Question

The mean and standard deviation of six observation 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

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 $4$.
$\Rightarrow$ Mean $=\overline{x}=\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}=8$ ....(1)
If each observation is multiplied by $3$ and the resulting observation are $y_i$, then
$y_i=3x_i,\text{for}i=1\text{to}6$
$\therefore$ New mean $\overline{y}=\frac{y_1+y_2+y_3+y_4+y_5+y_6}{6}$
$=\frac{3(x_1+x_2+x_3+x_4+x_5+x_6)}{6}$
$=3\times 8$ ........[Using (1) ]
$=24$
Standard deviation $\sigma =\sqrt{\frac{1}{n}\sum _{i=l}^6(x_i-\overline{x}{)}^2}$
$\therefore {\left(4\right)}^2=\frac{1}{6}\sum _{i=l}^6(x_i-\overline{x}{)}^2$
$\sum _{i=l}^6{(x_1-\overline{x})}^2=96$.......(2)
From (1) and (2) it can be observed that
$\overline{y}=3\overline{x}$
$\overline{x}=\frac{1}{3}\overline{y}$
Substituting the values of $x_i$ and $\overline{x}$ in (2) we obtain
$\sum _{i=l}^6{(\frac{1}{3}{y}_i-\frac{1}{3}\overline{y})}^2=96$
$\Rightarrow \sum _{i=l}^6{(y_i-\overline{y})}^2=864$
Therefore, variance of new observation $=$ $(\frac{1}{6}\times 864)=144$
Hence, the standard deviation of new observations is $\sqrt{144}=12$