Question

If mean and variance for the following series of numbers:
$15,30,a,25,27,b,13,20$ is $20\text{and}39.5$ then the value of $\sqrt[3]{3a}b$ is

• A. $216$
• B. $36$
• C. $6$
• D. $9$

Answer 1

The correct option is C $6$

$\text{Mean}=\frac{\sum x_i}{n}=\frac{15+30+a+25+27+b+13+20}{8}$
$\therefore \overline{X}=\frac{130+a+b}{8}=20$
$\Rightarrow a+b=\mathrm{30...}\left(i\right)$
$\text{Variance}=\frac{\sum _{i=1}^n(x_i-\overline{X}{)}^2}{n}=39.5$
${\sigma }^2=\frac{(20-a{)}^2+(20-b{)}^2+248}{8}$
${\sigma }^2=\frac{a^2+b^2-40(a+b)+1048}{8}$
$\Rightarrow 39.5\times 8=a^2+b^2-40\times 30+1048$
$\Rightarrow a^2+b^2=468...\left(ii\right)$
$\because (a+b{)}^2=(a^2+b^2+2ab)\Rightarrow 2ab={30}^2-468\Rightarrow ab=\frac{432}{2}=216$
$\therefore \sqrt[3]{3a}b=\sqrt[3]{3216}=6$