## 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

`$$ \begin{aligned} m_4 &=\frac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^4\\ &=\frac{116}{7}\\ &=16.5714 \end{aligned} $$`