Calculating standard deviation step by step Introduction
In this article, we'll learn how to calculate standard deviation "by hand".
Interestingly, in the real world no statistician would ever calculate standard deviation by hand. The calculations involved are somewhat complex, and the risk of making a mistake is high. Also, calculating by hand is slow. Very slow. This is why statisticians rely on spreadsheets and computer programs to crunch their numbers.
So what's the point of this article? Why are we taking time to learn a process statisticians don't actually use? The answer is that learning to do the calculations by hand will give us insight into how standard deviation really works. This insight is valuable. Instead of viewing standard deviation as some magical number our spreadsheet or computer program gives us, we'll be able to explain where that number comes from.
Overview of how to calculate standard deviation
The formula for standard deviation (SD) is
\Large\text{SD} = \sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\mu\rvert^2}}}{N}}SD=
N

​ ∣x−μ∣
2

​ standard deviation calculator

​ S, D, equals, square root of, start fraction, sum, start subscript, end subscript, start superscript, end superscript, open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript, divided by, N, end fraction, end square root
where \sum∑sum means "sum of", xxx is a value in the data set, \muμmu is the mean of the data set, and NNN is the number of data points in the population.
The standard deviation formula may look confusing, but it will make sense after we break it down. In the coming sections, we'll walk through a step-by-step interactive example. Here's a quick preview of the steps we're about to follow:
Step 1: Find the mean.
Step 2: For each data point, find the square of its distance to the mean.
Step 3: Sum the values from Step 2.
Step 4: Divide by the number of data points.
Step 5: Take the square root.
An important note
The formula above is for finding the standard deviation of a population. If you're dealing with a sample, you'll want to use a slightly different formula (below), which uses n-1n−1n, minus, 1 instead of NNN. The point of this article, however, is to familiarize you with the the process of computing standard deviation, which is basically the same no matter which formula you use.
\text{SD}_\text{sample} = \sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}{n-1}}SD
sample
​ =
n−1

​ ∣x−
x
ˉ
∣
2

​
​
[Why are there two formulas?]
Step-by-step interactive example for calculating standard deviation
First, we need a data set to work with. Let's pick something small so we don't get overwhelmed by the number of data points. Here's a good one:
6, 2, 3, 16,2,3,16, comma, 2, comma, 3, comma, 1
Step 1: Finding \goldD{\mu}μstart color goldD, mu, end color goldD in \sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\goldD{\mu}\rvert^2}}}{N}}
N

​ ∣x−μ∣
2

​
​ square root of, start fraction, sum, start subscript, end subscript, start superscript, end superscript, open vertical bar, x, minus, start color goldD, mu, end color goldD, close vertical bar, start superscript, 2, end superscript, divided by, N, end fraction, end square root
In this step, we find the mean of the data set, which is represented by the variable \muμmu.
Fill in the blank.
\mu = μ=mu, equals
[Explain]
\mu = \dfrac{6+2 + 3 + 1}{4} = \dfrac{12}{4} = \blueD3

mu, equals, start fraction, 6, plus, 2, plus, 3, plus, 1, divided by, 4, end fraction, equals, start fraction, 12, divided by, 4, end fraction, equals, start color blueD, 3, end color blueD
Step 2: Finding \goldD{\lvert x - \mu \rvert^2}∣x−μ∣
2
start color goldD, open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript, end color goldD in \sqrt{\dfrac{\sum\limits_{}^{}{\goldD{{\lvert x-\mu}\rvert^2}}}{N}}
N

​ ∣x−μ∣
2

​
​ square root of, start fraction, sum, start subscript, end subscript, start superscript, end superscript, start color goldD, open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript, end color goldD, divided by, N, end fraction, end square root
In this step, we find the distance from each data point to the mean (i.e., the deviations) and square each of those distances.
For example, the first data point is 666 and the mean is 333, so the distance between them is 333. Squaring this distance gives us 999.
Complete the table below.
Data point xxx Square of the distance from the mean \lvert x - \mu \rvert^2∣x−μ∣
2
open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript
666 999
222
333
111
[Explain]
xx \lvert x - \mu \rvert^2
open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript
66 \lvert6-\blueD{3}\rvert^2 = 3^2 = 9

open vertical bar, 6, minus, start color blueD, 3, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 3, start superscript, 2, end superscript, equals, 9
22 \lvert2-\blueD{3}\rvert^2 = 1^2 = 1

open vertical bar, 2, minus, start color blueD, 3, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 1, start superscript, 2, end superscript, equals, 1
33 \lvert3-\blueD{3}\rvert^2 = 0^2 = 0

open vertical bar, 3, minus, start color blueD, 3, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0
11 \lvert1-\blueD{3}\rvert^2 = 2^2 = 4

open vertical bar, 1, minus, start color blueD, 3, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 2, start superscript, 2, end superscript, equals, 4
Step 3: Finding \goldD{\sum\lvert x - \mu \rvert^2}∑∣x−μ∣
2
start color goldD, sum, open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript, end color goldD in \sqrt{\dfrac{\goldD{\sum\limits_{}^{}{{\lvert x-\mu}\rvert^2}}}{N}}
N

​ ∣x−μ∣
2

​
​ square root of, start fraction, start color goldD, sum, start subscript, end subscript, start superscript, end superscript, open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript, end color goldD, divided by, N, end fraction, end square root
The symbol \sum∑sum means "sum", so in this step we add up the four values we found in Step 2.
Fill in the blank.
\sum\lvert x - \mu \rvert^2 = ∑∣x−μ∣
2
=sum, open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript, equals
[Explain]
\sum\lvert x - \mu \rvert^2 = 9 + 1 + 0 + 4 = 14
sum, open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript, equals, 9, plus, 1, plus, 0, plus, 4, equals, 14
Step 4: Finding \goldD{\dfrac{\sum\lvert x - \mu \rvert^2}{N}}
N
∑∣x−μ∣
2

