Bowley's Coefficient of Skewness for grouped data
Bowley's coefficient of skewness is based on quartiles of the data. It is based on the middle 50 percent of the observations of data set. It means the Bowley's coefficient of skewness leaves the 25 percent observations in each tail of the data set.
For a symmetric distribution, the two quartiles namely $Q_1$ and $Q_3$ are equidistant from the median i.e. $Q_2$. Thus, $Q_3 - Q_2 = Q_2 - Q_1$.
The Bowley's coefficient of skewness is defined as
$$S_b = \dfrac{Q_3+Q_1 - 2Q_2}{Q_3 - Q_1}$$
The formula for $i^{th}$ quartile is
$$ \begin{aligned} Q_i=l + \bigg(\frac{\frac{iN}{4} - F_<}{f}\bigg)\times h; \quad i=1,2,3 \end{aligned} $$
where,
- $l :$ the lower limit of the $i^{th}$ quartile class
- $N=\sum f :$ total number of observations
- $f :$ frequency of the $i^{th}$ quartile class
- $F_< :$ cumulative frequency of the class previous to $i^{th}$ quartile class
- $h :$ the class width
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 Caculator for grouped data
Use this calculator to find the Bowley's Coefficient of Skewness for grouped data.
| Bowley's Coeff. 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): | |
| First Quartile : ($Q_1$) | |
| Median : ($Q_2$) | |
| Third Quartile : ($Q_3$) | |
| Bowley's Coeff. of Skewness : | |
How to find Bowley'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" button for decile calculation
Step 5 - Gives output as number of observation (N)
Step 6 - Gives output as $Q_1$, $Q_2$ and $Q_3$
Step 7 - Gives output as Bowley's Coefficient of Skewness
Bowley's Coefficient of Skewness Example 1
The following table gives the number of children of 80 families in a village
| No.of children | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| No. of families | 12 | 23 | 16 | 9 | 10 | 10 |
Find the Bowley's coefficient of skewness.
Solution
Let $X$ denote no. of children.
The Bowley's coefficient of skewness is
$$ \begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1} \end{aligned} $$
| $x_i$ | $f_i$ | $cf$ | |
|---|---|---|---|
| 0 | 12 | 12 | |
| 1 | 23 | 35 | |
| 2 | 16 | 51 | |
| 3 | 9 | 60 | |
| 4 | 10 | 70 | |
| 5 | 10 | 80 | |
| Total | 80 |
Quartiles
The formula for $i^{th}$ quartile is
$Q_i =\bigg(\dfrac{i(N)}{4}\bigg)^{th}$ value, $i=1,2,3$
where $N$ is the total number of observations.
First Quartile $Q_1$
$$ \begin{aligned} Q_{1} &=\bigg(\dfrac{1(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(80)}{4}\bigg)^{th}\text{ value}\\ &=\big(20\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $20$ is $35$. The corresponding value of $X$ is the $1^{st}$ quartile. That is, $Q_1 =1$ days.
Thus, lower $25$ % of the families had no. of children less than or equal to $1$.
Second Quartile $Q_2$
$$ \begin{aligned} Q_{2} &=\bigg(\dfrac{2(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{2(80)}{4}\bigg)^{th}\text{ value}\\ &=\big(40\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $40$ is $51$. The corresponding value of $X$ is the $2^{nd}$ quartile. That is, $Q_2 =2$ days.
Thus, lower $50$ % of the families had no. of children less than or equal to $2$.
Third Quartile $Q_3$
$$ \begin{aligned} Q_{3} &=\bigg(\dfrac{3(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{3(80)}{4}\bigg)^{th}\text{ value}\\ &=\big(60\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $60$ is $70$. The corresponding value of $X$ is the $3^{rd}$ quartile. That is, $Q_3 =4$ days.
Thus, lower $75$ % of the families had no. of children less than or equal to $4$.
