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# Bowley’s Coefficient of Skewness Calculator for Ungrouped data

## Bowley’s Coefficient of Skewness for Ungrouped data

Skewness is a measure of symmetry. The meaning of skewness is "lack of symmetry". Skewness gives us an idea about the concentration of higher or lower data values around the central value of the data.

For a symmetric distribution, the two quartiles namely $Q_1$ and $Q_3$ are equidistant from the median (i.e. $Q_2$). That is for symmetric distribution $Q_3 – Q_2 = Q_2 -Q_1$.

If the distribution is not symmetric (i.e., skewed) then the distance $Q_3-Q_2$ is not equal to the distance $Q_2-Q_1$. That is for asymmetric distribution $Q_3-Q_2\neq Q_2-Q1$.

The absolute measure of skewness is $(Q_3-Q2)-(Q_2-Q1)= Q_3+Q_1-2*Q2$.

Bowley’s coefficient of skewness is the relative measure of skewness. It is denoted by $S_b$ and is defined as

$$S_b = \dfrac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1}$$

where,

• $Q_1$ is the first quartile,
• $Q_2$ is the second quartile,
• $Q_3$ is the third quartile,

## Types of Skewness

• If $S_b < 0$, i.e., $Q_3-Q_2 < Q_2-Q1$ then the distribution is negatively skewed.
• If $S_b = 0$, i.e., $Q_3-Q_2 = Q_2-Q1$ then the distribution is Symmetric or not skewed.
• If $S_b > 0$, i.e., $Q_3-Q_2 > Q_2-Q1$ then the distribution is positively skewed.

## Bowley’s Coefficient of Skewness Calculator for ungrouped data

Use this calculator to find the Bowley’s Coefficient of Skewness for ungrouped (raw) data.

Bowley’s Coeff. of Skewness Calculator
Enter the X Values (Separated by comma,)
Results
Number of Obs. (n):
Ascending order of X values :
First Quartile : ($Q_1$)
Median : ($Q_2$)
Third Quartile : ($Q_3$)
Bowley’s Coeff. of Skewness :

## How to calculate Bowley’s Coefficient of Skewness for ungrouped data?

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

Step 2 – Click on "Calculate" button to get Decile for ungrouped data

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

Step 4 – Gives the output as ascending order data

Step 5 – Gives the Quartiles $Q_1$,$Q_2$ and $Q_3$.

Step 6 – Gives output as Bowley’s Coefficient of Skewness

## Range for Bowley’s coefficient of skewness ranges

The Bowley’s coefficient of skewness ranges from -1 to +1.

#### Proof

We know that, if $a > 0$ and $b > 0$, then $|a-b|\leq |a+b|$,

$$$$\label{sb} \text{i.e., } \bigg|\dfrac{a-b}{a+b} \bigg| \leq 1$$$$

Now, taking $a= Q_3 – Q_2$ and $b= Q_2-Q_1$ in inequality \eqref{sb} we get

\begin{aligned} & \bigg|\dfrac{(Q_3 - Q_2)-(Q_2-Q_1)}{(Q_3 - Q_2)+(Q_2-Q_1)}\bigg| \leq 1\\ &\Rightarrow \bigg|\dfrac{Q_3 + Q_1-2Q_2}{Q_3 -Q_1}\bigg| \leq 1\\ & \Rightarrow |S_b|\leq 1\\ & \Rightarrow -1\leq S_b\leq 1. \end{aligned}
Thus, Bowley’s coefficient of skewness ranges from -1 and +1.

## Bowley’s coefficient of skewness Example 1

A random sample of 15 patients yielded the following data on the length of stay (in days) in the hospital.

5, 6, 9, 10, 15, 10, 14, 12, 10, 13, 13, 9, 8, 10, 12.

Find Bowley’s coefficient of skewness.

#### Solution

Bowley’s coefficient of skewness is

\begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1} \end{aligned}

The formula for $i^{th}$ quartile is

$Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$

where $n$ is the total number of observations.