​ start color goldD, start fraction, sum, open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript, divided by, N, end fraction, end color goldD in \sqrt{\goldD{\dfrac{\sum\limits_{}^{}{{\lvert x-\mu}\rvert^2}}{N}}}
N

​ ∣x−μ∣
2

​
​ square root of, start color goldD, start fraction, sum, start subscript, end subscript, start superscript, end superscript, open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript, divided by, N, end fraction, end color goldD, end square root
In this step, we divide our result from Step 3 by the variable NNN, which is the number of data points.
Fill in the blank.
\dfrac{\sum\lvert x - \mu \rvert^2}{N} =
N
∑∣x−μ∣
2

​ =start fraction, sum, open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript, divided by, N, end fraction, equals
[Explain]
( N =4)left parenthesis, N, equals, 4, right parenthesis
\dfrac{\sum\lvert x - \mu \rvert^2}{N} =\dfrac{{14}}4 = {3.5}

start fraction, sum, open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript, divided by, N, end fraction, equals, start fraction, 14, divided by, 4, end fraction, equals, 3, point, 5
Step 5: Finding the standard deviation \sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\mu\rvert^2}}}{N}}
N

​ ∣x−μ∣
2

​
​ square root of, start fraction, sum, start subscript, end subscript, start superscript, end superscript, open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript, divided by, N, end fraction, end square root
We're almost finished! Just take the square root of the answer from Step 4 and we're done.
Fill in the blank.
\text{SD} = \sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\mu\rvert^2}}}{N}} \approx SD=
N

​ ∣x−μ∣
2

​
​ ≈S, D, equals, square root of, start fraction, sum, start subscript, end subscript, start superscript, end superscript, open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript, divided by, N, end fraction, end square root, approximately equals
[Explain]
\sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\mu\rvert^2}}}{N}} = \sqrt{{3.5}} \approx 1.87

square root of, start fraction, sum, start subscript, end subscript, start superscript, end superscript, open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript, divided by, N, end fraction, end square root, equals, square root of, 3, point, 5, end square root, approximately equals, 1, point, 87
1.871, point, 87
Yes! We did it! We successfully calculated the standard deviation of a small data set.
Summary of what we did
We broke down the formula into five steps:
Step 1: Find the mean \muμmu.
\mu = \dfrac{6+2 + 3 + 1}{4} = \dfrac{12}{4} = \blueD3μ=
4
6+2+3+1
​ =
4
12
​ =3mu, equals, start fraction, 6, plus, 2, plus, 3, plus, 1, divided by, 4, end fraction, equals, start fraction, 12, divided by, 4, end fraction, equals, start color blueD, 3, end color blueD
Step 2: Find the square of the distance from each data point to the mean \lvert x-\mu\rvert^2∣x−μ∣
2
open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript.
xxx \lvert x - \mu \rvert^2∣x−μ∣
2
open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript
666 \lvert6-\blueD{3}\rvert^2 = 3^2 = 9∣6−3∣
2
=3
2
=9open vertical bar, 6, minus, start color blueD, 3, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 3, start superscript, 2, end superscript, equals, 9
222 \lvert2-\blueD{3}\rvert^2 = 1^2 = 1∣2−3∣
2
=1
2
=1open vertical bar, 2, minus, start color blueD, 3, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 1, start superscript, 2, end superscript, equals, 1
333 \lvert3-\blueD{3}\rvert^2 = 0^2 = 0∣3−3∣
2
=0
2
=0open vertical bar, 3, minus, start color blueD, 3, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0
111 \lvert1-\blueD{3}\rvert^2 = 2^2 = 4∣1−3∣
2
=2
2
=4open vertical bar, 1, minus, start color blueD, 3, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 2, start superscript, 2, end superscript, equals, 4
Steps 3, 4, and 5:
\begin{aligned} \text{SD} &= \sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\mu\rvert^2}}}{N}}\\\\\\\\ &= \sqrt{\dfrac{9 + 1 + 0 + 4}{4}} \\\\\\\\ &= \sqrt{\dfrac{{14}}{4}} ~~~~~~~~\small \text{Sum the squares of the distances (Step 3).} \\\\\\\\ &= \sqrt{{3.5}} ~~~~~~~~\small \text{Divide by the number of data points (Step 4).} \\\\\\\\ &\approx 1.87 ~~~~~~~~\small \text{Take the square root (Step 5).} \end{aligned}
SD
​

N

​ ∣x−μ∣
2

​
​

4
9+1+0+4
​
​

4
14
​
​         Sum the squares of the distances (Step 3).

3.5
​         Divide by the number of data points (Step 4).
≈1.87        Take the square root (Step 5).
​
Try it yourself
Here's a reminder of the formula:
\Large\text{SD} = \sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\mu\rvert^2}}}{N}}SD=
N

​ ∣x−μ∣
2

​
​ S, D, equals, square root of, start fraction, sum, start subscript, end subscript, start superscript, end superscript, open vertical bar, x, minus, mu, close vertical bar, start superscript, 2, end superscript, divided by, N, end fraction, end square root
And here's a data set:
1, 4, 7, 2,6 1,4,7,2,61, comma, 4, comma, 7, comma, 2, comma, 6
Find the standard deviation of the data set.