Mathematics - Statistics Question with Solution | TestHub

We have,

$X=\left\{11,12,13,\dots .,40,41\right\}$

$\Rightarrow n\left(X\right)=31$

$Y=\left\{61,62,63,...,90,91\right\}$

$\Rightarrow n\left(Y\right)=31$

So,

$\overline{x}=\frac{{\sum }_{\mathrm{i}=11}^{41}\mathrm{i}}{31}=\frac{1}{31}\left[\frac{41\times 42}{2}-\frac{10\times 11}{2}\right]=26$

$\overline{y}=\frac{{\sum }_{\mathrm{i}=61}^{91}\mathrm{i}}{31}=\frac{1}{31}\left[\frac{91\times 92}{2}-\frac{60\times 61}{2}\right]=73$

Combined mean,

 $\mu =\frac{31\times 26+31\times 76}{31+31}$

$\Rightarrow \mu =\frac{26+76}{2}=51$

Also,

${\sigma }^2=\frac{1}{62}\times \left(\sum _{i=1}^{31}{\left(x_{\mathrm{i}}-\mu \right)}^2+\sum _{i=1}^{31}{\left(y_{\mathrm{i}}-\mu \right)}^2\right)$

Since, $x_{\mathrm{i}}\in X$ are in A.P. with $31$ elements  common difference $1$ , same is ${\mathrm{y}}_{\mathrm{i}}\in \mathrm{y}$, when written in increasing order.

$\therefore \sum _{\mathrm{i}=1}^{31}{\left({\mathrm{x}}_{\mathrm{i}}-\mu \right)}^2=\sum _{\mathrm{i}=1}^{31}{\left({\mathrm{y}}_{\mathrm{i}}-\mu \right)}^2={10}^2+{11}^2+\dots .+{40}^2$

$={10}^2+{11}^2+\dots .+{40}^2$

$=\frac{40\times 41\times 81}{6}-\frac{9\times 10\times 19}{6}=21855$

Hence, ${\sigma }^2=\frac{21855}{62}+\frac{21855}{62}=705$

$\therefore \left|\overline{\mathit{x}}+\overline{\mathit{y}}-{\sigma }^2\right|=|26+76-705|=603$