Tag: measures of dispersion and skewness

$N=12$
${ Q } _{ 1 }=$ Size of $\displaystyle\frac { N+1 }{ 4 } th=3.25th$ item
$=$ size of $3rd$ item $+0.25$(size of $4th$ item)$-$(size of $3rd$ item)
$=151+0.25(151-151)=$Rs.$151$
${ Q } _{ 3 }=$ Size of $\displaystyle\frac { 3(N+1) }{ 4 } th=9.75th$ item
$=$ size of $9th$ item $+0.75$(size of $10th$ item)$-$(size of $9th$ item)
$=162+0.75(162-162)=$Rs.$162$
$\therefore$ Quartile Deviation $\displaystyle=\frac { 1 }{ 2 } \left( { Q } _{ 3 }-{ Q } _{ 1 } \right) =\frac { 11 }{ 2 } =$Rs.$5.5$

Writing the series in ascending order $7,10,12,15,17,19,25$
$\displaystyle{ Q } _{ 1 }=\frac { 15+7 }{ 2 } =11,{ Q } _{ 3 }=\frac { 25+15 }{ 2 } =20$
$\therefore$ Quartile Deviation
$\displaystyle=\frac { { Q } _{ 3 }-{ Q } _{ 1 } }{ 2 } =\frac { 20-11 }{ 2 } =4.5$

The cumulative frequency table is as given below

Lower Quartile: We have, $\displaystyle\frac{N}{4} = \displaystyle\frac{543}{4} = 135.75$
Cumulative frequency just greater than $N/4$ is $196$ and the corresponding value of the variable is $40$.
$\therefore\quad Q _1 = $ Lower Quartile = $40$ years
Middle Quartile (Medium):
We have, $\displaystyle\frac{N}{2} = \displaystyle\frac{543}{2} = 271.5$
Cumulative frequency just greater than $N/2$ is $349$ and the corresponding value of the variable is $50$.
$\therefore\quad Q _2 = $ Middle quartile = $50$ years
Upper Quartile:
We have $\displaystyle\frac{3N}{4} = \displaystyle\frac{3\times543}{4} = 407.25$
Cumulative frequency just greater than $407.25$ is $489$ and the corresponding value of the variable is $60$.
$\therefore\quad Q _3= $ Upper quartile  $=60$ years
$\therefore\quad Q _1 + Q _3 - Q _2 = 40+60-50 = 50 = Q _2$

Quartile deviation is the half of the inter quartile deviation.

Given series of numbers is already arranged in ascending order.
Thus median is the centre value $ = 20 $
First Quartile $ {Q} _{1} $ will be the mid value of the observations to the left of $ 20 $ which is $ 9 $. 
Similarly, Third Quartile $ {Q} _{3} $ will be the mid value of the observations to the right of $ 20 $ which is $ 34 $.
And Quartile deviation will be $ \dfrac {{Q} _{3} - {Q} _{1}}{2} = \dfrac {34-9}{2} = \dfrac {25}{2} = 12.5 $.

use:  M.D.$=\displaystyle \frac{4}{4}\sigma ,Q.D=\frac{2}{3}\sigma $ $\displaystyle \Rightarrow \frac{M.D.}{Q.D}=\frac{6}{5}\Rightarrow Q.D.=\frac{5}{6}\left ( M.D. \right )$


$MD=\frac { 4 }{ 5 } \times \sigma \quad ,\quad QD=\frac { 2 }{ 3 } \times \sigma \quad ,\quad \ \frac { MD }{ QD } =\frac { 6 }{ 5 } ,\quad QD\quad =\quad \frac { 5\times 15 }{ 6 } =12.5$

Lower Quartile $ {Q} _{1} = 4 $

Upper Quartile $ {Q} _{3} = 12 $

And Quartile deviation will be $ \frac {{Q} _{3} - {Q} _{1}}{2} = \frac {12-4}{2} = \frac {8}{2} = 4 $