## 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 $\text{platy-kurtic}$.