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

Given, $\overline{x}$ = 8 and ${\sigma }^2=4$ and n = 6

Let the observation are $x_1,x_2,x_3,x_4,x_5,x_6.$ Then ,

Mean $\overline{x}=8$

i.e., Mean $=\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}=8$

$\therefore x_1,x_2,x_3,x_4,x_5,x_6.=48$

Now, if and each observation is multiplied by 3, then mean is

$3x_1+3x_2,3x_3,3x_4,3x_5,3x_6=48\times 3$

$\Rightarrow \sum x_i=144$

Now, new mean $\overline{x}=\frac{144}{6}=24$

Since, Variance =$4^2$ (Given)

i.e., $\frac{\sum x_i^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2=4^2$

$\Rightarrow \frac{\sum x_i^2}{6}-(8{)}^2=4^2\Rightarrow \frac{\sum x_i^2}{6}=16+64\Rightarrow \sum x^2=80\times 6$

$\Rightarrow \sum x^2=480$ ...(i)

Now,

New $\sum x^2=(3x_1{)}^2+(3x_2{)}^2+(3x_3{)}^2+(3x_4{)}^2+(3x_5{)}^2+(3x_6{)}^2$

$=9(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2)$

=$9\times 480$ [From Eq. (i)]

$\therefore \sum x^2=4320$

$\therefore$ New variance $\frac{\sum x^2}{n}-(\overline{x}{)}^2=\frac{4320}{6}-(24{)}^2=720-576=144$

$\therefore$ New standard deviation = $\sqrt{newdeviation}=\sqrt{144}=12$