Calculate the quartile deviation and its coefficient from the data given. | Statistics | SciPiPupil

Question: Calculate the quartile deviation and its coefficient from the data given.

Marks	0-10	10-20	20-30	30-40	40-50	50-60
No. of students	4	3	7	22	7	2

Solution:

Given,

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

Marks (x)	No. (f)	c.f.
0-10	4	4
10-20	3	7
20-30	7	14
30-40	22	36
40-50	7	43
50-60	2	45
	N = 45

Now,

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

$= \dfrac{45}{4}^{th}\; class$

$= 11.25 ^{th} \; class$

In c.f. table, just greater value than 11.25 is 14 whose corresponding class is (20-30). 

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

For $Q_1$ we have;

l = 20, i = 10, f = 7, c.f. = 7, $\frac{N}{4}$ = 11.25

So,

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

$= 20 + \dfrac{10}{7} × (11.25-7)$

$= 20 + 6.07$

$= 26.07$

And,

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

$= \dfrac{3×45}{4}^{th}\; class$

$= 33.75 ^{th} \; class$

In c.f. table, just greater value than 33.75 is 36 whose corresponding class is (30-40). 

So, $Q_3\; class= (30-40)$

For $Q_3$ we have;

l = 30, i = 10, f = 22, c.f. = 14, $\frac{3N}{4}$ = 33.75

So,

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

$= 30 + \dfrac{10}{22} × (33.75 - 14)$

$= 30 + 8.97$

$= 38.97$

We know,

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

$= \dfrac{38.97 - 26.07}{2}$

$= \dfrac{12.9}{2}$

$= 6.45$

Also,

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

$= \dfrac{38.97 - 26.07}{38.97+26.07}$

$= \dfrac{12.9}{65.04}$

$= 0.198$

Hence, the required quartile deviation of the above data is 6.45 and the coefficient of quartile deviation is 0.198.

Related Notes and Solutions:

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

#SciPiPupil