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

Question: Calculate the quartile deviation from the data given.

Age (in years)	10-20	20-30	30-40	40-50	50-60
No. of workers	2	6	22	13	7

Solution:

Given,

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

Age (x)	No. (f)	c.f.
10-20	2	2
20-30	6	8
30-40	22	30
40-50	13	43
50-60	7	50
	N = 50

Now,

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

$= \dfrac{50}{4}^{th}\; class$

$= 12.5 ^{th} \; class$

In c.f. table, just greater value than 12.5 is 30 whose corresponding class is (30-40). 

So, $Q_1\; class= (30-40)$

For $Q_1$ we have;

l = 30, i = 10, f = 22, c.f. = 8, $\frac{N}{4}$ = 12.5

So,

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

$= 30 + \dfrac{10}{22} × (12.5-8)$

$= 30 + 2.04$

$= 32.04$

And,

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

$= \dfrac{3×50}{4}^{th}\; class$

$= 37.5 ^{th} \; class$

In c.f. table, just greater value than 37.5 is 43 whose corresponding class is (40-50). 

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

For $Q_3$ we have;

l = 40, i = 10, f = 13, c.f. = 30, $\frac{3N}{4}$ = 37.5

So,

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

$= 40 + \dfrac{10}{13} × (37.5-30)$

$= 40 + 5.76$

$= 45.76$

We know,

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

$= \dfrac{45.76 - 32.04}{2}$

$= \dfrac{13.72}{2}$

$= 6.86 years$

Also,

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

$= \dfrac{45.76 - 32.04}{45.76+32.04}$

$= \dfrac{13.72}{77.8}$

$= 0.176$

$= 0.18$

Hence, the required quartile deviation of the above data is 6.86 years and the coefficient of quartile deviation is 0.18.

Related Notes and Solutions:

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

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