Bowley's Coefficient of Skewness
The coefficient of skewness based on quartiles is
$$ \begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1}\\ &= \frac{4 + 1 - 2*2}{4 - 1}\\ &=\frac{1}{3}\\ &= 0.3333 \end{aligned} $$
As the coefficient of skewness $S_b$ is $\text{greater than zero}$ (i.e., $S_b > 0$), the distribution is $\text{positively skewed}$.
Bowley's Coefficient of Skewness Example 2
The following table gives the frequency distribution of waiting time of 65 persons at a ticket counter to buy a movie ticket.
| Waiting time (in minutes) | 0-6 | 7-13 | 14-20 | 21-27 | 28- 34 |
|---|---|---|---|---|---|
| frequency | 5 | 12 | 18 | 30 | 10 |
Compute the Bowley's coefficient of skewness.
Solution
Let $X$ denote the waiting time in minutes.
The Bowley's coefficient of skewness is
$$ \begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1} \end{aligned} $$
Here the classes are inclusive. To make them exclusive type subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class.
| Class Interval | Class Boundries | $f_i$ | $cf$ | |
|---|---|---|---|---|
| 0-6 | -0.5-6.5 | 5 | 5 | |
| 7-13 | 6.5-13.5 | 12 | 17 | |
| 14-20 | 13.5-20.5 | 18 | 35 | |
| 21-27 | 20.5-27.5 | 20 | 55 | |
| 28-34 | 27.5-34.5 | 10 | 65 | |
| Total | 65 |
Quartiles
The formula for $i^{th}$ quartile is
$Q_i =\bigg(\dfrac{i(N)}{4}\bigg)^{th}$ value, $i=1,2,3$
where $N$ is the total number of observations.
First Quartile $Q_1$
$$ \begin{aligned} Q_{1} &=\bigg(\dfrac{1(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(65)}{4}\bigg)^{th}\text{ value}\\ &=\big(16.25\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $16.25$ is $17$. The corresponding class $6.5-13.5$ is the $1^{st}$ quartile class.
Thus
- $l = 6.5$, the lower limit of the $1^{st}$ quartile class
- $N=65$, total number of observations
- $f =12$, frequency of the $1^{st}$ quartile class
- $F_< = 5$, cumulative frequency of the class previous to $1^{st}$ quartile class
- $h =7$, the class width
The first quartile $Q_1$ can be computed as follows:
$$ \begin{aligned} Q_1 &= l + \bigg(\frac{\frac{1(N)}{4} - F_<}{f}\bigg)\times h\\ &= 6.5 + \bigg(\frac{\frac{1*65}{4} - 5}{12}\bigg)\times 7\\ &= 6.5 + \bigg(\frac{16.25 - 5}{12}\bigg)\times 7\\ &= 6.5 + \big(0.9375\big)\times 7\\ &= 6.5 + 6.5625\\ &= 13.0625 \text{ minutes} \end{aligned} $$
Thus, lower $25$ % of the persons had waiting time less than or equal to $13.0625$ minutes.
Second Quartile $Q_2$
$$ \begin{aligned} Q_{2} &=\bigg(\dfrac{2(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{2(65)}{4}\bigg)^{th}\text{ value}\\ &=\big(32.5\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $32.5$ is $35$. The corresponding class $13.5-20.5$ is the $2^{nd}$ quartile class.
Thus
- $l = 13.5$, the lower limit of the $2^{nd}$ quartile class
- $N=65$, total number of observations
- $f =18$, frequency of the $2^{nd}$ quartile class
- $F_< = 17$, cumulative frequency of the class previous to $2^{nd}$ quartile class
- $h =7$, the class width
The second quartile $Q_2$ can be computed as follows:
$$ \begin{aligned} Q_2 &= l + \bigg(\frac{\frac{2(N)}{4} - F_<}{f}\bigg)\times h\\ &= 13.5 + \bigg(\frac{\frac{2*65}{4} - 17}{18}\bigg)\times 7\\ &= 13.5 + \bigg(\frac{32.5 - 17}{18}\bigg)\times 7\\ &= 13.5 + \big(0.8611\big)\times 7\\ &= 13.5 + 6.0278\\ &= 19.5278 \text{ minutes} \end{aligned} $$
Thus, lower $50$ % of the persons had waiting time less than or equal to $19.5278$ minutes.
