# Moment coefficient of kurtosis calculator for grouped data

## 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

 $$\begin{eqnarray*} \overline{x}& =\frac{1}{N}\sum_{i=1}^{n}f_ix_i \end{eqnarray*}$$

The moment coefficient of kurtosis (also known as Pearson's moment coefficient of kurtosis) is denoted by $\beta_2$ and 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

## Interpretation of moment coefficient of kurtosis

• If $\gamma_2 > 0$ or $\beta_2 > 3$, then the frequency distribution is leptokurtic.
• If $\gamma_2 =0$ or $\beta_2 = 3$, then the frequency distribution is mesokurtic.
• If $\gamma_2 < 0$ or $\beta_2 < 3$, then the frequency distribution is platykurtic.

## Moment Coefficient of Kurtosis Calculator for grouped data

Use this calculator to find the Coefficient of Kurtosis based on moments for grouped data.

Moment coeff. of kurtosis
Type of Freq. Dist. DiscreteContinuous
Enter the Classes for X (Separated by comma,)
Enter the frequencies (f) (Separated by comma,)
Results
Number of Obs. (n):
Mean of X values:
First Central Moment :($m_1$)
Second Central Moment :($m_2$)
Third Central Moment :($m_3$)
Fourth Central Moment :($m_4$)
Coeff. of Kurtosis :($\beta_2$)
Coeff. of Kurtosis :($\gamma_2$)

## How to find moment coefficient of kurtosis for grouped data?

Step 1 - Select type of frequency distribution (Discrete or continuous)

Step 2 - Enter the Range or classes (X) seperated by comma (,)

Step 3 - Enter the Frequencies (f) seperated by comma

Step 4 - Click on "Calculate" button for moment coefficient of kurtosis calculation

Step 5 - Gives the output as number of observations $n$

Step 6 - Gives the mean, $m_1$,$m_2$,$m_3$,$m_4$, $\beta_2$ and $\gamma_2$.

Step 7 - Gives output as Moment Coefficient of Kurtosis

## Moment Coefficient of kurtosis 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}

$x_i$ $f_i$ $(x_i-\overline{x})^2$ $f_i*(x_i-\overline{x})^2$ $(x_i-\overline{x})^4$ $f_i*(x_i-\overline{x})^4$
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 $\text{platy-kurtic}$.

## Moment Coefficient of kurtosis 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}

$x_i$ $f_i$ $(x_i-\overline{x})^2$ $f_i(x_i-\overline{x})^2$ $(x_i-\overline{x})^4$ $f_i(x_i-\overline{x})^4$
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 $\text{platy-kurtic}$.

## Moment Coefficient of Kurtosis Example 3

The following table gives the frequency distribution of waiting time of 65 persons at a ticket counter to buy a movie ticket.

Waiting time (in minutes) 0-6 7-13 14-20 21-27 28- 34
frequency 5 12 18 30 10

Compute coefficient of kurtosis based on moments and interpret.

#### Solution

Class $x_i$ $f_i$ $f_i*x_i$
0-6 3 5 15
7-13 10 12 120
14-20 17 18 306
21-27 24 20 480
28-34 31 10 310
Total 65 1231

The mean of $X$ is

 \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{1231}{65}\\ &=18.9385 \end{aligned}

$x_i$ $f_i$ $(x_i-xb)^2$ $f_i(x_i-xb)^2$ $(x_i-xb)^4$ $f_i(x_i-xb)^4$
3 5 211.2866 1056.433 44642.0166 223210.083
10 12 56.7868 681.4416 3224.7378 38696.8536
17 18 0.287 5.166 0.0824 1.4832
24 20 41.7872 835.744 1746.168 34923.36
31 10 181.2874 1812.874 32865.1121 328651.121
Total 65 4391.6586 625482.9008

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{4391.6586}{65}\\ &=67.564 \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{625482.9008}{65}\\ &=9622.8139 \end{aligned}

The coefficient of kurtosis based on moments ($\beta_2$) is

 \begin{aligned} \beta_2 &=\frac{m_4}{m_2^2}\\ &=\frac{(9622.8139)}{(67.564)^2}\\ &=\frac{9622.8139}{4564.8941}\\ &=2.108 \end{aligned}
The coefficient of kurtosis based on moments ($\gamma_2$) is

 \begin{aligned} \gamma_2 &=\beta_2-3\\ &=2.108 -3\\ &=-0.892 \end{aligned}

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

## Moment Coefficient of Kurtosis Example 4

The Scores of students in a Math test is given in the table below :

Class Interval 10-20 20-30 30-40 40-50 50-60 60-70
Frequency ($f$) 6 8 12 10 5 4

Compute coefficient of Kurtosis based on moments and interpret.

