Question

Standard Deviation of $n$ observation $a_1,a_2,a_3,........,a_n$ is $\sigma .$ Then the standard deviation of the observation $\lambda a_1,\lambda a_2,........,\lambda a_n$ is

• A. $\lambda \sigma$
• B. $\lambda \sigma$
• C. $|\lambda |\sigma$
• D. ${\lambda }^n\sigma$

Answer 1

The correct option is C $|\lambda |\sigma$

Variance is,
${\sigma }^2=\frac{\sum _{i=1}^n{x}_i^2}{n}-(\overline{x}{)}^2$
New mean,
$\overline{x_1}=\lambda \overline{x}$
The new variance will be,
${\sigma }_1^2=\frac{\sum _{i=1}^n(\lambda x_i{)}^2}{n}-(\lambda \overline{x}{)}^2\Rightarrow {\sigma }_1^2={\lambda }^2(\frac{\sum _{i=1}^n{x}_i^2}{n}-(\overline{x}{)}^2)\Rightarrow {\sigma }_1^2={\lambda }^2{\sigma }^2\Rightarrow {\sigma }_1=|\lambda |\sigma$

So, if the standard deviation of $n$ observation $a_1,a_2,a_3,........,a_n$ is $\sigma ,$ then the standard deviation of the observation $\lambda a_1,\lambda a_2,........,\lambda a_n$ is $|\lambda |\sigma$