Karl Pearson coefficient of skewness for grouped data
Let $(x_i,f_i), i=1,2, \cdots , n$ be given frequency distribution.
Karl Pearson coefficient of skewness formula
The Karl Pearson’s coefficient skewness for grouped data is given by
$S_k =\dfrac{Mean-Mode)}{sd}=\dfrac{\overline{x}-\text{Mode}}{s_x}$
OR
$S_k =\dfrac{3(Mean-Median)}{sd}=\dfrac{\overline{x}-M}{s_x}$
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 coefficient of skewness formula with Example 1
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 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,Karl Pearson’s coefficient of skewness can be calculated by using empirical formula.
For asymmetric distribution,
$$ \begin{aligned} \text{Mean} - \text{Mode} &= 3(\text{Mean} - \text{Median}) \end{aligned} $$
Thus, Karl Pearson’s coefficient of skewness can be calculated 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 coefficient of skewness can be calculated by
$$ \begin{aligned} s_k &=\frac{3(Mean-Median)}{sd}\\ &=\frac{3\times(2.75-3)}{2.1602}\\ &= -0.5462 \end{aligned} $$
As the value of $s_k < 0$, the data is $\text{negatively skewed}$.
Karl Pearson coefficient of skewness formula with Example 2
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 |
Calculate Karl Pearson 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 coefficient of skewness can be calculated by
$$ \begin{aligned} s_k &=\frac{Mean-\text{Mode}}{sd}\\ &=\frac{7.92-6.8182}{3.1623}\\ &= 0.457 \end{aligned} $$
As the value of $s_k > 0$, the data is $\text{positively skewed}$.
Hope you like Karl Pearson coefficient of skewness for grouped data and step by step explanation about how to find Karl Pearson coefficient of skewness with examples.
Calculate Pearson coefficient of skewness for grouped data using Calculator link given below under resource section.
You can also refer Karl Pearson coefficient of skewness formula using formula link given below under resource section.
