Karl Pearson coefficient of skewness for grouped data
Let $(x_i,f_i), i=1,2, \cdots , n$ be given frequency distribution.
The Karl Pearson’s coefficient skewness is given by
$S_k =\dfrac{Mean-Mode}{sd}=\dfrac{\overline{x}-\text{Mode}}{s_x}$
OR
$S_k =\dfrac{3(Mean-Median)}{sd}=\dfrac{3(\overline{x}-M)}{s_x}$
where,
- $\overline{x}$ is the sample mean,
- $M$ is the median,
- $s_x$ is the sample standard deviation.
Sample mean
The sample mean $\overline{x}$ is given by
$$ \begin{eqnarray*} \overline{x}& =\frac{1}{N}\sum_{i=1}^{n}f_ix_i \end{eqnarray*} $$
Sample median
The median is given by
$\text{Median } = l + \bigg(\dfrac{\frac{N}{2} - F_<}{f}\bigg)\times h$
where,
- $N$, total number of observations
- $l$, the lower limit of the median class
- $f$, frequency of the median class
- $F_<$, cumulative frequency of the pre median class
- $h$, the class width
Sample mode
The mode of the distribution is given by
$\text{Mode } = l + \bigg(\dfrac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h$
where,
- $l$, the lower limit of the modal class
- $f_m$, frequency of the modal class
- $f_1$, frequency of the class pre-modal class
- $f_2$, frequency of the class post-modal class
- $h$, the class width
Sample Standard deviation
Sample standard deviation is given by
$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{\dfrac{1}{N-1}\bigg(\sum_{i=1}^{n}f_ix_i^2-\frac{\big(\sum_{i=1}^n f_ix_i\big)^2}{N}\bigg)} \end{aligned} $$
Karl Pearson’s Coefficient of Skewness Calculator for grouped data
Use this calculator to find the Karl Pearson’s coefficient of Skewness for grouped (raw) data.
| Pearson’s Coefficient of Skewness Calculator | |
|---|---|
| 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): | |
| Sample Mean : ($\overline{x}$) | |
| Sample Median : | |
| Sample std. deviation :($s_x$) | |
| Karl Pearson’s Coeff. of Skewness : | |
How to calculate Pearson’s 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" for Pearson’s coefficient of skewness calculation
Step 5 – Calculate sample mean, sample median and standard deviation
Step 6 – Calculate Pearson’s coefficient of skewness
Example 1 – Karl Pearson’s Coefficient of Skewness
The number of students absent in a class was recorded every day for 60 days and the information is given in the following frequency distribution.
| No.of Students absent (x) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| No.of days (f) | 3 | 6 | 18 | 18 | 8 | 5 | 2 |
Find the Karl Pearson’s coefficient of skewness.
Solution
| $x_i$ | $f_i$ | $f_i*x_i$ | $f_i*x_i^2$ | $cf$ | |
|---|---|---|---|---|---|
| 0 | 3 | 0 | 0 | 3 | |
| 1 | 6 | 6 | 6 | 9 | |
| 2 | 18 | 36 | 72 | 27 | |
| 3 | 18 | 54 | 162 | 45 | |
| 4 | 8 | 32 | 128 | 53 | |
| 5 | 5 | 25 | 125 | 58 | |
| 6 | 2 | 12 | 72 | 60 | |
| Total | 60 | 165 | 565 |
Sample mean
The sample mean of $X$ is
$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{165}{60}\\ &=2.75 \end{aligned} $$
The average of no. of students absent is $2.75$ students.
Since the given frequency distribution is bimodal, we use empirical formula to calculate Karl Pearson’s coefficient of skewness.