Third Quartile $Q_3$
$$ \begin{aligned} Q_{3} &=\bigg(\dfrac{3(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{3(65)}{4}\bigg)^{th}\text{ value}\\ &=\big(48.75\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $48.75$ is $55$. The corresponding class $20.5-27.5$ is the $3^{rd}$ quartile class.
Thus
- $l = 20.5$, the lower limit of the $3^{rd}$ quartile class
- $N=65$, total number of observations
- $f =20$, frequency of the $3^{rd}$ quartile class
- $F_< = 35$, cumulative frequency of the class previous to $3^{rd}$ quartile class
- $h =7$, the class width
The third quartile $Q_3$ can be computed as follows:
$$ \begin{aligned} Q_3 &= l + \bigg(\frac{\frac{3(N)}{4} - F_<}{f}\bigg)\times h\\ &= 20.5 + \bigg(\frac{\frac{3*65}{4} - 35}{20}\bigg)\times 7\\ &= 20.5 + \bigg(\frac{48.75 - 35}{20}\bigg)\times 7\\ &= 20.5 + \big(0.6875\big)\times 7\\ &= 20.5 + 4.8125\\ &= 25.3125 \text{ minutes} \end{aligned} $$
Thus, lower $75$ % of the persons had waiting time less than or equal to $25.3125$ minutes.
Bowley's Coefficient of Skewness
The coefficient of skewness based on quartiles is
$$ \begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1}\\ &= \frac{25.3125 + 13.0625 - 2*19.5278}{25.3125 - 13.0625}\\ &=\frac{-0.6806}{12.25}\\ &= -0.0556 \end{aligned} $$
As the coefficient of skewness $S_b$ is $\text{less than zero}$ (i.e., $S_b < 0$), the distribution is $\text{negatively skewed}$.
Bowley's Coefficient of Skewness Example 3
The following table gives the amount of time (in minutes) spent on the internet each evening by a group of 56 students.
| Time spent on Internet ($x$) | 10-12 | 13-15 | 16-18 | 19-21 | 22-24 |
|---|---|---|---|---|---|
| No. of students ($f$) | 3 | 12 | 15 | 24 | 2 |
Calculate Bowley's coefficient of skewness.
Solution
Let $X$ denote the amount of time (in minutes) spent on the internet.
The Bowley's coefficient of skewness is
$$ \begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1} \end{aligned} $$
Here the classes are inclusive. To make them exclusive type subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class.
| Class Interval | Class Boundries | $f_i$ | $cf$ | |
|---|---|---|---|---|
| 10-12 | 9.5-12.5 | 3 | 3 | |
| 13-15 | 12.5-15.5 | 12 | 15 | |
| 16-18 | 15.5-18.5 | 15 | 30 | |
| 19-21 | 18.5-21.5 | 24 | 54 | |
| 22-24 | 21.5-24.5 | 2 | 56 | |
| Total | 56 |
Quartiles
The formula for $i^{th}$ quartile is
$Q_i =\bigg(\dfrac{i(N)}{4}\bigg)^{th}$ value, $i=1,2,3$
where $N$ is the total number of observations.
First Quartile $Q_1$
$$ \begin{aligned} Q_{1} &=\bigg(\dfrac{1(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(56)}{4}\bigg)^{th}\text{ value}\\ &=\big(14\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $14$ is $15$. The corresponding class $12.5-15.5$ is the $1^{st}$ quartile class.
