Moment coefficient of skewness for grouped data
Skewness is a measure of symmetry, or more precisely, the lack of symmetry. A data set is symmetric if it looks the same to the left and right of the center point. A frequency distribution is said to be skewed if it is not symmetric. The data set is said to be positively (negatively) skewed if it has a longer tail towards right (left). The degree of skewness is measured by its coefficient.
Various measures of skewness are
- Karl Pearson's measure of skewness
- Bowley's measure of skewness
- Kelly's measure of skewness
- Moments measure of skewness
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 skewness $\beta_1$ is defined as
$$\beta_1=\dfrac{m_3^2}{m_2^3}$$
The moment coefficient of skewness $\gamma_1$ is defined as
$$\gamma_1=\sqrt{\beta_1}=\dfrac{m_3}{m_2^{3/2}}$$
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$m_3 =\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^3$is third central moment
Interpretation of coefficient of skewness
- If $\gamma_1 > 0$, then the data is positively skewed.
- If $\gamma_1 = 0$, then the data is symmetric (i.e., absence of skewness).
- If $\gamma_1 < 0$, then the data is negatively skewed.
Moment Coefficient of Skewness Calculator for grouped data
Use this calculator to find the Coefficient of Skewness based on moments for grouped data.
| Moment coeff. of Skewness | |
|---|---|
| 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 Skewness :($\beta_1$) | |
| Coeff. of Skewness :($\gamma_1$) | |
How to find moment coefficient of skewness 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 skewness calculation
Step 5 - Gives the output as number of observations $n$
Step 6 - Gives the mean, $m_1$,$m_2$,$m_3$,$m_4$, $\beta_1$ and $\gamma_1$.
Step 7 - Gives output as Moment Coefficient of Skewness
Moment Coefficient of Skewness 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 moment coefficient of skewness 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
$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{96}{30}\\ &=3.2 \end{aligned} $$
| $x_i$ | $f_i$ | $(x_i-xb)^2$ | $f_i*(x_i-xb)^2$ | $(x_i-xb)^3$ | $f_i*(x_i-xb)^3$ | |
|---|---|---|---|---|---|---|
| 2 | 9 | 1.44 | 12.96 | -1.728 | -15.552 | |
| 3 | 11 | 0.04 | 0.44 | -0.008 | -0.088 | |
| 4 | 6 | 0.64 | 3.84 | 0.512 | 3.072 | |
| 5 | 3 | 3.24 | 9.72 | 5.832 | 17.496 | |
| 6 | 1 | 7.84 | 7.84 | 21.952 | 21.952 | |
| Total | 96 | 34.8 | 26.88 |
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 third central moment is
$$ \begin{aligned} m_3 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^3\\ &=\frac{26.88}{30}\\ &=0.896 \end{aligned} $$
The coefficient of skewness based on moments ($\beta_1$) is
$$ \begin{aligned} \beta_1 &=\frac{m_3^2}{m_2^3}\\ &=\frac{(0.896)^2}{(1.16)^3}\\ &=\frac{0.8028}{1.5609}\\ &=0.5143 \end{aligned} $$
The coefficient of skewness based on moments ($\gamma_1$) is
$$ \begin{aligned} \gamma_1 &=\frac{m_3}{m_2^{3/2}}\\ &=\frac{0.896}{(1.16)^{3/2}}\\ &=\frac{0.896}{1.2494}\\ &=0.7172 \end{aligned} $$
As the value of $\gamma_1 > 0$, the data is $\text{positively skewed}$.
Moment Coefficient of Skewness 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 coefficient of skewness based on moments and interpret.
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
$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{982}{56}\\ &=17.5357 \end{aligned} $$
| $x_i$ | $f_i$ | $(x_i-xb)^2$ | $f_i(x_i-xb)^2$ | $(x_i-xb)^3$ | $f_i(x_i-xb)^3$ | |
|---|---|---|---|---|---|---|
| 11 | 3 | 42.7154 | 128.1462 | -279.1749 | -837.5247 | |
| 14 | 12 | 12.5012 | 150.0144 | -44.2004 | -530.4048 | |
| 17 | 15 | 0.287 | 4.305 | -0.1537 | -2.3055 | |
| 20 | 24 | 6.0728 | 145.7472 | 14.9651 | 359.1624 | |
| 23 | 2 | 29.8586 | 59.7172 | 163.1562 | 326.3124 | |
| Total | 56 | 487.93 | -684.7602 |
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 third central moment is
$$ \begin{aligned} m_3 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^3\\ &=\frac{-684.7602}{56}\\ &=-12.2279 \end{aligned} $$
The coefficient of skewness based on moments ($\beta_1$) is
$$ \begin{aligned} \beta_1 &=\frac{m_3^2}{m_2^3}\\ &=\frac{(-12.2279)^2}{(8.713)^3}\\ &=\frac{149.5215}{661.4593}\\ &=0.226 \end{aligned} $$
The coefficient of skewness based on moments ($\gamma_1$) is
$$ \begin{aligned} \gamma_1 &=\frac{m_3}{m_2^{3/2}}\\ &=\frac{-12.2279}{(8.713)^{3/2}}\\ &=\frac{-12.2279}{25.7189}\\ &=-0.4754 \end{aligned} $$
As the value of $\gamma_1 < 0$, the data is $\text{negatively skewed}$.
