If the mean and variance of the data $65,68,58,44$, $48,45,60,\alpha ,\beta ,60$ where $\alpha >\beta$ are $56$ and $66.2$ respectively, then ${\alpha }^2+{\beta }^2$ is equal to

Given: $\overline{\mathrm{x}}=56$ and ${\mathrm{\sigma }}^2=66.2$

We know that, ${\sigma }^2=\frac{{\displaystyle \sum _{}^{}}{\left(x_i\right)}^2}{n}-{\left(\overline{x}\right)}^2$

$\Rightarrow \frac{{65}^2+{68}^2+{58}^2+{44}^2+{48}^2+{45}^2+{60}^2+{\alpha }^2+{\beta }^2+{60}^2}{10}-{\left(56\right)}^2=66.2$

$\Rightarrow \frac{{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+25678}{10}-(56{)}^2=66.2$

$\Rightarrow \frac{{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+25678}{10}=\left(3136+66.2\right)$

$\Rightarrow {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=32022-25678$

$\Rightarrow {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=6344$