Thus
- $l = 12.5$, the lower limit of the $1^{st}$ quartile class
- $N=56$, total number of observations
- $f =12$, frequency of the $1^{st}$ quartile class
- $F_< = 3$, cumulative frequency of the class previous to $1^{st}$ quartile class
- $h =3$, the class width
The first quartile $Q_1$ can be computed as follows:
$$ \begin{aligned} Q_1 &= l + \bigg(\frac{\frac{1(N)}{4} - F_<}{f}\bigg)\times h\\ &= 12.5 + \bigg(\frac{\frac{1*56}{4} - 3}{12}\bigg)\times 3\\ &= 12.5 + \bigg(\frac{14 - 3}{12}\bigg)\times 3\\ &= 12.5 + \big(0.9167\big)\times 3\\ &= 12.5 + 2.75\\ &= 15.25 \text{ minutes} \end{aligned} $$
Thus, lower $25$ % of the students spent less than or equal to $15.25$ minutes on the internet.
Second Quartile $Q_2$
$$ \begin{aligned} Q_{2} &=\bigg(\dfrac{2(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{2(56)}{4}\bigg)^{th}\text{ value}\\ &=\big(28\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $28$ is $30$. The corresponding class $15.5-18.5$ is the $2^{nd}$ quartile class.
Thus
- $l = 15.5$, the lower limit of the $2^{nd}$ quartile class
- $N=56$, total number of observations
- $f =15$, frequency of the $2^{nd}$ quartile class
- $F_< = 15$, cumulative frequency of the class previous to $2^{nd}$ quartile class
- $h =3$, the class width
The second quartile $Q_2$ can be computed as follows:
$$ \begin{aligned} Q_2 &= l + \bigg(\frac{\frac{2(N)}{4} - F_<}{f}\bigg)\times h\\ &= 15.5 + \bigg(\frac{\frac{2*56}{4} - 15}{15}\bigg)\times 3\\ &= 15.5 + \bigg(\frac{28 - 15}{15}\bigg)\times 3\\ &= 15.5 + \big(0.8667\big)\times 3\\ &= 15.5 + 2.6\\ &= 18.1 \text{ minutes} \end{aligned} $$
Thus, lower $50$ % of the students spent less than or equal to $18.1$ minutes on the internet.
Third Quartile $Q_3$
$$ \begin{aligned} Q_{3} &=\bigg(\dfrac{3(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{3(56)}{4}\bigg)^{th}\text{ value}\\ &=\big(42\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $42$ is $54$. The corresponding class $18.5-21.5$ is the $3^{rd}$ quartile class.
Thus
- $l = 18.5$, the lower limit of the $3^{rd}$ quartile class
- $N=56$, total number of observations
- $f =24$, frequency of the $3^{rd}$ quartile class
- $F_< = 30$, cumulative frequency of the class previous to $3^{rd}$ quartile class
- $h =3$, the class width
The third quartile $Q_3$ can be computed as follows:
$$ \begin{aligned} Q_3 &= l + \bigg(\frac{\frac{3(N)}{4} - F_<}{f}\bigg)\times h\\ &= 18.5 + \bigg(\frac{\frac{3*56}{4} - 30}{24}\bigg)\times 3\\ &= 18.5 + \bigg(\frac{42 - 30}{24}\bigg)\times 3\\ &= 18.5 + \big(0.5\big)\times 3\\ &= 18.5 + 1.5\\ &= 20 \text{ minutes} \end{aligned} $$
Thus, lower $75$ % of the students spent less than or equal to $20$ minutes on the internet.
Bowley's Coefficient of Skewness
The coefficient of skewness based on quartiles is
$$ \begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1}\\ &= \frac{20 + 15.25 - 2*18.1}{20 - 15.25}\\ &=\frac{-0.95}{4.75}\\ &= -0.2 \end{aligned} $$
As the coefficient of skewness $S_b$ is $\text{less than zero}$ (i.e., $S_b < 0$), the distribution is $\text{negatively skewed}$.
