Question

Consider three finite sets given as
$S_1=\{2,5,8,11,\dots \text{up to}100\text{terms}\}$
$S_2=\{1,10,19,28,\dots \text{up to}100\text{terms}\}$
$S_3=\{4,10,16,22,\dots \text{up to}100\text{terms}\}$

If ${\sigma }_1^2,{\sigma }_2^2$ and ${\sigma }_3^2$ are the variance of $S_1,S_2$ and $S_3$ respectively, then the value of $\frac{{\sigma }_1^2+{\sigma }_2^2-{\sigma }_3^2}{{\sigma }_1^2}$ is

• A. $3$
• B. $6$
• C. $8$
• D. $12$

Answer 1

The correct option is B $6$

$S_1=\{2,5,8,11,\dots \text{up to}100\text{terms}\}$
$S_2=\{1,10,19,28,\dots \text{up to}100\text{terms}\}$
$S_3=\{4,10,16,22,\dots \text{up to}100\text{terms}\}$

Let $n^{\text{th}}$ term of $S_1$ be $t_n$.
Then, $n^{\text{th}}$ term of $S_2$ will be $3t_n-5$.
And $n^{\text{th}}$ term of $S_3$ will be $2t_n$.

We know that,
$\text{Var}(aX+b)=a^2\text{Var}\left(X\right)$
$\Rightarrow {\sigma }_2^2=\left(3^2\right){\sigma }_1^2=9{\sigma }_1^2$
and ${\sigma }_3^2=\left(2^2\right){\sigma }_1^2=4{\sigma }_1^2$

$\therefore \frac{{\sigma }_1^2+{\sigma }_2^2-{\sigma }_3^2}{{\sigma }_1^2}=\frac{{\sigma }_1^2+9{\sigma }_1^2-4{\sigma }_1^2}{{\sigma }_1^2}=6$