The mean and standard deviation of six 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.
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, $\bar{x}=\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}}{6}=8$ .......$(1)$
If each observation is multiplied by $3$ and the resulting observations are $y_{i},$ then
$y_{1}=3 x_{1}$ i.e., $x_{1}=\frac{1}{3} y_{1},$ for $i=1$ to $6$
New Mean, $\bar{y}=\frac{y_{1}+y_{2}+y_{3}+y_{4}+y_{5}+y_{6}}{6}$
$=\frac{3\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}\right)}{6}$
$=3 \times 8$ .......[ Using $(1)$ ]
$=28$
Standard deviation, $\sigma = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^6 {{{\left( {{x_1} - \bar x} \right)}^2}} } $
$\therefore {\left( 4 \right)^2} = \frac{1}{6}\sum\limits_{i = 1}^6 {{{\left( {{x_i} - \bar x} \right)}^2}} $
$\sum\limits_{i = 1}^6 {{{\left( {{x_i} - \bar x} \right)}^2}} = 96$ ........$(2)$
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 $x_{1}$ and $\bar{x}$ in $(2),$ we obtain
$\sum\limits_{i = 1}^6 {{{\left( {\frac{1}{3}{y_1} - \frac{1}{3}\bar y} \right)}^2} = 96} $
$ \Rightarrow \sum\limits_{i = 1}^6 {{{\left( {{y_1} - \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 $\sqrt{144}=12$
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