Bowley's Coefficient of Skewness Example 4
The Scores of students in a Math test is given in the table below :
| Class Interval | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
|---|---|---|---|---|---|---|
| Frequency ($f$) | 6 | 8 | 12 | 10 | 5 | 4 |
Find Bowley's Coefficient of Skewness.
Solution
Let $X$ denote the scores in Math Test.
The Bowley's coefficient of skewness is
$$ \begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1} \end{aligned} $$
| Class Interval | Class Boundries | $f_i$ | $cf$ | |
|---|---|---|---|---|
| 10-20 | 10-20 | 6 | 6 | |
| 20-30 | 20-30 | 8 | 14 | |
| 30-40 | 30-40 | 12 | 26 | |
| 40-50 | 40-50 | 10 | 36 | |
| 50-60 | 50-60 | 5 | 41 | |
| 60-70 | 60-70 | 4 | 45 | |
| Total | 45 |
Quartiles
The formula for $i^{th}$ quartile is
$Q_i =\bigg(\dfrac{i(N)}{4}\bigg)^{th}$ value, $i=1,2,3$
where $N$ is the total number of observations.
First Quartile $Q_1$
$$ \begin{aligned} Q_{1} &=\bigg(\dfrac{1(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(45)}{4}\bigg)^{th}\text{ value}\\ &=\big(11.25\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $11.25$ is $14$. The corresponding class $20-30$ is the $1^{st}$ quartile class.
Thus
- $l = 20$, the lower limit of the $1^{st}$ quartile class
- $N=45$, total number of observations
- $f =8$, frequency of the $1^{st}$ quartile class
- $F_< = 6$, cumulative frequency of the class previous to $1^{st}$ quartile class
- $h =10$, the class width
The first quartile $Q_1$ can be computed as follows:
$$ \begin{aligned} Q_1 &= l + \bigg(\frac{\frac{1(N)}{4} - F_<}{f}\bigg)\times h\\ &= 20 + \bigg(\frac{\frac{1*45}{4} - 6}{8}\bigg)\times 10\\ &= 20 + \bigg(\frac{11.25 - 6}{8}\bigg)\times 10\\ &= 20 + \big(0.6562\big)\times 10\\ &= 20 + 6.5625\\ &= 26.5625 \text{ Scores} \end{aligned} $$
Thus, lower $25$ % of the students scores less than or equal to $26.5625$ marks.
Second Quartile $Q_2$
$$ \begin{aligned} Q_{2} &=\bigg(\dfrac{2(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{2(45)}{4}\bigg)^{th}\text{ value}\\ &=\big(22.5\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $22.5$ is $26$. The corresponding class $30-40$ is the $2^{nd}$ quartile class.
Thus
- $l = 30$, the lower limit of the $2^{nd}$ quartile class
- $N=45$, total number of observations
- $f =12$, frequency of the $2^{nd}$ quartile class
- $F_< = 14$, cumulative frequency of the class previous to $2^{nd}$ quartile class
- $h =10$, the class width
The second quartile $Q_2$ can be computed as follows:
$$ \begin{aligned} Q_2 &= l + \bigg(\frac{\frac{2(N)}{4} - F_<}{f}\bigg)\times h\\ &= 30 + \bigg(\frac{\frac{2*45}{4} - 14}{12}\bigg)\times 10\\ &= 30 + \bigg(\frac{22.5 - 14}{12}\bigg)\times 10\\ &= 30 + \big(0.7083\big)\times 10\\ &= 30 + 7.0833\\ &= 37.0833 \text{ Scores} \end{aligned} $$
Thus, lower $50$ % of the students scores less than or equal to $37.0833$ marks.
Third Quartile $Q_3$
$$ \begin{aligned} Q_{3} &=\bigg(\dfrac{3(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{3(45)}{4}\bigg)^{th}\text{ value}\\ &=\big(33.75\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $33.75$ is $36$. The corresponding class $40-50$ is the $3^{rd}$ quartile class.
