If $\sum _{i=1}^9 \left(x_i-5\right)=9$ and $\sum _{i=1}^9 {\left(x_i-5\right)}^2=45$, then the standard deviation of the $9$ items $x_i,x_2,\dots ,x_9$ is

Solution:

We have,

  $\therefore \sum _{i=1}^9 \left(x_i-5\right)=9$ and  $\sum _{i=1}^9 {\left(x_i-5\right)}^2=45$

$\Rightarrow \sum _{i=1}^9 x_i-45=9$ and $\sum _{i=1}^9 \left({x_i}^2-10x_i+25\right)=45$

$\Rightarrow \sum _{i=1}^9 x_i=54\text{ and }\sum _{i=1}^9 x_i^2-10\sum _{i=1}^9 x_i+25\times 9=45$

$\Rightarrow \sum _{i=1}^9 x_i=54$ and  $\sum _{i=1}^9 {x_i}^2-10\times 54+225=45$

$\Rightarrow \sum _{i=1}^9 x_i=54$ and  $\sum _{i=1}^9 x_i^2=360$

$\therefore$ Variance $=\frac{1}{9}\sum _{i=1}^9 x_i^2-{\left(\frac{1}{9}\sum _{i=1}^9 x_i\right)}^2$

$\Rightarrow$ Variance $=\frac{360}{9}-{\left(\frac{54}{9}\right)}^2=40-36=4$

Hence, standard deviation $=\sqrt{4}=2.$

ALITER Let the variable $U$ take values $u_1,u_2,\dots ,u_9$ such that $u_i=x_i-5,i=1,2,\dots ,9.$ Then

      $\overline{\overline{U}}=\overline{\overline{X}}-5$ and $\mathrm{Var}\left(U\right)=\mathrm{Var}\left(X\right)$

Now,

   $\mathrm{Var}\left(U\right)=\frac{1}{9}\sum _{i=1}^9 u_i^2-{\left(\frac{1}{9}\sum _{i=1}^9 u_i\right)}^2$

                   $=\frac{1}{9}\sum _{i=1}^9 {\left(x_i-5\right)}^2-{\left\{\frac{1}{9}\sum _{i=1}^9 \left(x_i-5\right)\right\}}^2=\frac{45}{9}-{\left(\frac{9}{9}\right)}^2=5-1=4$

$\therefore {\sigma }_U=\sqrt{\mathrm{Var}\left(U\right)}=2$