Measures of Dispersion — Maths STD 11 Science — Question

Combined mean $\left(\bar{x}_{ e }\right)=\frac{ n _x \bar{x}+ n _{ y } \bar{y}}{ n _x+ n _{ y }}=\frac{20(13)+30(17)}{20+30}$

$\begin{aligned} & =\frac{260+510}{50} \\ & =\frac{770}{50}=15.4\end{aligned}$

Now, $d _x=\bar{x}-\bar{x}_c=13-15.4=-2.4$

$d _y=\bar{y}-\bar{x}_{ c }=17-15.4=1.6$

$\therefore \quad$ Combined standard deviation $\left(\sigma_c\right)=\sqrt{\frac{ n _x\left(\sigma_x{ }^2+ d _x{ }^2\right)+ n _y\left(\sigma_y{ }^2+ d _y{ }^2\right)}{ n _x+ n _y}}$

$\begin{aligned} & =\sqrt{\frac{20\left[3^2+(-2.4)^2\right]+30\left(2^2+1.6^2\right)}{20+30}} \\ & =\sqrt{\frac{20(9+5.76)+30(4+2.56)}{50}} \\ & =\sqrt{\frac{20(14.76)+30(6.56)}{50}} \\ & =\sqrt{\frac{295.2+196.8}{50}}\end{aligned}$

$=\sqrt{\frac{492}{50}}=\sqrt{9.84}=3.14$