For asymmetric distribution,
$$ \begin{aligned} \text{Mean} - \text{Mode} &= 3(\text{Mean} - \text{Median}) \end{aligned} $$
Thus, Karl Pearson’s coefficient of skewness is given by
$$ \begin{aligned} S_k &=\dfrac{3(Mean-Median)}{sd}\\ &=\dfrac{\overline{x}-M}{s_x} \end{aligned} $$
Sample Median
Median no. of students absent is
$$ \begin{aligned} \text{Median} &=\bigg(\dfrac{N}{2}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{60}{2}\bigg)^{th}\text{ value}\\ &=\big(30\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $30$ is $45$. The corresponding value of $x$ is median. That is, $M =3$.
Thus, median number of accidents $M$ = $3$.
Sample variance
Sample variance of $X$ is
$$ \begin{aligned} s_x^2 &=\dfrac{1}{N-1}\bigg(\sum_{i=1}^{n}f_ix_i^2-\frac{\big(\sum_{i=1}^n f_ix_i\big)^2}{N}\bigg)\\ &=\dfrac{1}{59}\bigg(565-\frac{(165)^2}{60}\bigg)\\ &=\dfrac{1}{59}\big(565-\frac{27225}{60}\big)\\ &=\dfrac{1}{59}\big(565-453.75\big)\\ &= \frac{111.25}{59}\\ &=1.8856 \end{aligned} $$
Sample standard deviation
The standard deviation is the positive square root of the variance.
The sample standard deviation is
$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{1.8856}\\ &=1.3732 \end{aligned} $$
Thus the standard deviation of no. of students absent is $1.3732$ students.
Karl Pearson’s coefficient of skewness
The Karl Pearson’s coefficient skewness is
$$ \begin{aligned} s_k &=\frac{3(Mean-Median)}{sd}\\ &=\frac{3\times(2.75-3)}{2.1602}\\ &= -0.5462 \end{aligned} $$
As the value of $s_k < 0$, the data is $\text{negatively skewed}$.
Example 2 – Using Karl Pearson’s Coefficient of Skewness Method
The following table gives the distribution of weight (in pounds) of 100 newborn babies at certain hospital in 2012.
| Weight (in pounds) | 3-5 | 5-7 | 7-9 | 9-11 | 11-13 |
|---|---|---|---|---|---|
| No.of babies | 10 | 30 | 28 | 18 | 14 |
Compute Karl Pearson’s coefficient of skewness.
Solution
| Class Interval | mid-value ($x$) | $f$ | $f*x$ | $f*x^2$ | |
|---|---|---|---|---|---|
| 3-5 | 4 | 10 | 40 | 160 | |
| 5-7 | 6 | 30 | 180 | 1080 | |
| 7-9 | 8 | 28 | 224 | 1792 | |
| 9-11 | 10 | 18 | 180 | 1800 | |
| 11-13 | 12 | 14 | 168 | 2016 | |
| Total | 100 | 792 | 6848 |
Mean
The mean weight of babies is
$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{792}{100}\\ &=7.92 \text{ pounds} \end{aligned} $$
Sample Mode
The maximum frequency is $30$, the corresponding class $5-7$ is the modal class.
Mode of the given frequency distribution is:
$$ \begin{aligned} \text{Mode } &= l + \bigg(\frac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h\\ \end{aligned} $$
where,
- $l = 5$, the lower limit of the modal class
- $f_m =30$, frequency of the modal class
- $f_1 = 10$, frequency of the pre-modal class
- $f_2 = 28$, frequency of the post-modal class
- $h =2$, the class width
Thus mode of a frequency distribution is
$$ \begin{aligned} \text{Mode } &= l + \bigg(\frac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h\\ &= 5 + \bigg(\frac{30 - 10}{2\times30 - 10 - 28}\bigg)\times 2\\ &= 5 + \bigg(\frac{20}{22}\bigg)\times 2\\ &= 5 + \big(0.9091\big)\times 2\\ &= 5 + \big(1.8182\big)\\ &= 6.8182 \text{ pounds} \end{aligned} $$
Sample variance
Sample variance of $X$ is
$$ \begin{aligned} s_x^2 &=\dfrac{1}{N-1}\bigg(\sum_{i=1}^{n}f_ix_i^2-\frac{\big(\sum_{i=1}^n f_ix_i\big)^2}{N}\bigg)\\ &=\dfrac{1}{99}\bigg(6848-\frac{(792)^2}{100}\bigg)\\ &=\dfrac{1}{99}\big(6848-\frac{627264}{100}\big)\\ &=\dfrac{1}{99}\big(6848-6272.64\big)\\ &= \frac{575.36}{99}\\ &=5.8117 \end{aligned} $$
Sample standard deviation
The sample standard deviation is
$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{5.8117}\\ &=2.4107 \text{ pounds} \end{aligned} $$
Thus the standard deviation of weight of babies is $2.4107$ pounds.
