The frequency distribution:
$\begin{array}{|l|l|l|l|l|l|l|} \hline X & A & 2 A & 3 A & 4 A & 5 A & 6 A \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$
where $A$ is a positive integer,has a variance of $160$. Determine the value of $A$.

(7) The total frequency $N = \sum f_i = 2 + 1 + 1 + 1 + 1 + 1 = 7$.
Calculate $\sum f_i x_i = (2 \times A) + (1 \times 2A) + (1 \times 3A) + (1 \times 4A) + (1 \times 5A) + (1 \times 6A) = 2A + 2A + 3A + 4A + 5A + 6A = 22A$.
Calculate $\sum f_i x_i^2 = (2 \times A^2) + (1 \times 4A^2) + (1 \times 9A^2) + (1 \times 16A^2) + (1 \times 25A^2) + (1 \times 36A^2) = 2A^2 + 4A^2 + 9A^2 + 16A^2 + 25A^2 + 36A^2 = 92A^2$.
The variance $\sigma^2$ is given by $\sigma^2 = \frac{\sum f_i x_i^2}{N} - \left(\frac{\sum f_i x_i}{N}\right)^2$.
Substituting the values: $160 = \frac{92A^2}{7} - \left(\frac{22A}{7}\right)^2$.
$160 = \frac{92A^2}{7} - \frac{484A^2}{49} = \frac{644A^2 - 484A^2}{49} = \frac{160A^2}{49}$.
$160 = \frac{160A^2}{49} \implies A^2 = 49$.
Since $A$ is a positive integer,$A = 7$.