Thus
- $l = 40$, the lower limit of the $3^{rd}$ quartile class
- $N=45$, total number of observations
- $f =10$, frequency of the $3^{rd}$ quartile class
- $F_< = 26$, cumulative frequency of the class previous to $3^{rd}$ quartile class
- $h =10$, the class width
The third quartile $Q_3$ can be computed as follows:
$$ \begin{aligned} Q_3 &= l + \bigg(\frac{\frac{3(N)}{4} - F_<}{f}\bigg)\times h\\ &= 40 + \bigg(\frac{\frac{3*45}{4} - 26}{10}\bigg)\times 10\\ &= 40 + \bigg(\frac{33.75 - 26}{10}\bigg)\times 10\\ &= 40 + \big(0.775\big)\times 10\\ &= 40 + 7.75\\ &= 47.75 \text{ Scores} \end{aligned} $$
Thus, lower $75$ % of the students scores less than or equal to $47.75$ marks.
Bowley's Coefficient of Skewness
The coefficient of skewness based on quartiles is
$$ \begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1}\\ &= \frac{47.75 + 26.5625 - 2*37.0833}{47.75 - 26.5625}\\ &=\frac{0.1459}{21.1875}\\ &= 0.0069 \end{aligned} $$
As the coefficient of skewness $S_b$ is $\text{greater than zero}$ (i.e., $S_b > 0$), the distribution is $\text{positively skewed}$.
Bowley's Coefficient of Skewness Example 5
The following data shows the distribution of maximum loads in short tons supported by certain cables produced by a company:
| Maximum load | No. of Cables |
|---|---|
| 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 |
Find Bowley's Coefficient of Skewness.
Solution
Let $X$ denote the maximum load.
The Bowley's coefficient of skewness is
$$ \begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1} \end{aligned} $$
| Class Interval | Class Boundries | $f_i$ | $cf$ | |
|---|---|---|---|---|
| 9.25-9.75 | 9.25-9.75 | 2 | 2 | |
| 9.75-10.25 | 9.75-10.25 | 5 | 7 | |
| 10.25-10.75 | 10.25-10.75 | 12 | 19 | |
| 10.75-11.25 | 10.75-11.25 | 17 | 36 | |
| 11.25-11.75 | 11.25-11.75 | 14 | 50 | |
| 11.75-12.25 | 11.75-12.25 | 6 | 56 | |
| 12.25-12.75 | 12.25-12.75 | 3 | 59 | |
| 12.75-13.25 | 12.75-13.25 | 1 | 60 | |
| Total | 60 |
Quartiles
The formula for $i^{th}$ quartile is
$Q_i =\bigg(\dfrac{i(N)}{4}\bigg)^{th}$ value, $i=1,2,3$
where $N$ is the total number of observations.
First Quartile $Q_1$
$$ \begin{aligned} Q_{1} &=\bigg(\dfrac{1(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(60)}{4}\bigg)^{th}\text{ value}\\ &=\big(15\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $15$ is $19$. The corresponding class $10.25-10.75$ is the $1^{st}$ quartile class.
Thus
- $l = 10.25$, the lower limit of the $1^{st}$ quartile class
- $N=60$, total number of observations
- $f =12$, frequency of the $1^{st}$ quartile class
- $F_< = 7$, cumulative frequency of the class previous to $1^{st}$ quartile class
- $h =0.5$, the class width
The first quartile $Q_1$ can be computed as follows:
$$ \begin{aligned} Q_1 &= l + \bigg(\frac{\frac{1(N)}{4} - F_<}{f}\bigg)\times h\\ &= 10.25 + \bigg(\frac{\frac{1*60}{4} - 7}{12}\bigg)\times 0.5\\ &= 10.25 + \bigg(\frac{15 - 7}{12}\bigg)\times 0.5\\ &= 10.25 + \big(0.6667\big)\times 0.5\\ &= 10.25 + 0.3333\\ &= 10.5833 \text{ tons} \end{aligned} $$
Thus, lower $25$ % of the cables had maximum load less than or equal to $10.5833$ tons.
