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}