Question

The mean and variance of seven observations are 8 and 16 respectively. If each observation is multiplied by 3, find the new mean and new variance of the resulting observations.

Answer 1

Let the given observations be $x_1,x_2,x_3,x_4,x_5,x_6$.

Then, mean $=8\Rightarrow \frac{1}{6}(x_1+x_2+x_3+x_4+x_5+x_6)=8$

$\Rightarrow x_1+x_2+x_3+x_4+x_5+x_6=48$ . . . (i)

Also, variance = 16

$\Rightarrow \frac{1}{6}(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2)-8^2=16[\because {\sigma }^2=\frac{\sum x_i^2}{n}-(\overline{x}{)}^2]$

$\Rightarrow x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2=480$ . . . (ii)

When each observation is multiplied by 3, then new observations are $3x_1,3x_2,3x_3,3x_4,3x_{5}and3x_6$.

$\therefore$ $newmean$ $=\frac{1}{6}(3x_1+3x_2+3x_3+3x_4+3x_5+3x_6)$

$=\frac{3}{6}(x_1+x_2+x_3+x_4+x_5+x_6)=(\frac{1}{2}\times 48)=24$ [using (i)]

$\therefore$ new variance $=\frac{(3x_1{)}^2+(3x_2{)}^2+(3x_3{)}^2+(3x_4{)}^2+(3x_5{)}^2+(3x_6{)}^2}{6}-(24{)}^2$

$=\frac{9}{6}(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2)-576$

$=(\frac{9}{6}\times 480)-576=(720-576)=144$ [using (ii)].

Hence, new mean = 24 and new variance = 144.