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.
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