How to calculate standard deviation

The steps

1. Find the mean. Add every value and divide by how many there are.
2. Subtract the mean from each value. These are the deviations. Some will be negative — that is expected.
3. Square each deviation. Every result is now zero or positive.
4. Add the squared deviations together.
5. Divide by n − 1 if the data is a sample, or by n if it is the whole population. This is the variance.
6. Take the square root. This is the standard deviation.

Written as a formula, for a sample of n values with mean x̄:

s = √[ Σ(x − x̄)² / (n − 1) ]

The population version replaces x̄ with μ and n − 1 with N; nothing else changes. If any of the symbols is unfamiliar, the formula page takes the expression apart term by term. Here the point is the arithmetic, so on to a real dataset.

Worked example 1: sample standard deviation

Seven pupils sit a quiz marked out of 10 and score 4, 8, 6, 5, 3, 7, 9. They are seven pupils out of a much larger year group, so this is a sample and the denominator will be n − 1.

Step 1 — the mean. The scores add to 4 + 8 + 6 + 5 + 3 + 7 + 9 = 42, and there are 7 of them, so x̄ = 42 ÷ 7 = 6.

Steps 2 and 3 — deviations and squares. Subtract 6 from each score, then square the result. Laying this out as a table is the single best habit for doing the calculation by hand, because each column checks the one before it.

The middle column totals zero. It always does — the deviations above the mean exactly balance the ones below it — so if yours does not, the mean was wrong and there is no point going further. The squared column is what you want: Σ(x − x̄)² = 28.

Step 5 — divide. This is a sample of 7, so divide by n − 1 = 6: 28 ÷ 6 = 4.6667. That is the sample variance, s².

Step 6 — square root. √4.6667 = 2.1602. The sample standard deviation of the seven scores is s ≈ 2.16 marks.

s = √(28 / 6) = √4.6667 = 2.1602

Had these seven been the entire class rather than a sample, step 5 would divide by 7 instead: 28 ÷ 7 = 4 and √4 = 2 exactly. Same data, same working, different denominator, and the answer moves from 2.16 to 2.00 — which is why the sample-or-population question is settled before the arithmetic starts, not after.

The calculator below is loaded with the same seven numbers. Toggle it between sample and population to see the denominator change, or replace them with your own data and compare its working against yours line by line.

Worked example 2: population standard deviation

A committee has five members, aged 34, 41, 29, 38, 33. Every member is included and there is no larger group being estimated, so this is a population and the denominator is N.

1. Sum = 34 + 41 + 29 + 38 + 33 = 175. Mean μ = 175 ÷ 5 = 35.
2. Deviations from 35: −1, 6, −6, 3, −2. (Check: they sum to 0.)
3. Squared: 1, 36, 36, 9, 4.
4. Sum of squares = 1 + 36 + 36 + 9 + 4 = 86.
5. Divide by N = 5: 86 ÷ 5 = 17.2. This is the population variance, σ².
6. Square root: √17.2 = 4.1473.

σ = √(86 / 5) = √17.2 = 4.1473

The population standard deviation is σ ≈ 4.15 years. For comparison, the sample formula on the same ages would give √(86 ÷ 4) = √21.5 = 4.6368 — about 12% higher, which is typical of the gap at n = 5.

If you already know the mean

Search for "standard deviation from the mean" and you will find two different questions hiding behind the phrase. The first is: the mean has been given to me, what do I do with it? The answer is that you skip step 1 and nothing else. You still need every individual value, because steps 2 to 6 are done on the values, not on the mean.

Suppose you are told a set of five readings has a mean of 20, and the readings are 17, 22, 19, 24, 18. Deviations from 20 are −3, 2, −1, 4, −2. Squared they are 9, 4, 1, 16, 4, which sum to 34. As a sample: 34 ÷ 4 = 8.5, and √8.5 = 2.9155. As a population: 34 ÷ 5 = 6.8, and √6.8 = 2.6077.

The second question is: I have only the mean, can I work out the standard deviation from it? No. The mean tells you where the data is centred and nothing about how far it is spread. The sets 19, 20, 21 and 0, 20, 40 both have a mean of 20; their sample standard deviations are 1 and 20. You need the values, or at the very least the variance, the sum of squared deviations, or a standard error together with n.

