The initial variance is given as:
\[
\sigma^2 = 5
\]
When each observation \( x_i \) is multiplied by a constant \( k = 2 \), the new observations are \( y_i = kx_i \). The relationship between the original variance and the sum of squared deviations is established as:
\[
\sigma^2 = \frac{1}{n} \sum_{i=1}^{20} (x_i - \overline{x})^2
\]
Given \( \sigma^2 = 5 \) and \( n = 20 \):
\[
5 = \frac{1}{20} \sum_{i=1}^{20} (x_i - \overline{x})^2
\]
This leads to:
\[
\sum_{i=1}^{20} (x_i - \overline{x})^2 = 100 \quad \text{(i)}
\]
For the new observations \( y_i = 2x_i \), the mean is \( \overline{y} = 2\overline{x} \). We can express \( x_i \) and \( \overline{x} \) in terms of \( y_i \) and \( \overline{y} \) as \( x_i = \frac{1}{2} y_i \) and \( \overline{x} = \frac{1}{2} \overline{y} \). Substituting these into equation (i):
\[
\sum_{i=1}^{20} \left(\frac{1}{2} y_i - \frac{1}{2} \overline{y}\right)^2 = 100
\]
Simplifying the expression:
\[
\sum_{i=1}^{20} \left(\frac{1}{2}(y_i - \overline{y})\right)^2 = 100
\]
\[
\sum_{i=1}^{20} \frac{1}{4} (y_i - \overline{y})^2 = 100
\]
\[
\frac{1}{4} \sum_{i=1}^{20} (y_i - \overline{y})^2 = 100
\]
\[
\sum_{i=1}^{20} (y_i - \overline{y})^2 = 400
\]
The variance of the new observations \( y_i \) is therefore:
\[
\frac{1}{n} \sum_{i=1}^{20} (y_i - \overline{y})^2 = \frac{1}{20} \times 400 = 20
\]
This new variance can also be expressed as \( 20 = 2^2 \times 5 = k^2 \sigma^2 \).