in a distribution the difference of the lower and upper quartiles is 1

1.	in a distribution the difference of the lower and upper quartiles is 15 and their sum is 35 the median is 20 then coefficient of skewness is
Answer» $\begin{gathered}\begin{gathered}\bf \:<klux>GIVEN</klux> - \begin{cases} &\sf{Q_{3} - Q_{1} = 15} \\ &\sf{Q_{3} + Q_{1} = 35}\\ &\sf{median \: Q_{2} \: or \: M \: = 20} \end{cases}\end{gathered}\end{gathered}$ $\begin{gathered}\begin{gathered}\bf \: To\:find - \begin{cases} &\sf{coefficient \: of \: skewness} \end{cases}\end{gathered}\end{gathered}$ $\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}$ $\sf \: Coefficient \: of \: skewness, S_{k} = \dfrac{Q_{3} + Q_{1} - 2M}{Q_{3} - Q_{1}}$ where, $\: \: \: \: \: \: \: \: \bull \sf \: Q_{1} = lower \: quartile$ $\: \: \: \: \: \: \: \: \bull \sf \: Q_{3} = upper \: quartile$ $\: \: \: \: \: \: \: \: \bull \sf \: Q_{2} \: or \: M= median$ $\large\underline{\bold{Solution-}}$ Given that $\: \: \: \: \: \: \: \: \bull \sf \: Q_{3} + Q_{1} = 35$ $\: \: \: \: \: \: \: \: \bull \sf \: Q_{3} - Q_{1} = 15$ $\: \: \: \: \: \: \: \: \bull \sf \: M = 20$ Now, $\sf \: Coefficient \: of \: skewness, S_{k} = \dfrac{Q_{3} + Q_{1} - 2M}{Q_{3} - Q_{1}}$ On substituting all the values, we get $\sf \: Coefficient \: of \: skewness, S_{k} = \dfrac{35 - 2 \times 20}{15}$ $\sf \: Coefficient \: of \: skewness, S_{k} = \dfrac{35 - 40}{15}$ $\sf \: Coefficient \: of \: skewness, S_{k} = - \: \dfrac{5}{15}$ $\sf \: Coefficient \: of \: skewness, S_{k} = - \: \dfrac{1}{3}$ Additional Information 1. Pearson’s Coefficient of Skewness USING the mode. The formula is: $\boxed{ \sf{Coefficient \: of \: skewness, S_{k} = \dfrac{mean - mode}{standard \: deviation} }}$ 2. Pearson’s Coefficient of Skewness using the median. The formula is: $\boxed{ \sf{Coefficient \: of \: skewness, S_{k} = \dfrac{3(mean - median)}{standard \: deviation} }}$ 3. Properties of Skewness :- The direction of skewness is given by the sign. The coefficient COMPARES the sample distribution with a normal distribution. The larger the value, the larger the distribution differs from a normal distribution. A value of ZERO means no skewness at all. A large negative value means the distribution is negatively skewed. A large positive value means the distribution is positively skewed.

in a distribution the difference of the lower and upper quartiles is 15 and their sum is 35 the median is 20 then coefficient of skewness is

Answer»

$\begin{gathered}\begin{gathered}\bf \:<klux>GIVEN</klux> - \begin{cases} &\sf{Q_{3} - Q_{1} = 15} \\ &\sf{Q_{3} + Q_{1} = 35}\\ &\sf{median \: Q_{2} \: or \: M \: = 20} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\:find - \begin{cases} &\sf{coefficient \: of \: skewness} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}$

$\sf \: Coefficient \: of \: skewness, S_{k} = \dfrac{Q_{3} + Q_{1} - 2M}{Q_{3} - Q_{1}}$

where,

$\: \: \: \: \: \: \: \: \bull \sf \: Q_{1} = lower \: quartile$

$\: \: \: \: \: \: \: \: \bull \sf \: Q_{3} = upper \: quartile$

$\: \: \: \: \: \: \: \: \bull \sf \: Q_{2} \: or \: M= median$

$\large\underline{\bold{Solution-}}$

Given that

$\: \: \: \: \: \: \: \: \bull \sf \: Q_{3} + Q_{1} = 35$

$\: \: \: \: \: \: \: \: \bull \sf \: Q_{3} - Q_{1} = 15$

$\: \: \: \: \: \: \: \: \bull \sf \: M = 20$

Now,

$\sf \: Coefficient \: of \: skewness, S_{k} = \dfrac{Q_{3} + Q_{1} - 2M}{Q_{3} - Q_{1}}$

On substituting all the values, we get

$\sf \: Coefficient \: of \: skewness, S_{k} = \dfrac{35 - 2 \times 20}{15}$

$\sf \: Coefficient \: of \: skewness, S_{k} = \dfrac{35 - 40}{15}$

$\sf \: Coefficient \: of \: skewness, S_{k} = - \: \dfrac{5}{15}$

$\sf \: Coefficient \: of \: skewness, S_{k} = - \: \dfrac{1}{3}$

Additional Information

1. Pearson’s Coefficient of Skewness USING the mode. The formula is:

$\boxed{ \sf{Coefficient \: of \: skewness, S_{k} = \dfrac{mean - mode}{standard \: deviation} }}$

2. Pearson’s Coefficient of Skewness using the median. The formula is:

$\boxed{ \sf{Coefficient \: of \: skewness, S_{k} = \dfrac{3(mean - median)}{standard \: deviation} }}$

3. Properties of Skewness :-

The direction of skewness is given by the sign.

The coefficient COMPARES the sample distribution with a normal distribution. The larger the value, the larger the distribution differs from a normal distribution.

A value of ZERO means no skewness at all.

A large negative value means the distribution is negatively skewed.

A large positive value means the distribution is positively skewed.