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:
| Maximum load | No. of Cables |
|---|---|
| 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.
To learn more about other descriptive statistics, please refer to the following tutorial:
Let me know in the comments if you have any questions on Moment measure of kurtosis calculator for grouped data with examples and your thought on this article.
