You may have seen the Greek symbol sigma (\(\Sigma\)) used in maths, but what does it mean? It is a way to simply express long numbers, using something called summation notation. A good understanding of summation notation is important in statistics for calculating averages and sums. Use this resource to learn how to interpret and use summation notation.
Summation (or sigma) notation is a shorthand way to write the sum or addition of a sequence of numbers or mathematical terms. It is a great way to express the sum of a large data set that has a mathematical relationship.
The Greek symbol sigma \(\Sigma\) is used to denote summation. It is called the summation sign.
\(n\) is the index of summation. It is the value that you start at. For example, if \(n=2\), then first value you use is \(x_{2}\).
\(N\) is the upper limit of summation. It is the very last value of \(n\) you include in your sum. For example, if \(n=1\) and \(N=3\), then your sum will only include \(x_{1}+x_{2}+x_{3}\).
\(x\) tells you what to sum. It can be a single value or an algebraic expression.
\(x_{1}\) is the first item in a data set, \(x_{2}\) is the second item, \(x_{3}\) is the third, and so on. They are added in increments of \(1\).
Using summation notation
Consider summation notation of a data set where \(x\) is a value.
\[\begin{align*} \sum_{n=1}^{5}n & = 1+2+3+4+5\\
& = 15
\end{align*}\]
\(n=1\), so we start adding from \(x_{1}=1\). \(N=5\), so we add in increments of \(1\) until we have added the value \(x_{5}=5\).
This is just a very simple summation; summation notation can involve more complex sequences.
Let's consider the same notation, but where \(x_{n}=2n\).
\[\begin{align*} \sum_{n=1}^{5}2n & = \underbrace{(2\times1)}_{n=1}+\underbrace{(2\times2)}_{n=2}+\underbrace{(2\times3)}_{n=3}+\underbrace{(2\times4)}_{n=4}+\underbrace{(2\times5)}_{n=5}\\
& = 30
\end{align*}\]
We start from \(x_{1}=2\times1=2\) and add in increments of \(1\) until we have added the value \(x_{5}=2\times5=10\).
Example 1 – using summation notation
Expand and evaluate \(\sum_{n=0}^{3}(n^{2}-3)\).
We start from \(x_{0}=0^{2}-3=-3\) and add in increments sof \(1\) until we have added the value \(x_{3}=3^{2}-3=6\).
\[\begin{align*} \sum_{n=0}^{3}(n^{2}-3) & = \underbrace{(0^{2}-3)}_{n=0}+\underbrace{(1^{2}-3)}_{n=1}+\underbrace{(2^{2}-3)}_{n=2}+\underbrace{(3^{2}-3)}_{n=3}\\
& = (-3)+(-2)+1+6\\
& = 2
\end{align*}\]
Given the set of data where \(x_{1}=1\), \(x_{2}=2\), \(x_{3}=4\) and \(x_{4}=5\), evaluate \(\overline{x}=\dfrac{\sum_{n=1}^{N}x_{n}}{N}\) where \(N\) is the number of items in the data set.
There are \(4\) items in the data set, so \(N=4\).
\[\begin{align*} \overline{x} & = \frac{\sum_{n=1}^{4}x_{n}}{4}\\
& = \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}\\
& = \frac{1+2+4+5}{4}\\
& = \frac{12}{4}\\
& = 3
\end{align*}\]
This formula calculates the mean or average of a data set. See Mean, mode and median for more.
Given the set of data where \(x_{1}=1\), \(x_{2}=2\), \(x_{3}=4\) and \(x_{4}=5\), evaluate \(s^{2}=\dfrac{\sum_{n=1}^{4}\left(x_{n}-\overline{x}\right)^{2}}{N-1}\).
We have \(\overline{x}=3\) from Example 2, so we don't need to calculate it again. \(N=4\) as the final item is \(x_{4}\).
\[\begin{align*} s^{2} & = \frac{\sum_{n=1}^{4}\left(x_{n}-3\right)^{2}}{4-1}\\
& = \frac{(x_{1}-3)^{2}+(x_{2}-3)^{2}+(x_{3}-3)^{2}+(x_{4}-3)^{2}+(x_{5}-3)^{2}}{3}\\
& = \frac{(1-3)^{2}+(2-3)^{2}+(4-3)^{2}+(5-3)^{2}}{3}\\
& = \frac{4+1+1+4}{3}\\
& = \frac{10}{3}
\end{align*}\]
This formulas calculates the variance or spread of data around the mean. See Measures of spread for more.
Exercise – using summation notation
Evaluate the following.
\(\sum_{n=1}^{5}3n\)
\(\sum_{n=1}^{3}(5n-2)\)
\(\sum_{n=1}^{3}(5n)-2\)
Given \(x_{1}=-2\), \(x_{2}=0\), \(x_{3}=1\), \(x_{4}=3\) and \(x_{5}=3\), evaluate the following.