Arrange the data in ascending order

5, 6, 8, 9, 9, 10, 10, 10, 10, 12, 12, 13, 13, 14, 15

First Quartile $Q_1$

The first quartile $Q_1$ can be computed as follows:

\begin{aligned} Q_{1} &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{1(15+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(4\big)^{th} \text{ obs.}\\ &=9 \text{ days}. \end{aligned}

Thus, lower $25$ % of the patients had length of stay in the hospital less than or equal to $9$ days.

Second Quartile $Q_2$

The second quartile $Q_2$ can be computed as follows:

\begin{aligned} Q_{2} &=\text{Value of }\bigg(\dfrac{2(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{2(15+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(8\big)^{th} \text{ obs.}\\ &=10 \text{ days}. \end{aligned}
Thus, lower $50$ % of the patients had length of stay in the hospital less than or equal to $10$ days.

Third Quartile $Q_3$

The third quartile $Q_3$ can be computed as follows:

\begin{aligned} Q_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{3(15+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(12\big)^{th} \text{ obs.}\\ &=13 \text{ days}. \end{aligned}
Thus, lower $75$ % of the patients had length of stay in the hospital less than or equal to $13$ days.

Bowley’s coefficient of skewness

Bowley’s coefficient of skewness is

\begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1}\\ &=\frac{13+9 - 2* 10}{13 - 9}\\ &=0.5 \end{aligned}

As the coefficient of skewness $S_b > 0$, the data is $\text{positively skewed}$.

## Bowley’s coefficient of skewness Example 2

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.

Find the Bowley’s coefficient of skewness.

#### Solution

Bowley’s coefficient of skewness is

\begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1} \end{aligned}

The formula for $i^{th}$ quartile is

$Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$

where $n$ is the total number of observations.

Arrange the data in ascending order

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

First Quartile $Q_1$

The first quartile $Q_1$ can be computed as follows:

\begin{aligned} Q_{1} &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{1(20+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(5.25\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(5\big)^{th} \text{ obs.}\\ &\quad +0.25 \big(\text{Value of } \big(6\big)^{th}\text{ obs.}-\text{Value of }\big(5\big)^{th} \text{ obs.}\big)\\ &=75+0.25\big(75 -75\big)\\ &=75 \text{ mg/dl}. \end{aligned}
Thus, lower $25$ % of the patients had blood sugar level less than or equal to $75$ mg/dl.

Second Quartile $Q_2$

The second quartile $Q_2$ can be computed as follows:

\begin{aligned} Q_{2} &=\text{Value of }\bigg(\dfrac{2(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{2(20+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(10.5\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(10\big)^{th} \text{ obs.}\\ & \quad +0.5 \big(\text{Value of } \big(11\big)^{th}\text{ obs.}-\text{Value of }\big(10\big)^{th} \text{ obs.}\big)\\ &=78+0.5\big(79 -78\big)\\ &=78.5 \text{ mg/dl}. \end{aligned}

Thus, lower $50$ % of the patients had blood sugar level less than or equal to $78.5$ mg/dl.

Third Quartile $Q_3$

The third quartile $Q_3$ can be computed as follows:

\begin{aligned} Q_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{3(20+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(15.75\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(15\big)^{th} \text{ obs.}\\ &\quad +0.75 \big(\text{Value of } \big(16\big)^{th}\text{ obs.}-\text{Value of }\big(15\big)^{th} \text{ obs.}\big)\\ &=84+0.75\big(85 -84\big)\\ &=84.75 \text{ mg/dl}. \end{aligned}

Thus, lower $75$ % of the patients had blood sugar level less than or equal to $84.75$ mg/dl.

Bowley’s coefficient of skewness

Bowley’s coefficient of skewness is

\begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1}\\ &=\frac{84.75+75 - 2* 78.5}{84.75 - 75}\\ &=0.2820513 \end{aligned}

As the coefficient of skewness $S_b > 0$, the data is $\text{positively skewed}$.

## Bowley’s coefficient of skewness Example 3

The following data gives the hourly wage rates (in dollars) of 25 employees of a company.

