Question

If $\sum _{i=1}^{10}(x_1-15)=12$ and $\sum _{i=1}^{10}(x_i-15{)}^2=18$, then the S.D. of observations $x_1,x_2.............x_{10}$ is

• A. $\frac{2}{5}$
• B. $\frac{3}{5}$
• C. $\frac{4}{5}$
• D. none of these

Answer 1

Answer: (B)

$\frac{3}{5}$

Given $\sum _{i=1}^{10}(x_i-15)=12$ and $\sum _{i=1}^{10}{(x_i-15)}^2=18$
$\Rightarrow$
$\sum _{i=1}^{10}{x}_i-\sum _{i=1}^{10}15=12$ and $\sum _{i=1}^{10}{x}_i^2-30\sum _{i=1}^{10}{x}_i+\sum _{i=1}^{10}225=18$
$\Rightarrow \sum _{i=1}^{10}{x}_i=12+150=162$
Now $\sum _{i=1}^{10}{x}_i^2=30\sum _{i=1}^{10}{x}_i-\sum _{i=1}^{10}225+18=30\times 162+18-225\times 10=2628$
Hence required S.D $=\sqrt{\frac{2628}{10}-{\left(\frac{162}{10}\right)}^2}=\sqrt{\frac{9}{25}}=\frac{3}{5}$