Question

If the S.D. of $x_1,x_2,.....x_n$is $5$, then the S.D. of $x_1+5,x_2+5,x_3+5,....x_n+5$ is

• A. $0$
• B. $10$
• C. $5$
• D. cannot say

Answer 1

The correct option is C $5$

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