$The mean and s\tan dard deviation of 6 observations are 8 and 4 respectively. If each observation is multiplied by 3, find the new s\tan dard deviation of the resulting observations.$

$\begin{array}{l}\text{ Let the observations be }{x}_1,x_2,x_3,x_4,x_5\text{ and }{x}_6,\text{ so their mean }\overline{\text{x}}=\frac{{\displaystyle \sum _{\text{i}=1}^6{\text{x}}_{\text{i}}}}{6}=8\Rightarrow \sum _{\text{i}=1}^6{\text{x}}_{\text{i}}=48\\ \text{ On multiplying each observation by }3,\text{ we get the new observations as }3x_1,3x_2,3x_3,3x_4,3x_5\text{ and }3x_6\text{ . }\\ \text{ Now, their mean }=\overline{\text{x}}=\frac{{\displaystyle \sum _{\text{i}=1}^{6}3{\text{x}}_{\text{i}}}}{6}=\frac{3\times 48}{6}=24\\ \text{ Variance of new observations }=\frac{{\displaystyle \sum _{i=1}^6{\left(3x_i-24\right)}^2}}{6}=\frac{3^2{\displaystyle \sum _{i=1}^6{\left(x_i-8\right)}^2}}{6}\\ =\frac{9}{1}\times \text{ Variance of old observations }=9\times 4^2=144\\ \text{ Thus, standard deviation of new observations }=\sqrt{\text{ Variance }}=\sqrt{144}=12\end{array}$