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