Question

Let $X$ be a variate taking values $x_1,x_2,.....x_n$ and $Y$ be a variate taking values $y_1,y_2,.....y_n$ such that $y_i=6x_i+3;i=1,2,....n$. If $Var\left(Y\right)=30$, then ${\sigma }_X$ is equal to

• A. $\frac{5}{\sqrt{6}}$
• B. $\sqrt{\frac{5}{6}}$
• C. $30$
• D. $\sqrt{30}$

Answer 1

The correct option is B $\sqrt{\frac{5}{6}}$

If the S.D. of variate X is .
$\sigma \left(X\right)=\sqrt{\frac{{\sum }_{i=1}^N(X_i-\overline{X}{)}^2}{N}}$
$\sigma (aX+b)=\sqrt{\frac{{\sum }_{i=1}^N(x_i-\overline{x}{)}^2}{N}}$
= $\sqrt{\frac{{\left[\right(a{X}_1+b)-(a\overline{X}+b\left)\right]}^2+{\left[\right(a{X}_2+b)-(a\overline{X}+b\left)\right]}^2+....+{\left[\right(a{X}_N+b)-(a\overline{X}+b\left)\right]}^2}{N}}$
= $\sqrt{\frac{a^2[{(X_1-\overline{X})}^2+{(X_2-\overline{X})}^2+...+{(X_N-\overline{X})}^2]}{N}}$
= $\left|a\right|\sqrt{\frac{{\sum }_{i=1}^N{(X_i-\overline{X})}^2}{N}}$
= $\left|a\right|\sigma \left(X\right)$
Given $Var\left(Y\right)=30=>{\left(\sigma \right(Y\left)\right)}^2=30=>\sigma \left(Y\right)=\sqrt{30}=>\sigma (6X+3)=\sqrt{30}$
$\sigma \left(Y\right)=\sigma (aX+b)=\left|a\right|\sigma \left(X\right)=>\sigma (6X+3)=6\times \sigma \left(X\right)=>\sqrt{30}=6\times \sigma \left(X\right)=>\sigma \left(X\right)=\frac{\sqrt{30}}{6}=>\sigma \left(X\right)=\sqrt{\frac{5}{6}}$