Let $\mathrm{X}=\{11,12,13,\dots \dots ,40,41\}\text{ and }\mathrm{Y}=\{61,62,63,\dots \dots ,90,91\}$be the two sets of
observations. If $\overline{\mathrm{x}}\text{ and }\overline{\mathrm{y}}$ are their respective means and ${\sigma }^2$ is the variance of all the
observations in $\mathrm{X}\cup \mathrm{Y}\text{, then }\left|\overline{\mathrm{x}}+\overline{\mathrm{y}}-{\mathrm{\sigma }}^2\right|$is equal to __________.

$\begin{array}{l}\overline{\mathrm{x}}=\frac{\frac{31}{2}(11+41)}{31}=\frac{1}{2}\times 52=26\overline{\mathrm{y}}=\frac{\frac{31}{2}(61+91)}{31}=\frac{1}{2}\times 152=76\\ {\mathrm{\sigma }}^2=\frac{\sum {\mathrm{x}}_{\mathrm{i}}^2+\sum {\mathrm{y}}_{\mathrm{i}}^2}{62}-{\left(\frac{\sum {\mathrm{x}}_{\mathrm{i}}^2+\sum {\mathrm{y}}_{\mathrm{i}}^2}{62}\right)}^2\\ =\frac{\left({11}^2+{12}^2+\dots .+{41}^2\right)+\left({61}^2+{62}^2+\dots .+{91}^2\right)}{62}-(51{)}^2\\ =\frac{\left(\frac{41.42.83}{63}-\frac{10.11.21}{6}\right)+\left(\frac{91.92.183}{6}-\frac{60.61.121}{6}\right)}{62}-(51{)}^2\\ =\frac{(41.7.83-11.35)+(91.46.61-10.61.121)}{62}-(51{)}^2\\ =\frac{23436+181536}{62}-(51{)}^2=3306-2601=705\\ \overline{\mathrm{x}}+\overline{\mathrm{y}}-{\mathrm{\sigma }}^2=26+76-705=-603\end{array}$