares add up to

49 + 16 + 9 + 1 + 64 + 25 + 1 + 9 = 174

so $\sigma^2 = \frac{174}{8} = 21.75$ and $\sigma = 4.66$ again.

Why it works

The variance is the mean of the squared distances from the mean. Squaring stops the positive and negative deviations from cancelling out, but it also squares the units. The data are in hours, so the variance of 21.75 is in hours², which doesn’t describe the data directly.

Taking the square root undoes the squaring. It brings you back to hours, and gives a sensible “typical distance” from the mean. Here most students’ hours are within about 4.66 hours of 10.

That is also why the range check works. Every value is within 8 hours of the mean of 10 (the furthest is 18), so the typical distance from the mean can’t be more than that. The standard deviation is never more than half the range, here $\frac{15}{2} = 7.5$ . The answer of 4.66 passes; 21.75 doesn’t.

How to spot it next time

Write $\sigma^2 = \ldots$ for the variance and $\sigma = \ldots$ for the standard deviation, so you can see which one you have.
Read the question again before you give the final answer: does it ask for the variance or the standard deviation?
Compare your answer with the range. The standard deviation is never more than half the range, so an answer bigger than the range is certainly wrong.