JEE Main 2022 — Mathematics Probability & Statistics

Given,

$n=10$n=10 and mean $\displaystyle \bar{x}=\frac{\sum {x}_{i}}{10}=15$xˉ=10∑xi​​=15

$$\Rightarrow \sum _{i=1}^{10}{x}_{i}=150$$⇒i=1∑10​xi​=150

$$\Rightarrow \sum _{i=1}^{9}{x}_{i}+25=150$$⇒i=1∑9​xi​+25=150

$$\Rightarrow \sum _{i=1}^{9}{x}_{i}=125$$⇒i=1∑9​xi​=125

Now new sum will be $\displaystyle \sum _{i=1}^{9}{x}_{i}+15=140$i=1∑9​xi​+15=140

So, actual mean $=\frac{140}{10}=14={\bar{x}}_{\mathrm{new}}$=10140​=14=xˉnew​

Now finding variance

$$\frac{\sum {x}_{i}^{2}}{10}-{(\bar{x})}^{2}=15$$10∑xi2​​−(xˉ)2=15

$$\Rightarrow \frac{\sum {x}_{i}^{2}}{10}-{15}^{2}=15$$⇒10∑xi2​​−152=15

$$\Rightarrow \sum _{i=1}^{10}{x}_{i}^{2}=2400$$⇒i=1∑10​xi2​=2400

Removing observation of $25$25 we get,

$$\Rightarrow \sum _{i=1}^{9}{x}_{i}^{2}+625=2400$$⇒i=1∑9​xi2​+625=2400

$$\Rightarrow \sum _{i=1}^{9}{x}_{i}^{2}=1775$$⇒i=1∑9​xi2​=1775

Now adding observation $15$15 we get,

$$\sum _{i=1}^{9}{x}_{i}^{2}+{15}^{2}=2000={(\sum {x}_{i}^{2})}_{\mathrm{actual}}$$i=1∑9​xi2​+152=2000=(∑xi2​)actual​

Now calculating actual standard deviation we get,

$${\sigma }_{\mathrm{actual}}^{2}=\frac{{(\sum {x}_{i}^{2})}_{\mathrm{actual}}}{10}-{({\bar{x}}_{\mathrm{new}})}^{2}$$σactual2​=10(∑xi2​)actual​​−(xˉnew​)2

$$=\frac{2000}{10}-{14}^{2}$$=102000​−142

$$=200-196=4$$=200−196=4

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