Question

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

Answer 1

$$\begin{gathered} \mathrm{Mean}=\overline{\mathit{X}}\mathit{}=8 \\ n=6 \\ \mathrm{\sigma }=\mathrm{S}.\mathrm{D}=4 \\ If{x}_{1,}{x}_{2....}{x}_6\mathrm{are}\mathrm{the}\mathrm{given}\mathrm{observations} \\ \overline{X}=\frac{1}{n}\times \sum _{i=1}^6{x}_i \\ \Rightarrow 8=\frac{1}{6}\times \sum _{i=1}^6{x}_i \\ \mathrm{Let}{u}_{1,}{u}_2.....u_6\mathrm{be}\mathrm{the}\mathrm{new}\mathrm{observations} \\ \Rightarrow u_{\mathrm{i}}=3x_{\mathrm{i}}\left(\mathrm{for}i=1,2,3...6\right) \\ \Rightarrow \mathrm{Mean}\mathrm{of}\mathrm{new}\mathrm{observations}=\overline{\mathrm{U}}=\frac{1}{n}\times \underset{i=1}{\overset{6}{\sum u_i}} \\ =\frac{1}{6}\times \underset{i=1}{\overset{6}{\sum 3x_i}} \\ =3\times \frac{1}{6}\times \sum _{i=1}^6{x}_i \\ =3\overline{X} \\ =3\times 8 \\ =24 \end{gathered}$$

Thus, mean of the new observations is 24.


$$\begin{gathered} \mathrm{SD}={\mathrm{\sigma }}_{\mathrm{x}}=4 \\ {{\mathrm{\sigma }}_{\mathrm{x}}}^2=\mathrm{Variance}\mathrm{X} \\ \therefore \mathrm{Variance}X=16 \\ \Rightarrow \frac{1}{6}\sum _{i=1}^6{\left(x_i-\overline{X}\right)}^2=16...\left(1\right) \\ \mathrm{Variance}\left(\mathrm{U}\right)={{\sigma }_{\mathrm{u}}}^2=\frac{1}{6}\sum _{i=1}^6{\left(u_i-\overline{U}\right)}^2 \\ =\frac{1}{6}\times \sum _{i=1}^6{\left(3x_i-3\overline{X}\right)}^2 \\ =3^2\times \frac{1}{6}\sum _{i=1}^6{\left(x_i-\overline{X}\right)}^2 \\ =9\times 16 \\ {\mathrm{\sigma }}_u=\sqrt{\mathrm{Variance}\left(U\right)} \\ =\sqrt{9\times 16} \\ =12 \end{gathered}$$

Thus, standard deviation of the new observations is 12.