Question

The variance of 20 observations is 5 . If each observation is multiplied by 2 , then the new variance of the resulting observations is

• A. $2^3\times 5$
• B. $2^2\times 5$
• C. $2\times 5$
• D. $2^4\times 5$

Answer 1

Answer: (B)

Let the observations be $x_1,x_2,\dots x_{20}$ and $\bar{x}$ be their mean. Given that, variance $=5$ and $n=20$. We know that,
Variance$\left({\sigma }^2\right)=\frac{1}{n}\sum _{i=1}^{20}{\left(x_i-\bar{x}\right)}^2$.
i.e., $5=\frac{1}{20}\sum _{i=1}^{20}{\left(x_i-\bar{x}\right)}^2$ or $\sum _{i=1}^{20}{\left(x_i-\bar{x}\right)}^2=\mathrm{100.....}$(i)
If each observation is multiplied by 2 and the new resulting observations are $y_i$, then
$y_i=2x_i\text{ i.e., }{x}_i=\frac{1}{2}{y}_i$
Therefore, $\bar{y}=\frac{1}{n}\sum _{i=1}^{20}{y}_i=\frac{1}{20}\sum _{i=1}^{20}2x_i=2\cdot \frac{1}{20}\sum _{i=1}^{20}{x}_i$
i.e., $\bar{y}=2\bar{x}$ or $\bar{x}=\frac{1}{2}\bar{y}$
On substituting the values of $x_i$ and $\bar{x}$ in Eq. (i), we get
$\sum _{i=1}^{20}{\left(\frac{1}{2}{y}_i-\frac{1}{2}\bar{y}\right)}^2=100$
i.e., $\sum _{i=1}^{20}{\left(y_i-\bar{y}\right)}^2=400$
Thus, the variance of new observations
$=\frac{1}{20}\times 400=20=2^2\times 5$
Note The reader may note that, if each observation is multiplied by a constant $k$, the variance of the resulting observations becomes $k^2$ times the original variance.