The mean and standard deviation of six observations are 8 and 4, respectively.

Solution:

Let the observations be $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$, and $x_{6}$.

It is given that mean is 8 and standard deviation is 4.

Mean, $\bar{x}=\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}}{6}=8$

If each observation is multiplied by 3 and the resulting observations are yi, then

$y_{i}=3 x_{i}$ i.e., $x_{i}=\frac{1}{3} y_{,}$, for $i=1$ to 6

$\therefore$ New mean, $y=\frac{y_{1}+y_{2}+y_{3}+y_{4}+y_{5}+y_{6}}{6}$

$=\frac{3\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}\right)}{6}$

$=3 \times 8$ ...[Using (1)]

$=24$

Standard deviation, $\sigma=\sqrt{\frac{1}{n} \sum_{i=1}^{6}\left(x_{i}-\bar{x}\right)^{2}}$

$\therefore(4)^{2}=\frac{1}{6} \sum_{j=1}^{6}\left(x_{i}-\bar{x}\right)^{2}$

$\sum_{i=1}^{6}\left(x_{i}-\bar{x}\right)^{2}=96$ $\ldots(2)$

From (1) and (2), it can be observed that,

$\bar{y}=3 \bar{x}$

$\bar{x}=\frac{1}{3} \bar{y}$

Substituting the values of $x_{i}$ and $\bar{x}$ in (2), we obtain

$\sum_{i=1}^{6}\left(\frac{1}{3} y_{i}-\frac{1}{3} \bar{y}\right)^{2}=96$

$\Rightarrow \sum_{i=1}^{6}\left(y_{i}-\bar{y}\right)^{2}=864$

Therefore, variance of new observations $=\left(\frac{1}{6} \times 864\right)=144$

Hence, the standard deviation of new observations is $\sqrt{144}=12$