If the variance of the frequency distribution \(\begin{matrix}x & c & 2c & 3c & 4c & 5c & 6c \\ f & 2 & 1 & 1 & 1 & 1 & …

\(\begin{matrix}x & c & 2c & 3c & 4c & 5c & 6c \\ f & 2 & 1 & 1 & 1 & 1 & 1\end{matrix}\)

Mean \(\overset{¯}{x}=\frac{(2+2+3+4+5+6)C}{7}=\frac{22C}{7}\)

variance \({\sigma }^{2}=\frac{{c}^{2}(2+{2}^{2}+{3}^{2}+{4}^{2}+{5}^{2}+{6}^{2})}{7}-{\left(\frac{22c}{7}\right)}^{2}\)

\(=\frac{92{c}^{2}}{7}-{c}^{2}\times \frac{484}{49}\)

\(=\frac{(644-484){c}^{2}}{49}=\frac{160{c}^{2}}{49}\)

\(160=\frac{160\times {c}^{2}}{49}\Rightarrow c=7\)