## Moment coefficient of kurtosis 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 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
• $m_2 =\dfrac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^2$ is second sample central moment
• $m_4 =\dfrac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^4$ is fourth sample central moment

## Example

The hourly earning (in dollars) of sample of 7 workers are :

26, 21, 24, 22, 25, 24, 23.

Compute coefficient of kurtosis based on moments.

### Solution

The 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}

$x$ $(x-xb)$ $(x-xb)^2$ $(x-xb)^4$
27 2 4 16
27 2 4 16
24 -1 1 1
26 1 1 1
25 0 0 0
24 -1 1 1
22 -3 9 81
Total 175 0 20 116

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}

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}

Coefficient of Kurtosis

The coefficient of kurtosis based on moments ($\beta_2$) is \begin{aligned} \beta_2 &=\frac{m_4}{m_2^2}\\ &=\frac{(16.5714)}{(2.8571)^2}\\ &=\frac{16.5714}{8.163}\\ &=2.0301 \end{aligned} The coefficient of kurtosis based on moments ($\gamma_2$) is \begin{aligned} \gamma_2 &=\beta_2-3\\ &=2.0301 -3\\ &=-0.9699 \end{aligned}

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