Question

If the following frequency distribution
$\begin{array}{ccccccc}x& A& 2A& 3A& 4A& 5A& 6A\\ f& 2& 1& 1& 1& 1& 1\end{array}$
, where $A$ is a positive integer has a variance of $160$, then the value of $A$ is

Answer 1

Here $n=2+1+1+1+1+1=7$
${\sigma }^2=160\sum f_i{x}_i=2A+2A+3A+4A+5A+6A\Rightarrow \sum f_i{x}_i=22A\sum f_i{x}_i^2=2A^2+4A^2+9A^2+16A^2+25A^2+36A^2\Rightarrow \sum f_i{x}_i^2=92A^2{\sigma }^2=\frac{\sum f_i{x}_i^2}{n}-{\left(\frac{\sum f_i{x}_i}{n}\right)}^2\Rightarrow 160=\frac{92A^2}{7}-{\left(\frac{22A}{7}\right)}^2\therefore A=7$