Quartile Deviation
We have seen that range is the simplest to understand and easiest to compute. But range as a measure of dispersion has certain limitations. The presence of even one extreme item (high or low) in a distribution can reduce the utility of range as a measure of dispersion. Since it is based on two extreme items (highest and lowest) it fails to take into account the scatter within the range. Hence we need a measure of dispersion to overcome these limitations of range. Such a measure of dispersion is called quartile deviation. In the previous chapter we studied quartiles. Quartiles are those values which divide the series into four equal parts. Hence we have three quartilesQ_{1}, Q_{2}, and Q_{3}. Q_{1} is the lower quartile wherein \( { \frac{{1}}{{4}}} \)^{th} of the total observations lie below it and \( { \frac{{3}}{{4}}} \)^{th} above it. Q_{2} is same as median which divides the series into two equal parts. Q_{3} is the upper quartile, \( { \frac{{3}}{{4}}} \)^{th} of the value falls below it and \( { \frac{{1}}{{4}}} \)^{th} above.We have already studied the value of Q_{1} and Q_{3} for individual, discrete and continuous series, hence not repeated.
Upper and lower quartile ( Q_{1} and Q_{3} ) are used to calculate interquartile range.
$$ \mathbf {Interquartile\, range \,= Q_3\,\,Q_1} $$ Half of interquartile range is called quartile deviation.
Quartile deviation (semi interquartile range) is defined as half the distance between the third and first quartiles.
Quartile Deviation and inter quartile range are absolute measures of dispersion. The relative measure is coefficient of Quartile Deviation (Q.D)
Individual Series
STEPS:

Arrange the data in ascending order.

Q_{1} = Size of \( \Biggl[{{{\frac{N+1}{4}} }}\Biggl]^{th} \) item.

Q_{3} = Size of \( 3\Biggl[{{{\frac{N+1}{4}} }}\Biggl]^{th} \) item.

Interquartile range = Q_{3}  Q_{1}.

Q.D = \( {{{\frac{Q_3  Q_1}{2}} }} \).

Coefficient of Q.D = \( {{{\frac{Q_3  Q_1}{Q_3 + Q_1}} }} \).
Table 6.6  

Roll No.  Marks 
1  20 
2  28 
3  40 
4  12 
5  30 
6  15 
7  50 
12, 15, 20, 28, 30, 40. 50
$$ Q_1 \,= \,Size \,of\,\Biggl[{{\frac{N + 1 }{4}}}\Biggl]^{th} item $$ $$ = \,Size \,of\,\Biggl[{{\frac{7 + 1 }{4}}}\Biggl]^{th} item $$ $$ = 2^{nd}\,item$$ Size of 2^{nd} item is 15. Thus Q_{1} = 15
$$ Q_3 \,= \,Size \,of\,3\Biggl[{{\frac{N + 1 }{4}}}\Biggl]^{th} item $$ $$ Q_3 \,= \,Size \,of\,3\Biggl[{{\frac{7 + 1 }{4}}}\Biggl]^{th} item $$ $$ = Size\, of\, 6^{th}\,item$$ Size of 6^{th} item = 40; Q_{3} = 40.
$$Q.D \,=\, {{{\frac{Q_3  Q_1}{2}} }} $$ $$ = \,{{{\frac{40  15}{2}} }} $$ $$ = \,{{{\frac{25}{2}} }} $$ $$ = \, 12.5 $$ $$Coefficient \,of \,Q.D \,=\, {{{\frac{Q_3  Q_1}{Q_3 + Q_1}} }} $$ $$=\, {{{\frac{40  15}{40 + 15}} }} $$ $$=\, {{{\frac{25}{55}} }} $$ $$ = \, 0.455 $$
Discrete Series
STEPS:

Arrange the data in ascending order.

Find out cumulative frequency.

Q_{1} = Size of \( \Biggl[{{{\frac{N+1}{4}} }}\Biggl]^{th} \) item.

Q_{3} = Size of \( 3\Biggl[{{{\frac{N+1}{4}} }}\Biggl]^{th} \) item.

Q.D = \( {{{\frac{Q_3  Q_1}{2}} }} \).

Interquartile range = Q_{3}  Q_{1}.

