We are given that,
The mean of six observations $ = 8$.
The standard deviation of six observations $ = 4$.
Let us consider the six observations as ${x_1}$, ${x_2}$, ${x_3}$, ${x_4}$, ${x_5}$ and ${x_6}$.
The mean of six observations is given by,
$\bar x = \frac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = 8$
When each observation is multiplied by $3$ then,
\[{y_i} = 3{x_i}\]
So, \[{x_i} = \frac{1}{3}{y_i}\], for $i = 1$ to $6$ …… (1)
Now, the new mean becomes,
$\bar y = \frac{{{y_1} + {y_2} + {y_3} + {y_4} + {y_5} + {y_6}}}{6}$
$\bar y = \frac{{3\left( {{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}} \right)}}{6}$
$\bar y = 3 \times 8$
$\bar y = 24$
Also, we have Standard deviation,
$\sigma = \sqrt {\frac{1}{{\text{n}}}\sum\limits_{{\text{i}} = 1}^6 {{{\left( {{{\text{x}}_{\text{i}}} – \overline {\text{x}} } \right)}^2}} } $
Substituting the values and squaring on both sides,
${\sigma ^2} = \frac{1}{6}\sum\limits_{{\text{i}} = 1}^6 {{{\left( {{{\text{x}}_{\text{i}}} – \overline {\text{x}} } \right)}^2}} $
${\left( 4 \right)^2} = \frac{1}{6}\sum\limits_{{\text{i}} = 1}^6 {{{\left( {{{\text{x}}_{\text{i}}} – \overline {\text{x}} } \right)}^2}} $
$\sum\limits_{{\text{i}} = 1}^6 {{{\left( {{{\text{x}}_{\text{i}}} – \overline {\text{x}} } \right)}^2}} = 96$ …… (2)
From equations (1) and (2),
$\sum\limits_{{\text{i}} = 1}^6 {{{\left( {\frac{1}{3}{y_{\text{i}}} – \frac{1}{3}\overline y } \right)}^2}} = 96$
$\sum\limits_{{\text{i}} = 1}^6 {\left( {\frac{1}{3}{y_{\text{i}}} – \frac{1}{3}\overline y } \right)} = 864$
So, the new variance of the resulting observation $ = \frac{1}{6} \times 864$
$ = 144$
Hence, the new standard deviation of the resulting observation $ = \sqrt {144} $
$ = 12$