Standard deviation vs mean

Two numbers, two questions

The mean answers "what is a typical value?" It is the sum of the values divided by how many there are. The standard deviation answers "how much do the values differ from that typical value?" It is the square root of the average squared distance from the mean.

x̄ = Σx / n s = √[Σ(x − x̄)² / (n − 1)]

The formulas show the relationship. The standard deviation is built on the mean, since every deviation is measured from x̄, but the mean contains no information about spread and the standard deviation contains none about location. Knowing that a class averaged 70 tells you nothing about whether the scores ran from 68 to 72 or from 50 to 90. That exact pair of datasets is worked through on the what is standard deviation page: same mean, one SD ten times the other.

The reverse holds too, and is less often shown. Here are commute times in minutes on two routes over five days:

Route B is simply route A plus 20 minutes. The standard deviations are identical because the day-to-day variation is identical; only the centre has moved. If all you had was "SD = 4.74 minutes", you could not say which route you were looking at.

A worked example with both

Five midday temperatures in °C: 12, 15, 18, 20, 25. They sum to 90, so the mean is 90 ÷ 5 = 18.

x̄ = 90 / 5 = 18 °C s² = 98 / (5 − 1) = 24.5 °C² s = √24.5 = 4.95 °C

Read together: the days centred on 18 °C, and a typical day was about 5 degrees either side of that. Both numbers are in °C, which is what lets you set them side by side. A value of 25 °C is (25 − 18) ÷ 4.95 = 1.41 standard deviations above the mean; that is a z-score, and it needs the mean and the SD at once.

Reporting mean ± SD, and when it should be mean ± SE

The conventional way to give both is "mean ± SD": 18.0 ± 4.9 °C (n = 5). Written that way it describes the data. It says how much individual days varied, and for roughly bell-shaped data it lets a reader apply the 68–95–99.7 rule in their head.

"Mean ± SE" describes something else: how precisely the mean has been pinned down. The standard error is the SD divided by √n, so for the temperatures it is 4.95 ÷ √5 = 2.21 °C. The same data can therefore be written as 18.0 ± 4.9 or 18.0 ± 2.2, and a reader who does not know which was used will misread the spread by more than a factor of two. With n = 100 the gap becomes a factor of ten. Use the SD when describing a sample, the SE (or a confidence interval) when comparing means, and say which one it is every time.

The coefficient of variation: SD as a fraction of the mean

Divide the standard deviation by the mean and you get the coefficient of variation, usually given as a percentage:

CV = s / x̄ × 100% CV = 4.95 / 18 × 100% = 27.5%

The CV is the one place the two numbers are combined into a single figure. It is unit-free, so it lets you compare spread across data with very different means: routes A and B above have the same SD but CVs of 16.9% and 9.9%, because 4.74 minutes matters more on a 28-minute trip than on a 48-minute one. The coefficient of variation calculator works it out from raw data. It only makes sense for values measured from a true zero, as the next section shows.

Shifting and scaling: how each one responds

Transforming the data changes the mean and the SD in different ways, and the difference is a good test of what each one measures. Starting from the temperatures (mean 18, SD 4.95):

The general rule, for y = a·x + b:

mean(y) = a · mean(x) + b SD(y) = |a| · SD(x)

Adding a constant moves the mean and leaves the SD alone, because every value moves the same distance and no value gets closer to or further from the others. Multiplying scales both, and the SD by the absolute value of the multiplier, since a spread cannot be negative. The Fahrenheit row checks out: 1.8 × 18 + 32 = 64.4 and 1.8 × 4.95 = 8.91.

The CV column is the warning. Scaling leaves it unchanged, but shifting does not, so the same five days have a CV of 27.5% in Celsius and 13.8% in Fahrenheit. Neither is meaningful, because 0 °C and 0 °F are arbitrary points. Keep the CV for lengths, weights, times, counts and money.

What one outlier does to each

Suppose the sensor glitched on the last day and logged 60 instead of 25. The data is now 12, 15, 18, 20, 60.

The mean moved 7 degrees; the SD quadrupled. The SD reacts harder because the outlier's deviation is squared: 60 sits 35 from the new mean, and its 35² = 1225 is 78% of the new sum of squares, 1568. The median did not move, and neither does the interquartile range (15 to 20 with the inclusive quartile method, in both cases). When the data has genuine extreme values or a long tail, report the median and IQR alongside, or instead of, the mean and SD. The outlier calculator flags suspect points before you commit to either.

Common mistakes

"The standard deviation can't be bigger than the mean." It can. Take the number of prescriptions filled by seven patients in a year: 0, 0, 1, 1, 2, 3, 14. The mean is 3 and the sample SD is 4.97. Nothing is wrong with the arithmetic. For data that cannot go below zero, an SD above the mean does tell you something, though: the mean minus one SD is −1.97, a value that cannot occur, so the data cannot be symmetric around its mean. It has a long right tail, here a single heavy user, and the median of 1 describes a typical patient better than the mean of 3. For data that can be negative, the comparison means nothing. A fund's daily returns might average 0.05% with an SD of 1.2%, and the mean could as easily be zero.

Writing mean ± 2 SD as a range for skewed data. With the prescription counts that range is 3 ± 9.93, running from −6.93 to 12.93. The lower end is impossible and the upper end excludes the one real extreme. The ± form assumes rough symmetry.

Comparing SDs across groups with very different means. An SD of 5 kg is large for newborns and small for adult cattle. Use the CV, or compare each SD with its own mean.

Reporting ± without saying whether it is SD or SE. As shown above, the two differ by a factor of √n, and a caption that says only "mean ± error" leaves the reader guessing.

Treating a low SD as a good result. The SD describes variation; it does not judge it. Whether variation is a problem depends on the mean and on what the data is for, which is the theme of the use cases page.