Karl Pearson’s coefficient of skewness
The Karl Pearson’s coefficient skewness is
$$ \begin{aligned} s_k &=\frac{Mean-\text{Mode}}{sd}\\ &=\frac{7.92-6.8182}{3.1623}\\ &= 0.457 \end{aligned} $$
As the value of $s_k > 0$, the data is $\text{positively skewed}$.
Example 3 – Karl Pearson’s Coefficient of Skewness
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 the Karl Pearson’s coefficient of skewness and interpret.
Solution
| Class Interval | mid-value ($x$) | $f$ | $f*x$ | $f*x^2$ | |
|---|---|---|---|---|---|
| 9.25-9.75 | 9.5 | 2 | 40 | 160 | |
| 9.75-10.25 | 10 | 5 | 180 | 1080 | |
| 10.25-10.75 | 10.5 | 12 | 224 | 1792 | |
| 10.75-11.25 | 11 | 17 | 180 | 1800 | |
| 11.25-11.75 | 11.5 | 14 | 168 | 2016 | |
| 11.75-12.25 | 12 | 6 | 40 | 160 | |
| 12.25-12.75 | 12.5 | 3 | 180 | 1080 | |
| 12.75-13.25 | 13 | 1 | 224 | 1792 | |
| Total | 60 | 792 | 6848 |
Mean
The mean maximum load is
$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{665.5}{60}\\ &=11.0917 \text{ tons} \end{aligned} $$
Sample Mode
The maximum frequency is $17$, the corresponding class $10.75-11.25$ is the modal class.
Mode of the given frequency distribution is:
$$ \begin{aligned} \text{Mode } &= l + \bigg(\frac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h\\ \end{aligned} $$
where,
- $l = 10.75$, the lower limit of the modal class
- $f_m =17$, frequency of the modal class
- $f_1 = 12$, frequency of the pre-modal class
- $f_2 = 14$, frequency of the post-modal class
- $h =0.5$, the class width
Thus mode of a frequency distribution is
$$ \begin{aligned} \text{Mode } &= l + \bigg(\frac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h\\ &= 10.75 + \bigg(\frac{17 - 12}{2\times17 - 12 - 14}\bigg)\times 0.5\\ &= 10.75 + \bigg(\frac{5}{8}\bigg)\times 0.5\\ &= 10.75 + \big(0.625\big)\times 0.5\\ &= 10.75 + \big(0.3125\big)\\ &= 11.0625 \text{ tons} \end{aligned} $$
Sample variance
Sample variance of $X$ is
$$ \begin{aligned} s_x^2 &=\dfrac{1}{N-1}\bigg(\sum_{i=1}^{n}f_ix_i^2-\frac{\big(\sum_{i=1}^n f_ix_i\big)^2}{N}\bigg)\\ &=\dfrac{1}{59}\bigg(7413.75-\frac{(665.5)^2}{60}\bigg)\\ &=\dfrac{1}{59}\big(7413.75-\frac{442890.25}{60}\big)\\ &=\dfrac{1}{59}\big(7413.75-7381.50417\big)\\ &= \frac{32.24583}{59}\\ &=0.5465 \end{aligned} $$
Sample standard deviation
The sample standard deviation is
$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{5.8117}\\ &=2.4107 \text{ tons} \end{aligned} $$
Thus the standard deviation of maximum load is $2.4107$ tons.
