Central moments for ungrouped data
Let $x_1, x_2,\cdots, x_n$ be $n$ observations. The mean of $X$ is denoted by $\overline{x}$ and is given by
$$ \begin{eqnarray*} \overline{x}& =\frac{1}{n}\sum_{i=1}^{n}x_i \end{eqnarray*} $$
Formula
The first four sample central moments are as follows
$m_1=0$ (always)
$m_2 =\dfrac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^2$
$m_3 =\dfrac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^3$
$m_4 =\dfrac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^4$
where,
$n$total number of observations$\overline{x}$sample mean
Example
The hourly earning (in dollars) of sample of 7 workers are : 26,21,24,22,25,24,23.
Compute first four sample central moments.
Solution
The sample mean of $X$ is
$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{175}{7}\\ &=25 \text{ dollars} \end{aligned} $$
| $x$ | $(x-xb)$ | $(x-xb)^2$ | $(x-xb)^3$ | $(x-xb)^4$ | |
|---|---|---|---|---|---|
| 27 | 2 | 4 | 8 | 16 | |
| 27 | 2 | 4 | 8 | 16 | |
| 24 | -1 | 1 | -1 | 1 | |
| 26 | 1 | 1 | 1 | 1 | |
| 25 | 0 | 0 | 0 | 0 | |
| 24 | -1 | 1 | -1 | 1 | |
| 22 | -3 | 9 | -27 | 81 | |
| Total | 175 | 0 | 20 | -12 | 116 |
The first sample central moment $m_1$ is always zero.
Second sample central moment
The second sample central moment is
$$ \begin{aligned} m_2 &=\frac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^2\\ &=\frac{20}{7}\\ &=2.8571 \end{aligned} $$
Third sample central moment
The third sample central moment is
$$ \begin{aligned} m_3 &=\frac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^3\\ &=\frac{-12}{7}\\ &=-1.7143 \end{aligned} $$
Fourth sample central moment
The fourth sample central moment is
$$ \begin{aligned} m_4 &=\frac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^4\\ &=\frac{116}{7}\\ &=16.5714 \end{aligned} $$
