JEE Main 2022 — Statistics Question with Solution

Given,

$n=10$ and mean $\overline{x}=\frac{\sum x_i}{10}=15$

$\Rightarrow \sum _{i=1}^{10}{x}_i=150$

$\Rightarrow \sum _{i=1}^9{x}_i+25=150$

$\Rightarrow \sum _{i=1}^9{x}_i=125$

Now new sum will be $\sum _{i=1}^9{x}_i+15=140$

So, actual mean $=\frac{140}{10}=14={\overline{x}}_{\mathrm{new} }$

Now finding variance 

$\frac{\sum x_i^2}{10}-{\left(\overline{x}\right)}^2=15$

$\Rightarrow \frac{\sum x_i^2}{10}-{15}^2=15$

$\Rightarrow \sum _{i=1}^{10}{x}_i^2=2400$

Removing observation of $25$ we get,

$\Rightarrow \sum _{i=1}^9{x}_i^2+625=2400$

$\Rightarrow \sum _{i=1}^9{x}_i^2=1775$

Now adding observation $15$ we get,

$\sum _{i=1}^9{x}_i^2+{15}^2=2000={\left(\sum x_i^2\right)}_{\mathrm{actual} }$

Now calculating actual standard deviation we get,

${\sigma }_{\mathrm{actual} }^2=\frac{{\left(\sum x_i^2\right)}_{\mathrm{actual} }}{10}-{\left({\overline{x}}_{\mathrm{new} }\right)}^2$

$=\frac{2000}{10}-{14}^2$

$=200-196=4$

Calculating standard deviation, ${\left(S.D\right)}_{\mathrm{actual} }=\sqrt{4}=2$