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. $32$
• B. $12$
• C. $24$
• D. $16$

Answer 1

The correct option is D $16$

Given: Mean $=9$
$\Rightarrow \frac{6+10+7+13+12+12+(a+b)}{8}=9$
$\Rightarrow 60+(a+b)=72$
$\Rightarrow a+b=12\cdots \left(1\right)$
and variance $=\frac{37}{4}$
$\Rightarrow \frac{a^2+b^2+36+100+49+169+144+144}{8}-81=\frac{37}{4}$
$\Rightarrow a^2+b^2+642-648=74$
$\Rightarrow a^2+b^2=80\cdots \left(2\right)$
We know that
$(a+b{)}^2=a^2+b^2+2ab$
$\Rightarrow 144=80+2ab$
$\Rightarrow 2ab=64$
Now, $(a-b{)}^2=a^2+b^2-2ab$
$\Rightarrow (a-b{)}^2=80-64$
$\therefore (a-b{)}^2=16$