20, 28, 30, 18, 27, 19, 22, 21, 24, 25,
18, 25, 20, 27, 24, 20, 23, 32, 20, 35,
22, 26, 25, 28, 31.

Find the Bowley’s coefficient of skewness.

#### Solution

Bowley’s coefficient of skewness is

\begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1} \end{aligned}

The formula for $i^{th}$ quartile is

$Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$

where $n$ is the total number of observations.

Arrange the data in ascending order

18, 18, 19, 20, 20, 20, 20, 21, 22, 22, 23, 24, 24, 25, 25, 25, 26, 27, 27, 28, 28, 30, 31, 32, 35

First Quartile $Q_1$

The first quartile $Q_1$ can be computed as follows:

\begin{aligned} Q_{1} &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{1(25+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(6.5\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(6\big)^{th} \text{ obs.}\\ &\quad +0.5 \big(\text{Value of } \big(7\big)^{th}\text{ obs.}-\text{Value of }\big(6\big)^{th} \text{ obs.}\big)\\ &=20+0.5\big(20 -20\big)\\ &=20 \text{ dollars}. \end{aligned}
Thus, lower $25$ % of the employees had hourly wage rate less than or equal to $20$ dollars.

Second Quartile $Q_2$

The second quartile $Q_2$ can be computed as follows:

\begin{aligned} Q_{2} &=\text{Value of }\bigg(\dfrac{2(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{2(25+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(13\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(13\big)^{th} \text{ obs.}\\ &\quad +0 \big(\text{Value of } \big(13\big)^{th}\text{ obs.}-\text{Value of }\big(13\big)^{th} \text{ obs.}\big)\\ &=24+0\big(24 -24\big)\\ &=24 \text{ dollars}. \end{aligned}

Thus, lower $50$ % of the employees had hourly wage rate less than or equal to $24$ dollars.

Third Quartile $Q_3$

The third quartile $Q_3$ can be computed as follows:

\begin{aligned} Q_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{3(25+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(19.5\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(19\big)^{th} \text{ obs.}\\ &\quad +0.5 \big(\text{Value of } \big(20\big)^{th}\text{ obs.}-\text{Value of }\big(19\big)^{th} \text{ obs.}\big)\\ &=27+0.5\big(28 -27\big)\\ &=27.5 \text{ dollars}. \end{aligned}

Thus, lower $75$ % of the employees had hourly wage rate less than or equal to $27.5$ dollars.

Bowley’s coefficient of skewness

Bowley’s coefficient of skewness is

\begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1}\\ &=\frac{27.5+20 - 2* 24}{27.5 - 20}\\ &=-0.0666667 \end{aligned}

As the coefficient of skewness $S_b < 0$, the data is $\text{negatively skewed}$.

## Bowley’s coefficient of skewness Example 4

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.

Find Bowley’s coefficient of skewness.

#### Solution

Bowley’s coefficient of skewness is

\begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1} \end{aligned}

The formula for $i^{th}$ quartile is

$Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$

where $n$ is the total number of observations.

Arrange the data in ascending order

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

First Quartile $Q_1$

The first quartile $Q_1$ can be computed as follows:

\begin{aligned} Q_{1} &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{1(18+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(4.75\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(4\big)^{th} \text{ obs.}\\ &\quad +0.75 \big(\text{Value of } \big(5\big)^{th}\text{ obs.}-\text{Value of }\big(4\big)^{th} \text{ obs.}\big)\\ &=65+0.75\big(66 -65\big)\\ &=65.75 \text{ mmHg}. \end{aligned}
Thus, lower $25$ % of the patients had diastolic blood pressure less than or equal to $65.75$ mmHg.

Second Quartile $Q_2$

The second quartile $Q_2$ can be computed as follows:

\begin{aligned} Q_{2} &=\text{Value of }\bigg(\dfrac{2(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{2(18+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(9.5\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(9\big)^{th} \text{ obs.}\\ &\quad +0.5 \big(\text{Value of } \big(10\big)^{th}\text{ obs.}-\text{Value of }\big(9\big)^{th} \text{ obs.}\big)\\ &=73+0.5\big(74 -73\big)\\ &=73.5 \text{ mmHg}. \end{aligned}

Thus, lower $50$ % of the patients had diastolic blood pressure less than or equal to $73.5$ mmHg.

