Let the observations be $x_{1}, x_{2}, \ldots, x_{20}$ 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\limits_{i = 1}^{20} {{{\left( {{x_i} – \bar x} \right)}^2}} $

i.e., $5 = \frac{1}{{20}}\sum\limits_{i = 1}^{20} {{{\left( {{x_i} – \bar x} \right)}^2}} $

or    $\sum\limits_{i = 1}^{20} {{{\left( {{x_i} – \bar x} \right)}^2}}  = 100$        …….$(1)$

If each observation is multiplied by $2,$ and the new resulting observations are $y_{i},$ then

$y_{i}=2 x_{i} \text { i.e., } x_{i}=\frac{1}{2} y_{i}$

Therefore $\bar y = \frac{1}{n}\sum\limits_{i = 1}^{20} {{y_i}}  = \frac{1}{{20}}\sum\limits_{i = 1}^{20} {2{x_i} = 2.\frac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}} } $

i.e.  $\bar{y}=2 \bar{x} \quad$ or $\quad \bar{x}=\frac{1}{2} \bar{y}$

Substituting the values of $x_{i}$ and $\bar{x}$ in $(1),$ we get

${\sum\limits_{i = 1}^{20} {\left( {\frac{1}{2}{y_i} – \frac{1}{2}\bar y} \right)} ^2} = 100$ i.e.,  $\sum\limits_{i = 1}^{20} {{{\left( {{y_i} – \bar y} \right)}^2} = 400} $

Thus the variance of new observations $=\frac{1}{20} \times 400=20=2^{2} \times 5$