# Moment coefficient of skewness Calculator for grouped data

## 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

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:

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.