Home » Statistics » What is Karl Pearson coefficient of skewness Calculator | formula | Example for ungrouped data

# What is Karl Pearson coefficient of skewness Calculator | formula | Example for ungrouped data

## Karl Pearson coefficient of skewness for ungrouped data

Let $x_i, i=1,2, \cdots , n$ be $n$ observations.

The Karl Pearson’s coefficient skewness is given by

$S_k =\dfrac{Mean-Mode}{sd}=\dfrac{\overline{x}-Mode}{s_x}$

OR

$S_k =\dfrac{3(Mean-Median)}{sd}=\dfrac{3(\overline{x}-M)}{s_x}$

where,

• $\overline{x}$ is the sample mean of the data,
• $M$ is the median of the data,
• $Mode$ is the mode of the data,
• $s_x$ is the sample standard deviation of the data.

Sample mean

The sample mean $\overline{x}$ is given by

 $$\begin{eqnarray*} \overline{x}& =\frac{1}{n}\sum_{i=1}^{n}x_i \end{eqnarray*}$$

Sample Mode

Sample mode is the value of data which occurs maximum number of times, i.e., most frequent value of the data.

Sample Median

Arrange the data in ascending order of magnitude.

Median of $X$ is given by

 $$\begin{equation*} Md= \left\{ \begin{array}{ll} \text{value of }\big(\frac{n+1}{2}\big)^{th}\text{ obs.}, & \hbox{if n is odd;} \\ \text{average of }\big(\frac{n}{2}\big)^{th}\text{ and }\big(\frac{n}{2}+1\big)^{th} \text{ obs.}, & \hbox{if n is even.} \end{array} \right. \end{equation*}$$

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}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)} \end{aligned}

## Karl Pearson’s Coefficient of Skewness Calculator for ungrouped data

Use this calculator to find the Karl Pearson’s coefficient of Skewness for ungrouped (raw) data.

Pearson’s Coeff. of Skewness Calculator
Enter the X Values (Separated by comma,)
Results
Number of Obs. (n):
Ascending order of X values :
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 ungrouped data?

Step 1 – Enter the $x$ values separated by commas

Step 2 – Click on "Calculate" button to get Pearson’s coefficient of skewness for ungrouped data

Step 3 – Gives the output as number of observations $n$

Step 4 – Display the data in ascending order

Step 5 – Calculate the sample mean, sample median and sample standard deviation

Step 6 – Calculate the Pearson’s coefficient of skewness

## Karl Pearson’s Coefficient of Skewness Example 1

The age (in years) of 6 randomly selected students from a class are

22,25,24,23,24,20.

Find the Karl Pearson’s coefficient of skewness.

#### Solution

$x_i$ $x_i^2$
22 484
25 625
24 576
23 529
24 576
20 400
Total 138 3190

Sample mean

The sample mean of $X$ is

 \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{138}{6}\\ &=23\text{ years} \end{aligned}

The average of age of students is $23$ years.

Sample Median
The data in ascending order of magnitude is $20, 22, 23, 24, 24, 25$.

Here $n = 6$ which is even.

Sample median = average of $(\frac{n}{2})^{th}$ and $(\frac{n}{2}+1)^{th}$ observations.

Thus the median age of students is

 \begin{aligned} M &= \frac{\big(\frac{6}{2}\big)^{th}\text{Obs.} +\big(\frac{6}{2}+1\big)^{th}\text{Obs.}}{2}\\ &= \frac{\big(3\big)^{th}\text{Obs.} +\big(4\big)^{th}\text{Obs.}}{2}\\ &=\frac{23 +24}{2} \\ &= 23.5 \text{ years}. \end{aligned}

The median age of students is $M=23.5$ years.

Sample variance

Sample variance of $X$ is

 \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{5}\bigg(3190-\frac{(138)^2}{6}\bigg)\\ &=\dfrac{1}{5}\big(3190-\frac{19044}{6}\big)\\ &=\dfrac{1}{5}\big(3190-3174\big)\\ &= \frac{16}{5}\\ &=3.2 \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{3.2}\\ &=1.7889 \text{ years} \end{aligned}

Thus the standard deviation of age of students is $1.7889$ years.

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(23-23.5)}{1.7889}\\ &= -0.8385 \end{aligned}

As the value of $s_k < 0$, the data is $\text{negatively skewed}$.

## Karl Pearson’s Coefficient of Skewness Example 2

A random sample of 11 patients yielded the following data on the length of stay (in days) in the hospital.

12,9,10,15,10,14,7,10,8,11,15

Find the Karl Pearson’s coefficient of skewness.

