Question

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.

Answer 1

Let the observations be x1, x2, x3, x4, x5, and x6.

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 yi, then

$y_i=3x_i,i.e,x_1=\frac{1}{3}y_i,fori=1\,to\,6$

$New mean,\bar{y}\frac{y_1+y_2+y_3+y_4+y_5+y_6}{6}$

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

$3×8$ $....[(Using(1)]$

$=24$

$Standard\,deviation\,σ=√\frac{1}{n}\sum_{ti1}^6(x_i-\bar{x})^2$

$\sum_{i=1}^6(x_i-\bar{x})^2=96$ $....(2)$

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 xi and $\bar{x}$ in (2), we obtain

$\sum_{i=1}^6(\frac{1}{3}y_i-\frac{1}{3}\bar{y})^2=96$

$\sum_{i=1}^6(y_i-\bar{y})^2=864$

Therefore, variance of new observations = $(\frac{1}{6}×864)=144$

Hence, the standard deviation of new observations is $√144=12$