JEE Main 2024 — 6 April, Shift 1 — Question 4

Mean $(\bar{x})=10$ $\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}=10$

$\Sigma \mathrm{x}_{\mathrm{i}}=10 \times 20=200$

If 8 is replaced by 12 , then $\Sigma x_{i}=200-8+12=204$

$\therefore$ Correct mean $(\overline{\mathrm{x}})=\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}$

$=\frac{204}{20}=10.2$

$\because$ Standard deviation $=2$

$\therefore$ Variance $=(\text { S.D. })^{2}=2^{2}=4$

$\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^{2}}{20}-\left(\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}\right)^{2}=4$

$\Rightarrow \frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{20}-(10)^{2}=4$

$\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^{2}}{20}=104$

$\Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}^{2}=2080$

Now, replaced ' 8 ' observations by ' 12 ' Then,

$\Sigma \mathrm{x}_{\mathrm{i}}^{2}=2080-8^{2}+12^{2}=2160$

$\therefore$ Variance of removing observations $\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^{2}}{20}-\left(\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}\right)^{2}$

$\Rightarrow \frac{2160}{20}-(10.2)^{2}$

$\Rightarrow 108-104.04$

$\Rightarrow 3.96$

Correct standard deviation

$=\sqrt{3.96}$