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	15	25	35	45	55	65
Frequency	12	10	8	10	7	5

Solution:

Given,

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

Class	Frequency (f)	c.f.
15 (10-20)	12	12
25 (20-30)	10	22
35 (30-40)	8	30
45 (40-50)	10	40
55 (50-60)	7	47
65 (60-70)	5	52
	N = 52

Now,

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

$= \dfrac{52}{4}^{th}\; class$

$= 13 ^{th} \; class$

In c.f. table, just greater value than 13 is 22 whose corresponding class is (30-40). 

So, $Q_1\; class= (20-30)$

For $Q_1$ we have;

l = 20, i = 10, f = 10, c.f. = 12, $\frac{N}{4}$ = 13

So,

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

$= 20 + \dfrac{10}{10} × (13-12)$

$= 20 + 1$

$= 21$

And,

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

$= \dfrac{3×52}{4}^{th}\; class$

$= 39 ^{th} \; class$

In c.f. table, just greater value than 39 is 40 whose corresponding class is (40-50). 

So, $Q_3\; class= (40-50)$

For $Q_3$ we have;

l = 40, i = 10, f = 10, c.f. = 30, $\frac{3N}{4}$ = 39

So,

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

$= 40 + \dfrac{10}{10} × (39-30)$

$= 40 + 9$

$= 49$

We know,

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

$= \dfrac{49 - 21}{2}$

$= \dfrac{28}{2}$

$= 14$

Also,

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

$= \dfrac{49- 21}{49+21}$

$= \dfrac{28}{70}$

$= 0.4$

Hence, the required quartile deviation of the above data is 14 and the coefficient of quartile deviation is 0.4.

Related Notes and Solutions:

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

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