Third Quartile $Q_3$

The third quartile $Q_3$ can be computed as follows:

\begin{aligned} Q_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{3(18+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(14.25\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(14\big)^{th} \text{ obs.}\\ &\quad +0.25 \big(\text{Value of } \big(15\big)^{th}\text{ obs.}-\text{Value of }\big(14\big)^{th} \text{ obs.}\big)\\ &=78+0.25\big(79 -78\big)\\ &=78.25 \text{ mmHg}. \end{aligned}

Thus, lower $75$ % of the patients had diastolic blood pressure less than or equal to $78.25$ mmHg.

Bowley’s coefficient of skewness

Bowley’s coefficient of skewness is

\begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1}\\ &=\frac{78.25+65.75 - 2* 73.5}{78.25 - 65.75}\\ &=-0.24 \end{aligned}

As the coefficient of skewness $S_b < 0$, the data is $\text{negatively skewed}$.

## Bowley’s coefficient of skewness Example 5

The following data are the heights, correct to the nearest centimeters, for a group of children:

126, 129, 129, 132, 132, 133, 133, 135, 136, 137,
137, 138, 141, 143, 144, 146, 147, 152, 154, 161

Find Bowley’s coefficient of skewness.

#### Solution

Bowley’s coefficient of skewness is

\begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1} \end{aligned}

The formula for $i^{th}$ quartile is

$Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$

where $n$ is the total number of observations.

Arrange the data in ascending order

126, 129, 129, 132, 132, 133, 133, 135, 136, 137, 137, 138, 141, 143, 144, 146, 147, 152, 154, 161

First Quartile $Q_1$

The first quartile $Q_1$ can be computed as follows:

\begin{aligned} Q_{1} &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{1(20+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(5.25\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(5\big)^{th} \text{ obs.}\\ &\quad +0.25 \big(\text{Value of } \big(6\big)^{th}\text{ obs.}-\text{Value of }\big(5\big)^{th} \text{ obs.}\big)\\ &=132+0.25\big(133 -132\big)\\ &=132.25 \text{ cm}. \end{aligned}
Thus, lower $25$ % of the children had height less than or equal to $132.25$ cm.

Second Quartile $Q_2$

The second quartile $Q_2$ can be computed as follows:

\begin{aligned} Q_{2} &=\text{Value of }\bigg(\dfrac{2(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{2(20+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(10.5\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(10\big)^{th} \text{ obs.}\\ &\quad +0.5 \big(\text{Value of } \big(11\big)^{th}\text{ obs.}-\text{Value of }\big(10\big)^{th} \text{ obs.}\big)\\ &=137+0.5\big(137 -137\big)\\ &=137 \text{ cm}. \end{aligned}

Thus, lower $50$ % of the children had height less than or equal to $137$ cm.

Third Quartile $Q_3$

The third quartile $Q_3$ can be computed as follows:

\begin{aligned} Q_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{3(20+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(15.75\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(15\big)^{th} \text{ obs.}\\ &\quad +0.75 \big(\text{Value of } \big(16\big)^{th}\text{ obs.}-\text{Value of }\big(15\big)^{th} \text{ obs.}\big)\\ &=144+0.75\big(146 -144\big)\\ &=145.5 \text{ cm}. \end{aligned}

Thus, lower $75$ % of the children had height less than or equal to $145.5$ cm.

Bowley’s coefficient of skewness

Bowley’s coefficient of skewness is

\begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1}\\ &=\frac{145.5+132.25 - 2* 137}{145.5 - 132.25}\\ &=0.2830189 \end{aligned}

As the coefficient of skewness $S_b > 0$, the data is $\text{positively skewed}$.

## Conclusion

In this tutorial, you learned about how to calculate Bowley’s coefficient of skewness for ungrouped data. You also learned about how to solve numerical problems based on Bowley’s coefficient of skewness for ungrouped data.