Mathematics - Statistics Question with Solution | TestHub

Mean $=\frac{6+10+7+13+\mathrm{a}+12+\mathrm{b}+12}{8}=9$

$\Rightarrow 60+a+b=72$

$\Rightarrow a+b=12 ...\left(1\right)$

variance $=\frac{\sum x_i^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2=\frac{37}{4}$

Here, $\Sigma x_i^2=6^2+{10}^2+7^2+{13}^2+a^2+b^2+{12}^2+{12}^2$

$=a^2+b^2+642$

So, variance is $\frac{a^2+b^2+642}{8}-{\left(9\right)}^2=\frac{37}{4}$

$\Rightarrow \frac{a^2+b^2}{8}+\frac{321}{4}-81=\frac{37}{4}$

$\Rightarrow \frac{a^2+b^2}{8}=81+\frac{37}{4}-\frac{321}{4}$

$\Rightarrow \frac{a^2+b^2}{8}=81-71$

$\therefore a^2+b^2=80$

From $\left(1\right)$ $a^2+b^2+2ab=144$

$\Rightarrow 80+2ab=144 \therefore 2ab=64$

So, ${\left(a-b\right)}^2=a^2+b^2-2ab=80-64=16.$