Standard deviation is a measurement of how spread out the numbers are of a set of data values. The closer the standard deviation is to zero, the closer the data points are to the mean. Large values of standard deviation is an indication the data is spread out away from the mean. This will show how to calculate the standard deviation of a set of data.

Standard deviation, represented by the lower case Greek letter, σ is calculated from the variance from the mean of each data point. Variance is simply the average of the squared difference of each data point from the mean.

There are three steps to calculating variance:

- Find the mean of the data.
- For each number in the data set, subtract the mean found in step 1 from each value and then square each value.
- Find the mean of the values found in step 2.

Example: Let’s take a set of test scores from a math class of nine students. The scores were:

**65, 95, 73, 88, 83, 92, 74, 83, and 94**

Step 1 is find the mean. To find the mean, add all these scores together.

65 + 95 + 73 + 88 + 83 + 92 + 74 + 83 + 94 = 747

Divide this value by the total number of tests (9 scores)

747 ÷ 9 = 83

The mean score on the test was a score of 83.

For step 2, we need to subtract the mean from each test score and square each result.

(65 – 83)² = (-18)² = 324

(95 – 83)² = (12)² =144

(73 – 83)² = (-10)² = 100

(88 – 83)² = (5)² = 25

(83 – 83)² = (0)² = 0

(92 – 83)² = (9)² = 81

(74 – 83)² = (-9)² = 81

(83 – 83)² = (0)² = 0

(94 – 83)² = (11)² = 121

Step 3 is find the mean of these values. Add them all together:

324 + 144 + 100 + 25 + 0 + 81 + 81 + 0 + 121 = 876

Divide this value by the total number of scores (9 scores)

876 ÷ 9 = 97 (rounded to the nearest whole score)

The variance of the test scores is 97.

The standard deviation is simply the square root of the variance.

σ = √97 = 9.8 (round to nearest whole test score = 10)

This means scores within one standard deviation, or 10 points of the average score could be all considered ‘average scores’ of the class. The two scores 65 and 73 would be considered ‘below average’ and the 94 would be ‘above average’.

This calculation of standard deviation is for population measurements. This is when you can account for all the data in the population of the set. This example had a class of nine students. We know all the scores of all the students in the class. What if these nine scores were randomly taken from a larger set of scores, say the entire 8th Grade. The set of nine test scores is considered a **sample** set from the population.

Sample standard deviations are calculated slightly different. The first two steps are identical. In step 3, instead of dividing by the total number of tests, you divide by one less than the total number.

In our example above, the total from step 2 added together was 876 for 9 test scores. To find the sample variance, divide this number by one less than 9, or 8

876 ÷ 8 = 109.5

The sample variance is 109.5. Take the square root of this value to get the sample standard deviation:

sample standard deviation = √109.5 = 10.5

**Review**

To find the population standard deviation:

- Find the mean of the data.
- For each number in the data set, subtract the mean found in step 1 from each value and then square each value.
- Find the mean of the values found in step 2.
- Divide the value of step 3 by the total number of values.
- Take the square root of the result of step 4.

To find the sample standard deviation:

- Find the mean of the data.
- For each number in the data set, subtract the mean found in step 1 from each value and then square each value.
- Find the mean of the values found in step 2.
- Divide the value of step 3 by the total number of values minus 1.
- Take the square root of the result of step 4.

95-83=12, Example, Step 2

The devil is in the details.

Thank you for pointing out the error. Updated and fixed.

BTW, thanks. Learned this in high school but that was 40 years ago. And hadn’t been able to find nice straightforward description of the calc. Much obliged

Thank you for an explanation I have been awaiting for.. well… since 1972 :). Indeed, I had not been acquainted with the notion of sample standard deviation prior to reading this article (or perhaps I forgot). Odd, I guess in that I now make a living proof-reading the translations of science papers too, and had a background in the practical and applied sciences. My assumption then, is that the data which are outside the set deviation norms are then kicked out, the rest retained and utilized. Is this so?

The outlying data should be retained. Researchers look for explanations for reasons why data didn’t fit the curve, but shouldn’t discount it. Sometimes it’s random chance, but sometimes there is a new discovery to be made there 🙂