Karl Pearson coefficient of skewness for ungrouped data
- 1 Karl Pearson coefficient of skewness for ungrouped data
- 2 Karl Pearson's Coefficient of Skewness Calculator for ungrouped data
- 3 How to calculate Pearson's coefficient of skewness for ungrouped data?
- 4 Karl Pearson's Coefficient of Skewness Example 1
- 5 Karl Pearson's Coefficient of Skewness Example 2
- 6 Karl Pearson's Coefficient of Skewness Example 3
- 7 Karl Pearson's Coefficient of Skewness Example 4
- 8 Karl Pearson's Coefficient of Skewness Example 5
- 9 Conclusion
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 = average of $(\frac{n}{2})^{th}$ and $(\frac{n}{2}+1)^{th}$ observations.
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 $7, 8, 9, 10, 10, 10, 11, 12, 14, 15, 15$.
Sample median = average of $(\frac{n}{2})^{th}$ and $(\frac{n}{2}+1)^{th}$ observations.
That is
$$ \begin{aligned} M &= \text{value of }\bigg(\frac{n+1}{2}\bigg)^{th}\text{ obs.}\\ &= \text{value of }\bigg(\frac{10+1}{2}\bigg)^{th}\text{ obs.}\\ &= \text{value of } \big(5.5\big)^{th}\text{Obs.}\\ &=10 \text{ mmHg} \end{aligned} $$
The median systolic blood pressure is $M=10$ 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
$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{242.4}\\ &=2.7203 \text{ mmHg} \end{aligned} $$
Thus the standard deviation of systolic blood pressure is $2.7203$ mmHg.
Karl Pearson's coefficient of skewness
The Karl Pearson's coefficient skewness is
$$ \begin{aligned} s_k &=\frac{3(Mean-Median)}{sd}\\ &=\frac{3\times(129.2-10)}{15.5692}\\ &= 131.4561 \end{aligned} $$
As the value of $s_k > 0$, the data is $\text{positively skewed}$.
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:
75,89,72,78,87, 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 $7, 8, 9, 10, 10, 10, 11, 12, 14, 15, 15$.
Sample median = average of $(\frac{n}{2})^{th}$ and $(\frac{n}{2}+1)^{th}$ observations.
That is
$$ \begin{aligned} M &= \text{value of }\bigg(\frac{n+1}{2}\bigg)^{th}\text{ obs.}\\ &= \text{value of }\bigg(\frac{20+1}{2}\bigg)^{th}\text{ obs.}\\ &= \text{value of } \big(10.5\big)^{th}\text{Obs.}\\ &=15 \text{ mg/dl} \end{aligned} $$
The median blood sugar level is $M=10$ 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
$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{57.1026}\\ &=2.7203 \text{ mg/dl} \end{aligned} $$
Thus the standard deviation of blood sugar level is $2.7203$ mg/dl.
Karl Pearson's coefficient of skewness
The Karl Pearson's coefficient skewness is
$$ \begin{aligned} s_k &=\frac{3(Mean-Median)}{sd}\\ &=\frac{3\times(80.55-10)}{7.5566}\\ &= 77.8039 \end{aligned} $$
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 $7, 8, 9, 10, 10, 10, 11, 12, 14, 15, 15$.
Sample median = average of $(\frac{n}{2})^{th}$ and $(\frac{n}{2}+1)^{th}$ observations.
That is
$$ \begin{aligned} M &= \text{value of }\bigg(\frac{n+1}{2}\bigg)^{th}\text{ obs.}\\ &= \text{value of }\bigg(\frac{18+1}{2}\bigg)^{th}\text{ obs.}\\ &= \text{value of } \big(9.5\big)^{th}\text{Obs.}\\ &=14 \text{ mmHg} \end{aligned} $$
The median diastolic blood pressure is $M=10$ 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
$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{44.2647}\\ &=2.7203 \text{ mmHg} \end{aligned} $$
Thus the standard deviation of diastolic blood pressure is $2.7203$ mmHg.
Karl Pearson's coefficient of skewness
The Karl Pearson's coefficient skewness is
$$ \begin{aligned} s_k &=\frac{3(Mean-Median)}{sd}\\ &=\frac{3\times(72.5-10)}{6.6532}\\ &= 68.9262 \end{aligned} $$
As the value of $s_k > 0$, the data is $\text{positively skewed}$.
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:
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.