#### Solution

$x_i$ $x_i^2$
12 144
9 81
10 100
15 225
10 100
14 196
7 49
10 100
8 64
11 121
15 225
Total 121 1405

Sample mean

The sample mean of $X$ is

 \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{121}{11}\\ &=11\text{ days} \end{aligned}

The average of length of stay in the hospital is $11$ days.

Sample Median

$n = 11$ which is odd.

The data in ascending order of magnitude is $7, 8, 9, 10, 10, 10, 11, 12, 14, 15, 15$.

Sample median = value of $(\frac{n+1}{2})^{th}$ observation.

That is

 \begin{aligned} M &= \text{value of }\bigg(\frac{n+1}{2}\bigg)^{th}\text{ obs.}\\ &= \text{value of }\bigg(\frac{11+1}{2}\bigg)^{th}\text{ obs.}\\ &= \text{value of } \big(6\big)^{th}\text{Obs.}\\ &=10 \text{ days} \end{aligned}
The median length of stay in the hospital is $M=10$ days.

Sample variance

Sample variance of $X$ is

 \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{10}\bigg(1405-\frac{(121)^2}{11}\bigg)\\ &=\dfrac{1}{10}\big(1405-\frac{14641}{11}\big)\\ &=\dfrac{1}{10}\big(1405-1331\big)\\ &= \frac{74}{10}\\ &=7.4 \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{7.4}\\ &=2.7203 \text{ days} \end{aligned}

Thus the standard deviation of length of stay in the hospital is $2.7203$ days.

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(11-10)}{2.7203}\\ &= 1.1028 \end{aligned}

As the value of $s_k > 0$, the data is $\text{positively skewed}$.

## Karl Pearson’s Coefficient of Skewness Example 3

The systolic blood pressure (in mmHg) of 10 randomly selected patients are :

123, 128, 136, 112, 143, 114, 104, 137, 145, 150.

Compute Pearson’s coefficient of skewness

#### Solution

$x_i$ $x_i^2$
123 15129
128 16384
136 18496
112 12544
143 20449
114 12996
104 10816
137 18769
145 21025
150 22500
Total 1292 169108

Sample mean

The sample mean of $X$ is

 \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{1292}{10}\\ &=129.2\text{ mmHg} \end{aligned}

The average of systolic blood pressure is $129.2$ mmHg.

Sample Median

$n = 10$ which is even.

The data in ascending order of magnitude is $104, 112, 114, 123, 128, 136, 137, 143, 145, 150$.

Sample median = average of $(\frac{n}{2})^{th}$ and $(\frac{n}{2}+1)^{th}$ observations.

That is

 \begin{aligned} M &= \text{Average of }\bigg(\frac{n}{2}\bigg)^{th}\text{ and } \bigg(\frac{n}{2}+1\bigg)^{th}\text{ obs.}\\ &= \text{Average of }\bigg(\frac{10}{2}\bigg)^{th}\text{ and } \bigg(\frac{10}{2}+1\bigg)^{th}\text{ obs.}\\ &= \text{Average of } \big(5\big)^{th}\text{ and }\big(6\big)^{th}\text{Obs.}\\ &=132 \text{ mmHg} \end{aligned}
The median systolic blood pressure is $M=132$ mmHg.

Sample variance

Sample variance of $X$ is

 \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{9}\bigg(169108-\frac{(1292)^2}{10}\bigg)\\ &=\dfrac{1}{9}\big(169108-\frac{1669264}{10}\big)\\ &=\dfrac{1}{9}\big(169108-166926.4\big)\\ &= \frac{2181.6}{9}\\ &=242.4 \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{242.4}\\ &=15.5692 \text{ mmHg} \end{aligned}

Thus the standard deviation of systolic blood pressure is $15.5692$ mmHg.

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(129.2-132)}{15.5692}\\ &= -0.5395 \end{aligned}

As the value of $s_k < 0$, the data is $\text{negatively skewed}$.

## Karl Pearson’s Coefficient of Skewness Example 4

Blood sugar level (in mg/dl) of a sample of 20 patients admitted to the hospitals are as follows:

75, 80, 72, 78, 82, 85, 73, 75, 97, 87,
84, 76, 73, 79, 99, 86, 83, 76, 78, 73.

Compute Pearson’s coefficient of skewness

#### Solution

$x_i$ $x_i^2$
75 5625
80 6400
72 5184
78 6084
82 6724
85 7225
73 5329
75 5625
97 9409
87 7569
84 7056
76 5776
73 5329
79 6241
99 9801
86 7396
83 6889
76 5776
78 6084
73 5329
Total 1611 130851

Sample mean

The sample mean of $X$ is

 \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{1611}{20}\\ &=80.55\text{ mg/dl} \end{aligned}

The average of blood sugar level is $80.55$ mg/dl.