Karl Pearson’s coefficient of skewness
The Karl Pearson’s coefficient skewness is
$$ \begin{aligned} s_k &=\frac{Mean-\text{Mode}}{sd}\\ &=\frac{7.92-11.0625}{1.2247}\\ &= -1.3036 \end{aligned} $$
As the value of $s_k < 0$, the data is $\text{negatively skewed}$.
Example 4 – Karl Pearson’s Coefficient of Skewness Calculator
Following table shows the weight of 100 pumpkin produced from a farm :
| Weight (’00 grams) | Frequency |
|---|---|
| $4 \leq x < 6$ | 4 |
| $6 \leq x < 8$ | 14 |
| $8 \leq x < 10$ | 34 |
| $10 \leq x < 12$ | 28 |
| $12 \leq x < 14$ | 20 |
Compute the Karl Pearson’s coefficient of skewness and interpret.
Solution
| Class Interval | mid-value ($x$) | $f$ | $f*x$ | $f*x^2$ | |
|---|---|---|---|---|---|
| 4-6 | 5 | 4 | 40 | 160 | |
| 6-8 | 7 | 14 | 180 | 1080 | |
| 8-10 | 9 | 34 | 224 | 1792 | |
| 10-12 | 11 | 28 | 180 | 1800 | |
| 12-14 | 13 | 20 | 168 | 2016 | |
| Total | 100 | 792 | 6848 |
Mean
The mean weight of pumpkin is
$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{992}{100}\\ &=9.92 \text{ ('00 grams)} \end{aligned} $$
Sample Mode
The maximum frequency is $34$, the corresponding class $8-10$ is the modal class.
Mode of the given frequency distribution is:
$$ \begin{aligned} \text{Mode } &= l + \bigg(\frac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h\\ \end{aligned} $$
where,
- $l = 8$, the lower limit of the modal class
- $f_m =34$, frequency of the modal class
- $f_1 = 14$, frequency of the pre-modal class
- $f_2 = 28$, frequency of the post-modal class
- $h =2$, the class width
Thus mode of a frequency distribution is
$$ \begin{aligned} \text{Mode } &= l + \bigg(\frac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h\\ &= 8 + \bigg(\frac{34 - 14}{2\times34 - 14 - 28}\bigg)\times 2\\ &= 8 + \bigg(\frac{20}{26}\bigg)\times 2\\ &= 8 + \big(0.7692\big)\times 2\\ &= 8 + \big(1.5385\big)\\ &= 9.5385 \text{ ('00 grams)} \end{aligned} $$
Sample variance
Sample variance of $X$ is
$$ \begin{aligned} s_x^2 &=\dfrac{1}{N-1}\bigg(\sum_{i=1}^{n}f_ix_i^2-\frac{\big(\sum_{i=1}^n f_ix_i\big)^2}{N}\bigg)\\ &=\dfrac{1}{99}\bigg(10308-\frac{(992)^2}{100}\bigg)\\ &=\dfrac{1}{99}\big(10308-\frac{984064}{100}\big)\\ &=\dfrac{1}{99}\big(10308-9840.64\big)\\ &= \frac{467.36}{99}\\ &=4.7208 \end{aligned} $$
Sample standard deviation
The sample standard deviation is
$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{5.8117}\\ &=2.4107 \text{ ('00 grams)} \end{aligned} $$
Thus the standard deviation of weight of pumpkin is $2.4107$ (’00 grams).
Karl Pearson’s coefficient of skewness
The Karl Pearson’s coefficient skewness is
$$ \begin{aligned} s_k &=\frac{Mean-\text{Mode}}{sd}\\ &=\frac{7.92-9.5385}{3.1623}\\ &= -0.6714 \end{aligned} $$
As the value of $s_k < 0$, the data is $\text{negatively skewed}$.
