Question

Consider the following frequency distribution :

Class :	0-6	6 -12	12-18	18 - 24	24 - 30
Frequency	a	b	12	9	5

If mean $=\frac{309}{22}$ and median $=14$, then the value $(a-b)^{2}$ is equal to _____.

Answer 1

Class	Frequency	$x_i$	$f_ix_i$
0-6	b	3	3a
6- 12	b	9	9b
12 -18	12	15	180
18 - 24	9	21	189
24 - 30	5	27	135
	N = (26 + a + b)		(504 + 3a + 9b)

Mean $=\frac{3 a+9 b+180+189+135}{a+b+26}=\frac{309}{22} $
$\Rightarrow 66 a+198 b+11088=309 a+309 b+8034 $
$\Rightarrow 243 a+111 b=3054 $
$\Rightarrow 81 a+37 b=1018 \rightarrow(1) $
Now, Median$=12+\frac{\frac{a+b+26}{2}-(a+b)}{12} \times 6=14$
$\Rightarrow \frac{13}{2}-\left(\frac{a+b}{4}\right)=2 $
$\Rightarrow \frac{a+b}{4}=\frac{9}{2} $
$\Rightarrow a+b=18 \rightarrow(2)$
From equation (1) $\&(2)$
$a=8, b=10 $
$\therefore (a-b)^{2}=(8-10)^{2}$