Second Quartile $Q_2$
$$ \begin{aligned} Q_{2} &=\bigg(\dfrac{2(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{2(60)}{4}\bigg)^{th}\text{ value}\\ &=\big(30\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $30$ is $36$. The corresponding class $10.75-11.25$ is the $2^{nd}$ quartile class.
Thus
- $l = 10.75$, the lower limit of the $2^{nd}$ quartile class
- $N=60$, total number of observations
- $f =17$, frequency of the $2^{nd}$ quartile class
- $F_< = 19$, cumulative frequency of the class previous to $2^{nd}$ quartile class
- $h =0.5$, the class width
The second quartile $Q_2$ can be computed as follows:
$$ \begin{aligned} Q_2 &= l + \bigg(\frac{\frac{2(N)}{4} - F_<}{f}\bigg)\times h\\ &= 10.75 + \bigg(\frac{\frac{2*60}{4} - 19}{17}\bigg)\times 0.5\\ &= 10.75 + \bigg(\frac{30 - 19}{17}\bigg)\times 0.5\\ &= 10.75 + \big(0.6471\big)\times 0.5\\ &= 10.75 + 0.3235\\ &= 11.0735 \text{ tons} \end{aligned} $$
Thus, lower $50$ % of the cables had maximum load less than or equal to $11.0735$ tons.
Third Quartile $Q_3$
$$ \begin{aligned} Q_{3} &=\bigg(\dfrac{3(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{3(60)}{4}\bigg)^{th}\text{ value}\\ &=\big(45\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $45$ is $50$. The corresponding class $11.25-11.75$ is the $3^{rd}$ quartile class.
Thus
- $l = 11.25$, the lower limit of the $3^{rd}$ quartile class
- $N=60$, total number of observations
- $f =14$, frequency of the $3^{rd}$ quartile class
- $F_< = 36$, cumulative frequency of the class previous to $3^{rd}$ quartile class
- $h =0.5$, the class width
The third quartile $Q_3$ can be computed as follows:
$$ \begin{aligned} Q_3 &= l + \bigg(\frac{\frac{3(N)}{4} - F_<}{f}\bigg)\times h\\ &= 11.25 + \bigg(\frac{\frac{3*60}{4} - 36}{14}\bigg)\times 0.5\\ &= 11.25 + \bigg(\frac{45 - 36}{14}\bigg)\times 0.5\\ &= 11.25 + \big(0.6429\big)\times 0.5\\ &= 11.25 + 0.3214\\ &= 11.5714 \text{ tons} \end{aligned} $$
Thus, lower $75$ % of the cables had maximum load less than or equal to $11.5714$ tons.
Bowley's Coefficient of Skewness
The coefficient of skewness based on quartiles is
$$ \begin{aligned} S_b &= \frac{Q_3+Q_1 - 2Q_2}{Q_3 -Q_1}\\ &= \frac{11.5714 + 10.5833 - 2*11.0735}{11.5714 - 10.5833}\\ &=\frac{0.0077}{0.9881}\\ &= 0.0078 \end{aligned} $$
As the coefficient of skewness $S_b$ is $\text{greater than zero}$ (i.e., $S_b > 0$), the distribution is $\text{positively skewed}$.
Conclusion
In this tutorial, you learned about how to calculate Bowley's coefficient of skewness for grouped data. You also learned about how to solve numerical problems based on Bowley's coefficient of skewness for grouped data.
To learn more about other descriptive statistics, please refer to the following tutorial:
Let me know in the comments if you have any questions on Bowley's coefficient of Skewness calculator for grouped data with examples and your thought on this article.
