Compute quartile deviation and its coefficient from the following data: | Statistics | SciPiPupil

Question: Compute quartile deviation and its coefficient from the following data:

Mid value (yrs.)	62.5	67.5	72.5	77.5	82.5	87.5
Frequency	7	5	8	4	3	3

Solution:

Given,

Arranging the given data in cumulative frequency (c.f.) table:

Class	Frequency (f)	c.f.
62.5 (60-65)	7	7
67.5 (65-70)	5	12
72.5 (70-75)	8	20
77.5 (75-80)	4	24
82.5 (80-85)	3	27
87.5 (85-90)	3	30
	N = 30

Now,

$Q_1 \; class = \dfrac{N}{4}^{th} \; class$

$= \dfrac{30}{4}^{th}\; class$

$= 7.5 ^{th} \; class$

In c.f. table, just greater value than 7.5 is 12 whose corresponding class is (65-70). 

So, $Q_1\; class= (65-70)$

For $Q_1$ we have;

l = 65, i = 5, f = 5, c.f. = 7, $\frac{N}{4}$ = 7.5

So,

$Q_1 = l + \dfrac{i}{f} × \left ( \dfrac{N}{4} - c.f. \right )$

$= 65 + \dfrac{5}{5} × (7.5-7)$

$= 65 + 0.5$

$= 65.5$

And,

$Q_3 \; class = \dfrac{3N}{4}^{th} \; class$

$= \dfrac{3×30}{4}^{th}\; class$

$= 22.5 ^{th} \; class$

In c.f. table, just greater value than 22.5 is 24 whose corresponding class is (75-80). 

So, $Q_3\; class= (75-80)$

For $Q_3$ we have;

l = 75, i = 5, f = 4, c.f. = 20, $\frac{3N}{4}$ = 22.5

So,

$Q_3 = l + \dfrac{i}{f} × \left ( \dfrac{3N}{4} - c.f. \right )$

$= 75 + \dfrac{5}{4} × (22.5-20)$

$= 75 + 3.125$

$= 78.125$

$= 78.13$

We know,

Quartile Deviation (Q.D.) = $\dfrac{Q_3 - Q_1}{2}$

$= \dfrac{78.13 - 65.5}{2}$

$= \dfrac{12.63}{2}$

$= 6.315 yrs$

Also,

Coefficient of Q.D. = $\dfrac{Q_3 - Q_1}{Q_3 + Q_1}$

$= \dfrac{78.13-65.5}{78.13+65.5}$

$= \dfrac{12.63}{143.63}$

$= 0.087$

$= 0.09$

Hence, the required quartile deviation of the above data is 6.315 years and the coefficient of quartile deviation is 0.09.

Related Notes and Solutions:

Here is the website link to the notes of Statistics of Class 10.

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