JEE Main 2024 — Mathematics Probability & Statistics

$\begin{aligned}
& \text { Mean }(\overline{\mathrm{x}})=10 \
& \Rightarrow \frac{\Sigma \mathrm{x}{\mathrm{i}}}{20}=10 \
& \Sigma \mathrm{x}{\mathrm{i}}=10 \times 20=200
\end{aligned}$

If 8 is replaced by 12, then $\Sigma x_i=200-8+12=204$Σxi​=200−8+12=204 $\therefore$∴ Correct mean $(\overline{\mathrm{x}})=\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}$(x)=20Σxi​​ $=\frac{204}{20}=10.2$=20204​=10.2 $\because$∵ Standard deviation $=2$=2 $\therefore$∴ Variance $=(\text { S.D. })^2=2^2=4$=( S.D. )2=22=4 $\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{20}-\left(\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}\right)^2=4$⇒20Σxi2​​−(20Σxi​​)2=4 $\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{20}-(10)^2=4$⇒20Σxi2​​−(10)2=4 $\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{20}=104$⇒20Σxi2​​=104 $\Rightarrow \Sigma x_i^2=2080$⇒Σxi2​=2080

Now, replaced ' 8 ' observations by ' 12 ' Then, $\Sigma \mathrm{x}_{\mathrm{i}}^2=2080-8^2+12^2=2160$Σxi2​=2080−82+122=2160 $\therefore$∴ Variance of removing observations $\begin{aligned}
& \Rightarrow \frac{\Sigma x_i^2}{20}-\left(\frac{\Sigma x_i}{20}\right)^2 \
& \Rightarrow \frac{2160}{20}-(10.2)^2 \
& \Rightarrow 108-104.04 \
& \Rightarrow 3.96
\end{aligned}$

Correct standard deviation $=\sqrt{3.96}$=3.96​