QJEEM24Maf43e: JEE Main 2024 Mathematics - Class 11 JEEM24Maf43e Solution: The mean and standard deviation of 15 observations were found…

Let the incorrect mean be $\mu^{\prime}$ and standard deviation be $\sigma^{\prime}$

We have

$\mu^{\prime}=\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{15}=12 \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}=180$

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

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

Also

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

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

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

Required value

$\begin{aligned}& =15\left(\mu+\mu^2+\sigma^2\right) \\& =15\left(\frac{182}{15}+\frac{182 \times 182}{15 \times 15}+\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}\right) \\& =15\left(\frac{182}{15}+\frac{2339}{15}\right) \\& =2521\end{aligned}$