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 \(√144=12\)