JEE Main 2024 — 27 January, Shift 2 — Question 21

Let the incorrect mean be $\mu^{\prime}$ and standard deviation be $\sigma^{\prime}$ We have $\mu^{\prime}=\frac{\sum\mathrm{x}_{\mathrm{i}}}{15}=12\Rightarrow\Sigma\mathrm{x}_{\mathrm{i}}=180$

As per given information correct $\Sigma \mathrm{x}_{\mathrm{i}}=180-10+12$

$\Rightarrow \mu($ correct mean $)=\frac{182}{15}$

Also $\sigma^{\prime}=\sqrt{\frac{\sum \mathrm{x}_{\mathrm{i}}{ }^{2}}{15}-144}=3 \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}{ }^{2}=2295$

Correct $\Sigma \mathrm{x}_{\mathrm{i}}^{2}=2295-100+144=2339$

$\sigma^{2}($ correct variance $)=\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}$

Required value $=15\left(\mu+\mu^{2}+\sigma^{2}\right)$

$=15\left(\frac{182}{15}+\frac{182 \times 182}{15 \times 15}+\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}\right)$

$=15\left(\frac{182}{15}+\frac{2339}{15}\right)$

$=2521$