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# Karl Pearson coefficient of skewness Calculator

## 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:

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

 \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.