Coefficient of Q.D = \( {{{\frac{Q_3  Q_1}{Q_3 + Q_1}} }} \).
Table 6.7  

Marks  No. of Students 
10  4 
20  7 
30  15 
40  8 
50  7 
60  2 
Table 6.8  

Marks  No. of Students  C.F 
10  4  4 
20  7  11 
30  15  26 
40  8  34 
50  7  41 
60  2  43 
$$ Q_3 \,= \,Size \,of\,3\Biggl[{{\frac{N + 1 }{4}}}\Biggl]^{th} item $$ $$ = \,Size \,of\,3\Biggl[{{\frac{43 + 1 }{4}}}\Biggl]^{th} item $$ $$ = \,Size \,of\Biggl[{{\frac{3 × 44 }{4}}}\Biggl]^{th} item $$ $$ = Size\, of\, 33^{rd}\,item$$ Size of 33^{rd} item = 40; Q_{3} = 40.
$$ Interquartile \,range \,=\, Q_3\,  \,Q_1 $$ $$ =\, 40\,  \,20 $$ $$ =\,20 $$ $$ Q.D \,=\, {{{\frac{Q_3  Q_1}{2}} }} $$ $$ = \,{{{\frac{40  20}{2}} }} $$ $$ = \,{{{\frac{20}{2}} }} $$ $$ = \, 10 $$ $$ Coefficient \,of \,Q.D \,=\, {{{\frac{Q_3  Q_1}{Q_3 + Q_1}} }} $$ $$ =\, {{{\frac{40  20}{40 + 20}} }} $$ $$ =\, {{{\frac{20}{60}} }} $$ $$ = \, 0.333 $$
Continuous Series
STEPS:

Find out cumulative frequency.

Find Q_{1} and Q_{3} classes as follows.
\( Q_1\,=\,Size\,of\,{{{\frac{N}{4}} }}^{th} item \)
\( Q_1 \,= \,{ L + \frac{\frac{N}{4}  {cf}}{f} × h} \)
\( Q_3\,=\,Size\,of\,{{{\frac{3N}{4}} }}^{th} item \)
\( Q_3 \,= \,{ L + \frac{\frac{3N}{4}  {cf}}{f} × h} \)

Interquartile range = Q_{3}  Q_{1}.

Q.D = \( {{{\frac{Q_3  Q_1}{2}} }} \).

Coefficient of Q.D = \( {{{\frac{Q_3  Q_1}{Q_3 + Q_1}} }} \).
Table 6.9  

Wages (₹)  No. of Workers 
20  25  2 
25  30  10 
30  35  25 
35  40  16 
40  45  7 
Table 6.10  

Wages (₹)  No. of Workers  C.F 
20  25  2  2 
25  30  10  12 
30  35  25  37 
35  40  16  53 
40  45  7  60 
N = 60 
$$ Q_1 \,= \,{ L + \frac{\frac{N}{4}  {cf}}{f} × h} $$ L = 30;
\( {\frac{N}{4}} \) = 15;
CF = 12;
f = 25;
h = 5
$$ Q_1 \,= \,{ 30 + \frac{{15}  {12}}{25} × 5} $$ $$ =\, 30 \,+\,0.6 $$ $$ =\,30.6 $$ $$ Q_3\,=\,Size\,of\,{{{\frac{3N}{4}} }}^{th} item $$ $$ =\,{{{\frac{3 × 60}{4}} }} $$ $$ =\,{{{\frac{180}{4}} }} $$ $$ =\,45^{th} item $$ Q_{3} lies in the class 35  40
$$ Q_3 \,= \,{ L + \frac{\frac{3N}{4}  {cf}}{f} × h} $$ L = 35;
\( {\frac{3N}{4}} \) = 45;
CF = 37;
f = 16;
h = 5
$$ Q_3 \,= \,{ 35 + \frac{{45}  {37}}{16} × 5} $$ $$ =\, 35 \,+\,2.5 $$ $$ =\, 37.5 $$ $$ Interquartile \,range \,=\, Q_3\,  \,Q_1 $$ $$ =\, 37.5\,  \,30.6 $$ $$ =\,6.9 $$ $$ Q.D \,=\, {{{\frac{Q_3  Q_1}{2}} }} $$ $$ = \,{{{\frac{37.5  30.6}{2}} }} $$ $$ = \,{{{\frac{6.9}{2}} }} $$ $$ = \, 3.45 $$ $$ Coefficient \,of \,Q.D \,=\, {{{\frac{Q_3  Q_1}{Q_3 + Q_1}} }} $$ $$ =\, {{{\frac{37.5  30.6}{37.5 + 30.6}} }} $$ $$ =\, {{{\frac{6.9}{68.1}} }} $$ $$ = \, 0.101 $$
MERITS OF QUARTILE DEVIATION
 It is easily computed and readily understood.
 It is not affected by extreme items.
 It can be computed even for an open end distribution.
 It is superior and more reliable than the range.
DEMERITS OF QUARTILE DEVIATION
 It is not based on all the items in a series.
 It is not based on all the observations.
 It is not capable of further algebraic treatment.
 It does not indicate variation of items from the average.
 Its value is very much affected by sampling fluctuations.