Question

The frequency distribution:

x	A	2A	3A	4A	5A	6A
f	2	1	1	1	1	1

where A is a positive integer, has a variance of 160. Determine the value of A?

Answer 1

(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	$\sum f_i=7$	$\sum f_i{x}_i=22A$	$\sum f_i{x}_i^2=92A^2$

Now, variance ${\sigma }^2=\frac{\sum f_i{x}_i^2}{N}-{\left(\frac{\sum f_i{x}_i}{N}\right)}^2$

Where $N=\sum f_i$

Substituting values from the table
and given ${\sigma }^2=160$, we get

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

$160=\frac{92A^2}{7}-\frac{484A^2}{49}$

$160=\frac{(92\times 7-484)A^2}{49}$

$\Rightarrow A^2=\frac{160\times 49}{644-484}=49$

$\Rightarrow A=7$ $[\because A\in Z^{+}]$

Hence, the value of A = 7