The mean and standard deviation of $15$ observations were found to be $12$ and $3$ respectively. On rechecking it was found that an observation was read as $10$ in place of $12.$ If $\mu$ and ${\sigma }^2$ denote the mean and variance of the correct observations respectively, then $15\left(\mu +{\mu }^2+{\sigma }^2\right)$ is equal to _________.

Given: Incorrect mean, $\overline{x}=12$ and incorrect standard deviation, ${\mathrm{\sigma }}^{\text{'}}=3$.

$\Rightarrow \frac{{\mathrm{\Sigma x}}_{\mathrm{i}}}{15}=12$

$\Rightarrow {\mathrm{\Sigma x}}_{\mathrm{i}}=180$

Correct mean is given by,

$\mathrm{\mu }=\frac{{\mathrm{\Sigma x}}_{\mathrm{i}}-10+12}{15}$

$\Rightarrow \mathrm{\mu }=\frac{182}{15}$

Also, ${\mathrm{\sigma }}^{\text{'}}=\sqrt{\frac{\sum {\mathrm{x}}_{\mathrm{i}}^2}{15}-144}=3$

$\Rightarrow \frac{\sum {\mathrm{x}}_{\mathrm{i}}^2}{15}=9+144$

$\Rightarrow \sum {\mathrm{x}}_{\mathrm{i}}^2=2295$

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

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

$\Rightarrow 15\left(\mathrm{\mu }+{\mathrm{\mu }}^2+{\mathrm{\sigma }}^2\right)=15\left(\frac{182}{15}+\frac{182\times 182}{15\times 15}+\frac{2339}{15}-\frac{182\times 182}{15\times 15}\right)$

$\Rightarrow 15\left(\mathrm{\mu }+{\mathrm{\mu }}^2+{\mathrm{\sigma }}^2\right)=15\left(\frac{182}{15}+\frac{2339}{15}\right)$

$\Rightarrow 15\left(\mathrm{\mu }+{\mathrm{\mu }}^2+{\mathrm{\sigma }}^2\right)=2521$