Question

If a is a positive integer and the frequency distribution.

x:	a	2a	3a	4a	5a	6a
f:	2	1	1	1	1	1

has a variance of 160. Determine the value of a.

Answer 1

$x_i$	$x_i^2$	$f_i$	$f_i{x}_i$	$f_i{x}_i^2$
$a$	$a^2$	$2$	$2a$	$2a^2$
$2a$	$4a^2$	$1$	$2a$	$4a^2$
$3a$	$9a^2$	$1$	$3a$	$9a^2$
$4a$	$16a^2$	$1$	$4a$	$16a^2$
$5a$	$25a^2$	$1$	$5a$	$25a^2$
$6a$	$36a^2$	$1$	$6a$	$36a^2$
Total			$22a$	$92a^2$

Variance$={\sigma }^2=\frac{\sum f_i{x}_i^2}{\sum f_i}-{\left(\frac{\sum f_i{x}_i}{\sum f_i}\right)}^2$

$\Rightarrow 160=\frac{92a^2}{7}-{\left(\frac{22a}{7}\right)}^2$

$\Rightarrow 160=\frac{92\times 7a^2-484a^2}{49}$

$\Rightarrow 160\times 49=644a^2-484a^2$

$\Rightarrow 160\times 49=160a^2$

$\Rightarrow a^2=49$

$\Rightarrow a=\pm 7$

Given that $a$ is positive.Hence $a=7$