One subtlety worth knowing. If the mean you were given is a rounded figure, the deviations inherit the rounding and the final answer drifts. Where you can recompute the mean exactly from the values, do so. And if the mean is a known population mean supplied from outside the data, rather than one calculated from these values, divide by n rather than n − 1: the correction exists to compensate for measuring spread around the sample's own mean, and it is not needed when you did not.

How do you estimate the standard deviation from the range?

If you only know the smallest value, the largest value and how many values there were, divide the range by a number that depends on the sample size. For about 25 to 30 values that number is close to 4, which is where the "range ÷ 4" rule of thumb comes from.

The divisor has to grow with the sample, because a bigger sample reaches further into the tails: ten values from a normal distribution typically span about 3.1 standard deviations, a hundred span about 5. So ÷ 4 understates the SD of a 10-value sample by about a quarter and overstates the SD of a 100-value sample by about a quarter. The estimator below adjusts for n. It assumes roughly normal data and is only an estimate; with the raw values, calculate the real thing.

Sample or population: how to decide

Ask one question: do these values make up the entire group I am describing, or are they a subset I am using to say something about a larger group? Twelve blood-pressure readings from a clinic are a sample of that clinic's patients. The heights of all twenty-three players in a squad, when the question is about that squad, are a population. The same numbers can be either — it depends on what claim you are making, not on the data itself.

When in doubt, treat the data as a sample. Almost every dataset in science, business and coursework is one, and the sample formula is what STDEV.S, a TI-84's Sx and the default on this site all compute. The choice matters most for small n, where dividing by n − 1 rather than n changes the answer by more than 10% at n = 5, and hardly at all above a few hundred values. The full guide to deciding is here, including a table of the gap at each n.

By hand, or with a calculator or spreadsheet

Working through the table by hand is worth doing a few times, because it is the only way to really understand what the number means, and it is what an exam expects. Beyond about ten values, though, hand arithmetic is slow and error-prone, and a tool is the right choice. Excel and Google Sheets do it in one cell: =STDEV.S(A1:A20) for a sample, =STDEV.P(A1:A20) for a population. The Excel guide covers the functions and the text-stored-as-number trap that silently drops rows.

On a TI-84, enter the data in a list and run 1-Var Stats; Sx is the sample standard deviation and σx the population one. The TI-84 guide has the keystrokes. And the calculator on this site shows each of the six steps with the intermediate numbers, which makes it a good way to check hand working before an exam.

Common mistakes

• Forgetting to square. The raw deviations sum to zero, so an unsquared version gives 0 for every dataset. If your total in step 4 is zero, this is why.
• Dividing by n for a sample. The most common error in coursework. It gives the population figure, which is always slightly too small for a sample. Decide sample or population before you start and write the denominator down.
• Taking the square root too early. Rooting each squared deviation individually just undoes step 3 and returns the absolute deviations. The root is taken once, at the end, of the whole average.
• Rounding intermediate steps. Keep at least four decimal places in the mean and the variance, and round only the final answer. A rounded mean nudges every deviation, and the errors compound through the squaring.
• Mixing units. All values must be in the same unit before you begin. Standard deviation carries the unit of the data, so a mix of centimetres and metres gives a meaningless number.
• Treating the range as the standard deviation. Largest minus smallest is the range, a different and much cruder measure. The SD is always smaller than the range and, for most data, roughly a quarter of it.

Practice problems

Each answer gives the mean and the sample standard deviation, with the population figure where asked. Work them on paper with the three-column table, then check. Rounded to three decimal places; if you rounded earlier your answer may differ in the last digit. For more, the printable standard deviation worksheets have four graded sets with answer keys, the online practice problems mark each answer as you go, and ten worked examples apply the method to real data.

Problem 1

Six delivery times, in minutes: 10, 12, 15, 11, 17, 13. Find the mean and the sample standard deviation.

Problem 2

Six measured lengths, in centimetres: 2.5, 3.1, 2.8, 3.4, 2.2, 3.0. Find the mean and the sample standard deviation.

Problem 3

Eight test scores: 45, 52, 38, 60, 49, 41, 55, 50. Find the mean, then both the sample and the population standard deviation.

Problem 4

Eight overnight temperatures, in °C: −5, 3, 0, 7, −2, 6, 1, 6. Find the mean and the sample standard deviation. Watch the signs.