Find the quartile deviation from the following data. | Statistics | SciPiPupil

Question: Find the quartile deviation from the following data:

Marks	<5	<10	<15	<20	<25	<30
Frequency (f)	2	4	7	13	27	40

Solution:

Given,

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

Marks (x)	No. (f)	c.f.
<5 (0-5)	2	2
<10 (5-10)	2	4
<15 (10-15)	3	7
<20 (15-20)	6	13
<25 (20-25)	14	27
<30 (25-30)	13	40
	N = 40

Now,

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

$= \dfrac{40}{4}^{th}\; class$

$= 10^{th} \; class$

In c.f. table, just greater value than 10 is 13 whose corresponding class is (15-20). 

So, $Q_1\; class= (15-20)$

For $Q_1$ we have;

l = 15, i = 5, f = 6, c.f. = 7, $\frac{N}{4}$ = 10

So,

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

$= 15 + \dfrac{5}{6} × (10-7)$

$= 15 + 2.5$

$= 17.5$

And,

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

$= \dfrac{3×40}{4}^{th}\; class$

$= 30 ^{th} \; class$

In c.f. table, just greater value than 30 is 40 whose corresponding class is (25-30). 

So, $Q_3\; class= (25-30)$

For $Q_3$ we have;

l = 25, i = 5, f = 13, c.f. = 27, $\frac{3N}{4}$ = 30

So,

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

$= 25 + \dfrac{5}{13} × (30 - 27)$

$= 25 + 1.15$

$= 26.15$

We know,

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

$= \dfrac{26.15 - 17.5}{2}$

$= \dfrac{8.65}{2}$

$= 4.325$

Hence, the required quartile deviation of the above data is 4.325.

Related Notes and Solutions:

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

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