#### Solution

Class $x_i$ $f_i$ $f_i*x_i$
10-20 15 6 90
20-30 25 8 200
30-40 35 12 420
40-50 45 10 450
50-60 55 5 275
60-70 65 4 260
Total 45 1695

The mean of $X$ is

 \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{1695}{45}\\ &=37.6667 \end{aligned}

$x_i$ $f_i$ $(x_i-xb)^2$ $f_i(x_i-xb)^2$ $(x_i-xb)^4$ $f_i(x_i-xb)^4$
15 6 6.4298 38.5788 41.342 248.052
25 8 55.7158 445.7264 3104.2475 24833.98
35 12 305.0018 3660.0216 93026.0824 1116312.9888
45 10 754.2878 7542.878 568950.0467 5689500.467
55 5 1403.5738 7017.869 1970019.3404 9850096.702
65 4 2252.8598 9011.4392 5075377.1635 20301508.654
Total 45 27716.513 36982500.8438

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{27716.513}{45}\\ &=615.9225 \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{36982500.8438}{45}\\ &=821833.3521 \end{aligned}

The coefficient of kurtosis based on moments ($\beta_2$) is

 \begin{aligned} \beta_2 &=\frac{m_4}{m_2^2}\\ &=\frac{(821833.3521)}{(615.9225)^2}\\ &=\frac{821833.3521}{379360.526}\\ &=2.1664 \end{aligned}
The coefficient of kurtosis based on moments ($\gamma_2$) is

 \begin{aligned} \gamma_2 &=\beta_2-3\\ &=2.1664 -3\\ &=-0.8336 \end{aligned}

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

## Moment Coefficient of kurtosis Example 5

The following data shows the distribution of maximum loads in short tons supported by certain cables produced by a company:

9.25-9.75 2
9.75-10.25 5
10.25-10.75 12
10.75-11.25 17
11.25-11.75 14
11.75-12.25 6
12.25-12.75 3
12.75-13.25 1

Compute coefficient of kurtosis based on moments and interpret.

#### Solution

Class $x_i$ $f_i$ $f_i*x_i$
9.25-9.75 9.5 2 19
9.75-10.25 10 5 50
10.25-10.75 10.5 12 126
10.75-11.25 11 17 187
11.25-11.75 11.5 14 161
11.75-12.25 12 6 72
12.25-12.75 12.5 3 37.5
12.75-13.25 13 1 13
Total 60 665.5

The mean of $X$ is

 \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{665.5}{60}\\ &=11.0917 \end{aligned}

$x_i$ $f_i$ $(x_i-xb)^2$ $f_i(x_i-xb)^2$ $(x_i-xb)^4$ $f_i(x_i-xb)^4$
9.5 2 64.5725 129.145 4169.6045 8339.209
10 5 56.7868 283.934 3224.7378 16123.689
10.5 12 49.5011 594.0132 2450.3564 29404.2768
11 17 42.7154 726.1618 1824.6032 31018.2544
11.5 14 36.4297 510.0158 1327.1212 18579.6968
12 6 30.644 183.864 939.0532 5634.3192
12.5 3 25.3583 76.0749 643.0421 1929.1263
13 1 20.5726 20.5726 423.2308 423.2308
Total 60 2523.7813 111451.8023

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{2523.7813}{60}\\ &=42.063 \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{111451.8023}{60}\\ &=1857.53 \end{aligned}

The coefficient of kurtosis based on moments ($\beta_2$) is

 \begin{aligned} \beta_2 &=\frac{m_4}{m_2^2}\\ &=\frac{(1857.53)}{(42.063)^2}\\ &=\frac{1857.53}{1769.296}\\ &=1.0499 \end{aligned}
The coefficient of kurtosis based on moments ($\gamma_2$) is

 \begin{aligned} \gamma_2 &=\beta_2-3\\ &=1.0499 -3\\ &=-1.9501 \end{aligned}

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

## Conclusion

In this tutorial, you learned about how to calculate moment coefficient of kurtosis. You also learned about how to solve numerical problems based on moment coefficient of kurtosis for grouped data. 