Question

Consider the following frequency distribution

x	A	2A	3A	4A	5A
f	2	1	1	1	1

where, $A$ is a positive integer and has variance $160$. Then the value of $A$ is.

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

Answer: (C)

$x_i$	$f_i$	$f_i{x}_i$	$f_i{x}_i^2$
A	2	2A	$2A^2$
2A	1	2A	$4A^2$
3A	1	3A	$9A^2$
4A	1	4A	$16A^2$
5A	1	5A	$25A^2$
6A	1	6A	$36A^2$
Total	7	22A	$92A^2$

$\because {\sigma }^2=\frac{\sum f_i{x}_i^2}{\sum f_i}-\left(\frac{\sum f_i{x}_i}{\sum f_i}\right)$
$\Rightarrow 160=\frac{92A^2}{7}-{\left(\frac{22A}{7}\right)}^2$
$\Rightarrow 160=\frac{92A^2}{7}-\frac{484A^2}{49}$
$\Rightarrow 160=\frac{92\times 7A^2-484A^2}{49}$
$\Rightarrow 160\times 49=644A^2-484A^2$
$\Rightarrow 160A^2=7840$
$\Rightarrow A^2=\frac{7840}{160}$
$\Rightarrow A^2=49$
$\Rightarrow A=\pm 7$
$A=7$ as $A$ is a positive integer.