Example 5 – Karl Pearson’s Coefficient of Skewness Calculator
The following table gives the amount of time (in minutes) spent on the internet each evening by a group of 56 students. Compute quartile deviation for the following frequency distribution.
| Time spent on Internet ($x$) | No. of Students ($f$) |
|---|---|
| 10-12 | 3 |
| 13-15 | 12 |
| 16-18 | 15 |
| 19-21 | 24 |
| 22-24 | 2 |
Compute the Karl Pearson’s coefficient of skewness and interpret.
Solution
| Class Interval | mid-value ($x$) | $f$ | $f*x$ | $f*x^2$ | |
|---|---|---|---|---|---|
| 9.5-12.5 | 11 | 3 | 40 | 160 | |
| 12.5-15.5 | 14 | 12 | 180 | 1080 | |
| 15.5-18.5 | 17 | 15 | 224 | 1792 | |
| 18.5-21.5 | 20 | 24 | 180 | 1800 | |
| 21.5-24.5 | 23 | 2 | 168 | 2016 | |
| Total | 56 | 792 | 6848 |
Mean
The mean amount of time (in minutes) spent on the internet is
$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{982}{56}\\ &=17.5357 \text{ minutes} \end{aligned} $$
Sample Mode
The maximum frequency is $24$, the corresponding class $18.5-21.5$ is the modal class.
Mode of the given frequency distribution is:
$$ \begin{aligned} \text{Mode } &= l + \bigg(\frac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h\\ \end{aligned} $$
where,
- $l = 18.5$, the lower limit of the modal class
- $f_m =24$, frequency of the modal class
- $f_1 = 15$, frequency of the pre-modal class
- $f_2 = 2$, frequency of the post-modal class
- $h =3$, the class width
Thus mode of a frequency distribution is
$$ \begin{aligned} \text{Mode } &= l + \bigg(\frac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h\\ &= 18.5 + \bigg(\frac{24 - 15}{2\times24 - 15 - 2}\bigg)\times 3\\ &= 18.5 + \bigg(\frac{9}{31}\bigg)\times 3\\ &= 18.5 + \big(0.2903\big)\times 3\\ &= 18.5 + \big(0.871\big)\\ &= 19.371 \text{ minutes} \end{aligned} $$
Sample variance
Sample variance of $X$ is
$$ \begin{aligned} s_x^2 &=\dfrac{1}{N-1}\bigg(\sum_{i=1}^{n}f_ix_i^2-\frac{\big(\sum_{i=1}^n f_ix_i\big)^2}{N}\bigg)\\ &=\dfrac{1}{55}\bigg(17708-\frac{(982)^2}{56}\bigg)\\ &=\dfrac{1}{55}\big(17708-\frac{964324}{56}\big)\\ &=\dfrac{1}{55}\big(17708-17220.07143\big)\\ &= \frac{487.92857}{55}\\ &=8.8714 \end{aligned} $$
Sample standard deviation
The sample standard deviation is
$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{5.8117}\\ &=2.4107 \text{ minutes} \end{aligned} $$
Thus the standard deviation of amount of time (in minutes) spent on the internet is $2.4107$ minutes.
Karl Pearson’s coefficient of skewness
The Karl Pearson’s coefficient skewness is
$$ \begin{aligned} s_k &=\frac{Mean-\text{Mode}}{sd}\\ &=\frac{7.92-19.371}{4.7434}\\ &= -4.7501 \end{aligned} $$
As the value of $s_k < 0$, the data is $\text{negatively skewed}$.
Conclusion
In this tutorial, you learned about formula for Pearson’s coefficient of skewness for grouped data and how to calculate Pearson’s coefficient of skewness for grouped data. You also learned about how to solve numerical problems based on Pearson’s coefficient of skewness for grouped data.
Compute the Karl Pearson’s coefficient of skewness and interpret.To learn more about other descriptive statistics measures, please refer to the following tutorials:
Descriptive Statistics
Empirical Rule Calculator
Let me know in the comments if you have any questions on Pearson’s coefficient of skewness calculator for grouped data with examples and your thought on this article.
