Question

For a distribution, Bowley’s coefficient of skewness is 0.6. The sum of upper and lower quartiles is 100 and median is 38. Find the upper and lower quartiles.

Answer 1

Hint: To solve this problem we have to know about the coefficient of skewness, and about the quartiles. The coefficient of skewness measures the skewness of a distribution, it is based on the notion of the moment of the distribution. A quartile is a statistical value, which divides the number of data points into four parts or quarters of equal size. Median is also a statistical value which is the middle value of the set of data, when arranged in the order of lowest to highest.

Complete step by step solution:
Given the median is 38.
Skewness refers to the distortion or asymmetry in a symmetrical bell curve, or normal distribution in a set of data. If the curve is shifted to the left or to the right, then it is said to be skewed. Negatively skewed distributions are also known as left-skewed distributions.
The Bowley’s skewness is also known as quartile skewness coefficient, which is given below.
The skewness can be measured as the ratio of the difference between the lengths of the upper quartile and the lower quartile to the length of the interquartile range, which is given by:
Upper quartile = ${Q_3} - {Q_1}$
Lower quartile = ${Q_2} - {Q_1}$
Interquartile = ${Q_3} - {Q_1}$
The Bowley’s skewness coefficient is given by:
$ \Rightarrow \dfrac{{({Q_3} - {Q_2}) - ({Q_2} - {Q_1})}}{{{Q_3} - {Q_1}}} = \dfrac{{{Q_1} + {Q_3} - 2{Q_2}}}{{{Q_3} - {Q_1}}}$
Here the $Q's$ denote the interquartile ranges.
Given the value of the sum of the upper quartile and the lower quartile is 100, as given below:
$ \Rightarrow {Q_3} + {Q_1} = 100$
Also given the value of the median is 38, which is given by:
$ \Rightarrow {Q_2} = 38$
Given the Bowley’s coefficient of skewness is 0.6., which is mathematically expressed as given below :
$ \Rightarrow 0.6 = \dfrac{{{Q_1} + {Q_3} - 2{Q_2}}}{{{Q_3} - {Q_1}}}$
$ \Rightarrow 0.6 = \dfrac{{100 - 2(38)}}{{{Q_3} - {Q_1}}}$
$ \Rightarrow {Q_3} - {Q_1} = \dfrac{{100 - 76}}{{0.6}}$
$ \Rightarrow {Q_3} - {Q_1} = 40$
Now we have 2 variables and 2 equations , hence solving for the values of ${Q_1},{Q_3}$:
$ \Rightarrow {Q_3} + {Q_1} = 100$
$ \Rightarrow {Q_3} - {Q_1} = 40$
$ \Rightarrow 2{Q_3} = 140$
$\therefore {Q_3} = 70$
Now solving ${Q_1}$ from the expression ${Q_3} + {Q_1} = 100$, as given below:
$ \Rightarrow 70 + {Q_1} = 100$
$\therefore {Q_1} = 30$
Thus the values of ${Q_1} = 30$ and ${Q_3} = 70$.

The upper and lower quartiles are 70 and 30 respectively.

Note: If the skewness coefficient is between 0.5 to 1, then the distribution is moderately skewed. Here given the skewness is 0.6, hence the distribution is moderately skewed. If the skewness coefficient is between -0.5 and 0.5, then the distribution is approximately symmetric. Else the distribution is highly skewed.