Sample Median

$n = 20$ which is even.

The data in ascending order of magnitude is $72, 73, 73, 73, 75, 75, 76, 76, 78, 78, 79, 80, 82, 83, 84, 85, 86, 87, 97, 99$.

Sample median = average of $(\frac{n}{2})^{th}$ and $(\frac{n}{2}+1)^{th}$ observations.

That is

 \begin{aligned} M &= \text{Average of }\bigg(\frac{n}{2}\bigg)^{th}\text{ and } \bigg(\frac{n}{2}+1\bigg)^{th}\text{ obs.}\\ &= \text{Average of }\bigg(\frac{20}{2}\bigg)^{th}\text{ and } \bigg(\frac{20}{2}+1\bigg)^{th}\text{ obs.}\\ &= \text{Average of } \big(10\big)^{th}\text{ and }\big(11\big)^{th}\text{Obs.}\\ &=78.5 \text{ mg/dl} \end{aligned}

The median blood sugar level is $M=78.5$ mg/dl.

Sample variance

Sample variance of $X$ is

 \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{19}\bigg(130851-\frac{(1611)^2}{20}\bigg)\\ &=\dfrac{1}{19}\big(130851-\frac{2595321}{20}\big)\\ &=\dfrac{1}{19}\big(130851-129766.05\big)\\ &= \frac{1084.95}{19}\\ &=57.1026 \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{57.1026}\\ &=7.5566 \text{ mg/dl} \end{aligned}

Thus the standard deviation of blood sugar level is $7.5566$ mg/dl.

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(80.55-78.5)}{7.5566}\\ &= 0.8139 \end{aligned}

As the value of $s_k > 0$, the data is $\text{positively skewed}$.

## Karl Pearson’s Coefficient of Skewness Example 5

Diastolic blood pressure (in mmHg) of a sample of 18 patients admitted to the hospitals are as follows:

65, 76, 64, 73, 74, 80, 71, 68, 66,
81, 79, 75, 70, 62, 83, 63, 77, 78.

Compute Pearson’s coefficient of skewness.

#### Solution

$x_i$ $x_i^2$
65 4225
76 5776
64 4096
73 5329
74 5476
80 6400
71 5041
68 4624
66 4356
81 6561
79 6241
75 5625
70 4900
62 3844
83 6889
63 3969
77 5929
78 6084
Total 1305 95365

Sample mean

The sample mean of $X$ is

 \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{1305}{18}\\ &=72.5\text{ mmHg} \end{aligned}

The average of diastolic blood pressure is $72.5$ mmHg.

Sample Median

$n = 18$ which is even.

The data in ascending order of magnitude is $62, 63, 64, 65, 66, 68, 70, 71, 73, 74, 75, 76, 77, 78, 79, 80, 81, 83$.

Sample median = average of $(\frac{n}{2})^{th}$ and $(\frac{n}{2}+1)^{th}$ observations.

That is

 \begin{aligned} M &= \text{Average of }\bigg(\frac{n}{2}\bigg)^{th}\text{ and } \bigg(\frac{n}{2}+1\bigg)^{th}\text{ obs.}\\ &= \text{Average of }\bigg(\frac{18}{2}\bigg)^{th}\text{ and } \bigg(\frac{18}{2}+1\bigg)^{th}\text{ obs.}\\ &= \text{Average of } \big(9\big)^{th}\text{ and }\big(10\big)^{th}\text{Obs.}\\ &=73.5 \text{ mmHg} \end{aligned}

The median diastolic blood pressure is $M=73.5$ mmHg.

Sample variance

Sample variance of $X$ is

 \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{17}\bigg(95365-\frac{(1305)^2}{18}\bigg)\\ &=\dfrac{1}{17}\big(95365-\frac{1703025}{18}\big)\\ &=\dfrac{1}{17}\big(95365-94612.5\big)\\ &= \frac{752.5}{17}\\ &=44.2647 \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{44.2647}\\ &=6.6532 \text{ mmHg} \end{aligned}

Thus the standard deviation of diastolic blood pressure is $6.6532$ mmHg.

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(72.5-73.5)}{6.6532}\\ &= -0.4509 \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 ungrouped data and how to calculate Pearson’s coefficient of skewness for ungrouped data. You also learned about how to solve numerical problems based on Pearson’s coefficient of skewness for ungrouped 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

Let me know in the comments if you have any questions on Pearson’s coefficient of skewness calculator for ungrouped data with examples and your thought on this article.