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

Answer: 4

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