Moment Coefficient of Skewness 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 skewness 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
$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{1231}{65}\\ &=18.9385 \end{aligned} $$
| $x_i$ | $f_i$ | $(x_i-xb)^2$ | $f_i(x_i-xb)^2$ | $(x_i-xb)^3$ | $f_i(x_i-xb)^3$ | |
|---|---|---|---|---|---|---|
| 3 | 5 | 211.2866 | 1056.433 | -3071.1983 | -15355.9915 | |
| 10 | 12 | 56.7868 | 681.4416 | -427.9281 | -5135.1372 | |
| 17 | 18 | 0.287 | 5.166 | -0.1537 | -2.7666 | |
| 24 | 20 | 41.7872 | 835.744 | 270.1248 | 5402.496 | |
| 31 | 10 | 181.2874 | 1812.874 | 2440.9076 | 24409.076 | |
| Total | 65 | 4391.6586 | 9317.6767 |
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 third central moment is
$$ \begin{aligned} m_3 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^3\\ &=\frac{9317.6767}{65}\\ &=143.3489 \end{aligned} $$
The coefficient of skewness based on moments ($\beta_1$) is
$$ \begin{aligned} \beta_1 &=\frac{m_3^2}{m_2^3}\\ &=\frac{(143.3489)^2}{(67.564)^3}\\ &=\frac{20548.9071}{308422.5047}\\ &=0.0666 \end{aligned} $$
The coefficient of skewness based on moments ($\gamma_1$) is
$$ \begin{aligned} \gamma_1 &=\frac{m_3}{m_2^{3/2}}\\ &=\frac{143.3489}{(67.564)^{3/2}}\\ &=\frac{143.3489}{555.358}\\ &=0.2581 \end{aligned} $$
As the value of $\gamma_1 > 0$, the data is $\text{positively skewed}$.
Moment Coefficient of Skewness 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 skewness 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
$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{1695}{45}\\ &=37.6667 \end{aligned} $$
| $x_i$ | $f_i$ | $(x_i-xb)^2$ | $f_i(x_i-xb)^2$ | $(x_i-xb)^3$ | $f_i(x_i-xb)^3$ | |
|---|---|---|---|---|---|---|
| 15 | 6 | 6.4298 | 38.5788 | -16.304 | -97.824 | |
| 25 | 8 | 55.7158 | 445.7264 | 415.8793 | 3327.0344 | |
| 35 | 12 | 305.0018 | 3660.0216 | 5326.6425 | 63919.71 | |
| 45 | 10 | 754.2878 | 7542.878 | 20715.9857 | 207159.857 | |
| 55 | 5 | 1403.5738 | 7017.869 | 52583.909 | 262919.545 | |
| 65 | 4 | 2252.8598 | 9011.4392 | 106930.4122 | 427721.6488 | |
| Total | 45 | 27716.513 | 964949.9712 |
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 third central moment is
$$ \begin{aligned} m_3 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^3\\ &=\frac{964949.9712}{45}\\ &=21443.3327 \end{aligned} $$
The coefficient of skewness based on moments ($\beta_1$) is
$$ \begin{aligned} \beta_1 &=\frac{m_3^2}{m_2^3}\\ &=\frac{(21443.3327)^2}{(615.9225)^3}\\ &=\frac{459816517.2829}{233656683.5791}\\ &=1.9679 \end{aligned} $$
The coefficient of skewness based on moments ($\gamma_1$) is
$$ \begin{aligned} \gamma_1 &=\frac{m_3}{m_2^{3/2}}\\ &=\frac{21443.3327}{(615.9225)^{3/2}}\\ &=\frac{21443.3327}{15285.8328}\\ &=1.4028 \end{aligned} $$
As the value of $\gamma_1 > 0$, the data is $\text{positively skewed}$.
Moment Coefficient of Skewness 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 skewness 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
$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{665.5}{60}\\ &=11.0917 \end{aligned} $$
| $x_i$ | $f_i$ | $(x_i-xb)^2$ | $f_i(x_i-xb)^2$ | $(x_i-xb)^3$ | $f_i(x_i-xb)^3$ | |
|---|---|---|---|---|---|---|
| 9.5 | 2 | 64.5725 | 129.145 | -518.885 | -1037.77 | |
| 10 | 5 | 56.7868 | 283.934 | -427.9281 | -2139.6405 | |
| 10.5 | 12 | 49.5011 | 594.0132 | -348.2747 | -4179.2964 | |
| 11 | 17 | 42.7154 | 726.1618 | -279.1749 | -4745.9733 | |
| 11.5 | 14 | 36.4297 | 510.0158 | -219.8786 | -3078.3004 | |
| 12 | 6 | 30.644 | 183.864 | -169.6358 | -1017.8148 | |
| 12.5 | 3 | 25.3583 | 76.0749 | -127.6967 | -383.0901 | |
| 13 | 1 | 20.5726 | 20.5726 | -93.311 | -93.311 | |
| Total | 60 | 2523.7813 | -16675.1965 |
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 third central moment is
$$ \begin{aligned} m_3 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^3\\ &=\frac{-16675.1965}{60}\\ &=-277.9199 \end{aligned} $$
The coefficient of skewness based on moments ($\beta_1$) is
$$ \begin{aligned} \beta_1 &=\frac{m_3^2}{m_2^3}\\ &=\frac{(-277.9199)^2}{(42.063)^3}\\ &=\frac{77239.4708}{74421.8963}\\ &=1.0379 \end{aligned} $$
The coefficient of skewness based on moments ($\gamma_1$) is
$$ \begin{aligned} \gamma_1 &=\frac{m_3}{m_2^{3/2}}\\ &=\frac{-277.9199}{(42.063)^{3/2}}\\ &=\frac{-277.9199}{272.8038}\\ &=-1.0188 \end{aligned} $$
As the value of $\gamma_1 < 0$, the data is $\text{negatively skewed}$.
Conclusion
In this tutorial, you learned about how to calculate moment coefficient of skewness. You also learned about how to solve numerical problems based on moment coefficient of skewness 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 Skewness calculator for grouped data with examples.
