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Q.
The variance of 20 observations is 5. If each observation is multiplied by 2, then the new variance of the resulting observation is
Statistics
Solution:
Let the observations be $x_1, x_2, ...., x_{20}$ and $\bar{x}$ be their mean. Given that, variance = 5 and n = 20. We know that,
Variance $\left(\sigma\right)^{2}=\frac{1}{n} \displaystyle \sum_{i=1}^{20}(x_i-\bar{x})$
i.e. $5=\frac{1}{20}\displaystyle \sum_{i=1}^{20}(x_i-\bar{x})^2$ or $\displaystyle \sum_{i=1}^{20}(x_i-\bar{x})^2=100\,...(i)$
If each observation is multiplied by 2 and the new resulting observations are $y_i$, then
$y_{i}=2x_{i}$ i.e., $x_{i}=\frac{1}{2}y_{i}$
Therefore, $\bar{y}=\frac{1}{n}\displaystyle \sum_{i=1}^{20}y_i=\frac{1}{20}\displaystyle \sum_{i=1}^{20}2x_i=2.\frac{1}{20}$$\displaystyle \sum_{i=1}^{20}x_i$
i.e., $\bar{y}=2\bar{x}$ or $\bar{x}\frac{1}{2}\bar{y}$
On substituting the values of $x_i$ and $\bar{x}$ in eq. (i), we get
$\displaystyle \sum_{i=1}^{20}$$\left(\frac{1}{2}y_{i}-\frac{1}{2}\bar{y}\right)^{^2}=100$ i.e. $\displaystyle \sum_{i=1}^{20}(y_i-\bar{y})^2=400$
Thus, the variance of new observations
$=\frac{1}{20}\times400=20=2^{2}\times5$