## Moment coefficient of kurtosis 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 moment coefficient of kurtosis $\beta_2$ is defined as

$\beta_2=\dfrac{m_4}{m_2^2}$

The moment coefficient of kurtosis $\gamma_2$ is defined as

$\gamma_2=\beta_2-3$

where

• $N$ total number of observations
• $\overline{x}$ sample mean
• $m_2 =\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^2$ is second central moment
• $m_4 =\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^4$ is fourth central moment

## 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 moments coefficient of kurtosis 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}

$x_i$ $f_i$ $(x_i-xb)^2$ $f_i*(x_i-xb)^2$ $(x_i-xb)^4$ $f_i*(x_i-xb)^4$
2 9 1.44 12.96 2.0736 18.6624
3 11 0.04 0.44 0.0016 0.0176
4 6 0.64 3.84 0.4096 2.4576
5 3 3.24 9.72 10.4976 31.4928
6 1 7.84 7.84 61.4656 61.4656
Total 96 34.8 114.096

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 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}

The coefficient of kurtosis based on moments ($\beta_2$) is \begin{aligned} \beta_2 &=\frac{m_4}{m_2^2}\\ &=\frac{(3.8032)}{(1.16)^2}\\ &=\frac{3.8032}{1.3456}\\ &=2.8264 \end{aligned} The coefficient of kurtosis based on moments ($\gamma_2$) is \begin{aligned} \gamma_2 &=\beta_2-3\\ &=2.8264 -3\\ &=-0.1736 \end{aligned}

As the value of $\gamma_2 < 0$, the data is $\text{platy-kurtic}$.

## Example 2

The following table gives the amount of time (in minutes) spent on the internet each evening by a group of 56 students.

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

Compute moments coefficient of kurtosis for the above frequency distribution.

### 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}

$x_i$ $f_i$ $(x_i-xb)^2$ $f_i(x_i-xb)^2$ $(x_i-xb)^4$ $f_i(x_i-xb)^4$
11 3 42.7154 128.1462 1824.6032 5473.8096
14 12 12.5012 150.0144 156.2794 1875.3528
17 15 0.287 4.305 0.0824 1.236
20 24 6.0728 145.7472 36.8786 885.0864
23 2 29.8586 59.7172 891.5345 1783.069
Total 56 487.93 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 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}

The coefficient of kurtosis based on moments ($\beta_2$) is \begin{aligned} \beta_2 &=\frac{m_4}{m_2^2}\\ &=\frac{(178.9027)}{(8.713)^2}\\ &=\frac{178.9027}{75.9164}\\ &=2.3566 \end{aligned} The coefficient of kurtosis based on moments ($\gamma_2$) is \begin{aligned} \gamma_2 &=\beta_2-3\\ &=2.3566 -3\\ &=-0.6434 \end{aligned}

As the value of $\gamma_2 < 0$, the data is $\text{platy-kurtic}$.