Question

If the mean and variance of the following data: $6,10,7,13, a, 12, b, 12$ are 9 and $\frac{37}{4}$ respectively, then $(a-b)^{2}$ is equal to:

• A. 24
• B. 12
• C. 32
• D. 16

Answer 1

Answer: (D)

Mean $=\frac{6+10+7+13+a+12+b+12}{8}=9$
$60+a+b=72$
$a +b=12 \ldots(1)$
variance $=\frac{\sum x_{i}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2}=\frac{37}{4}$
$\sum x_{i}^{2}=6^{2}+10^{2}+7^{2}+13^{2}+a^{2}+b^{2}+12^{2}+12^{2}$
$=a^{2}+b^{2}+642$
$\frac{a^{2}+b^{2}+642}{8}-(9)^{2}=\frac{37}{4}$
$\frac{a^{2}+b^{2}}{8}+\frac{321}{4}-81=\frac{37}{4}$
$\frac{a^{2}+b^{2}}{8}=81+\frac{37}{4}-\frac{321}{4}$
$\frac{a^{2}+b^{2}}{8}=81-71$
$\therefore a^{2}+b^{2}=80 \ldots \ldots .(2)$
From (1) $a^{2}+b^{2}+2 a b=144$
$80+2 a b=144$
$\therefore 2 a b=64$
$(a-b)^{2}=a^{2}+b^{2}-2 a b=80-64=16$