## Central moments for grouped data

Let $(x_i,f_i), i=1,2, \cdots , n$ be given frequency distribution. The mean of $X$ is denoted by $\overline{x}$ and is given by $$\begin{eqnarray*} \overline{x}& =\frac{1}{N}\sum_{i=1}^{n}f_ix_i \end{eqnarray*}$$

## Formula

The first four central moments are as follows

$m_1=0$

$m_2 =\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^2$

$m_3 =\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^3$

$m_4 =\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^4$

where,

• $N$ total number of observations
• $\overline{x}$ sample mean

## Example 1

Following tables shows a frequency distribution of daily number of car accidents at a particular cross road during a month of April.

No.of car accidents ($x$) 2 3 4 5 6
No. of days ($f$) 9 11 6 3 1

Compute first four central moments for the above frequency distribution.

### Solution

$x$ Freq ($f$) $f*x$
2 9 18
3 11 33
4 6 24
5 3 15
6 1 6
Total 30 96

The mean of $X$ is

\begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{96}{30}\\ &=3.2 \end{aligned}

$x_i$ $f_i$ $(x_i-xb)^2$ $f_i*(x_i-xb)^2$ $(x_i-xb)^3$ $f_i*(x_i-xb)^3$ $(x_i-xb)^4$ $f_i*(x_i-xb)^4$
2 9 1.44 12.96 -1.728 -15.552 2.0736 18.6624
3 11 0.04 0.44 -0.008 -0.088 0.0016 0.0176
4 6 0.64 3.84 0.512 3.072 0.4096 2.4576
5 3 3.24 9.72 5.832 17.496 10.4976 31.4928
6 1 7.84 7.84 21.952 21.952 61.4656 61.4656
Total 96 34.8 26.88 114.096

The first central moment $m_1$ is always zero.

The second central moment is

\begin{aligned} m_2 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^2\\ &=\frac{34.8}{30}\\ &=1.16 \end{aligned} The third central moment is

\begin{aligned} m_3 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^3\\ &=\frac{26.88}{30}\\ &=0.896 \end{aligned}

The fourth central moment is

\begin{aligned} m_4 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^4\\ &=\frac{114.096}{30}\\ &=3.8032 \end{aligned}

## Example 2

The following table gives the amount of time (in minutes) spent on the internet each evening by a group of 56 students. Compute five number summary for the following frequency distribution.

Time spent on Internet ($x$) 10-12 13-15 16-18 19-21 22-24
No. of students ($f$) 3 12 15 24 2

### Solution

Class $x_i$ $f_i$ $f_i*x_i$
10-12 11 3 33
13-15 14 12 168
16-18 17 15 255
19-21 20 24 480
22-24 23 2 46
Total 56 982

The mean of $X$ is

\begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{982}{56}\\ &=17.5357 \end{aligned}

$x_i$ $f_i$ $(x_i-xb)^2$ $f_i(x_i-xb)^2$ $(x_i-xb)^3$ $f_i(x_i-xb)^3$ $(x_i-xb)^4$ $f_i(x_i-xb)^4$
11 3 42.7154 128.1462 -279.1749 -837.5247 1824.6032 5473.8096
14 12 12.5012 150.0144 -44.2004 -530.4048 156.2794 1875.3528
17 15 0.287 4.305 -0.1537 -2.3055 0.0824 1.236
20 24 6.0728 145.7472 14.9651 359.1624 36.8786 885.0864
23 2 29.8586 59.7172 163.1562 326.3124 891.5345 1783.069
Total 56 487.93 -684.7602 10018.5538

The first central moment $m_1$ is always zero.

The second central moment is

\begin{aligned} m_2 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^2\\ &=\frac{487.93}{56}\\ &=8.713 \end{aligned} The third central moment is

\begin{aligned} m_3 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^3\\ &=\frac{-684.7602}{56}\\ &=-12.2279 \end{aligned}

The fourth central moment is

\begin{aligned} m_4 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^4\\ &=\frac{10018.5538}{56}\\ &=178.9027 \end{aligned}