If the mean and variance of the following data: 6, 10, 7, 13, a, 12, b, 12

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

$60+a+b=72$

$a+b=12$          ......(1)

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

$\Sigma x_{i}^{2}=6^{2}+10^{2}+7^{2}+13^{2}+a^{2}+b^{2}+12^{2}+12^{2}$

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

$\frac{\mathrm{a}^{2}+\mathrm{b}^{2}+642}{8}-(9)^{2}=\frac{37}{4}$

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

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

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

$\therefore \mathrm{a}^{2}+\mathrm{b}^{2}=80$      ....(2)

From (1) $a^{2}+b^{2}+2 a b=144$

$80+2 a b=144 \quad \therefore 2 a b=64$

$(a-b)^{2}=a^{2}+b^{2}-2 a b=80-64=16$