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 x_i 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 \(√144=12\)