Given that \(\bar{x}\) is the mean and \(σ2\) is the variance of n observations x1,x2,...,xn. Prove that the mean and variance of the observations ax1,ax2,ax3,...,axn are \(a\bar{x}\) and \(a^2σ2\), respectively,\((a ≠ 0).\)
The given n observations are x1, x2…xn
\(Mean=\bar{x}\)
\(Varience= if^2\)
\(σ^2=\frac{1}{n}\sum_{i=1}^n(x_i-x-\bar{x})^2\,..........(1)\)
\(3×8\)
\(=24 ....[(Using(1)]\)
\(Standard\,deviation\,σ=√\frac{1}{n}\sum_{ti1}^6(x_i-\bar{x})^2\)
If each observation is multiplied by a and the new observations are yi, then
\(y_i=ax,\,i.e.,\,x_i=\frac{1}{a}y_1\)
∴ \(\bar{y}\frac{1}{n}\sum_{i=1}^ny_i=\frac{1}{2}\sum_{i=1}^nax_i=\frac{a}{n}\sum_{i=1}^nx_i=a\bar{x}\,\,(\bar{x}=\frac{1}{n}\sum_{i=1}^nx_i)\)
Therefore, mean of the observations, ax1, ax2 ….axn, is \(a\bar{x}\),
Substituting the values of x and \(\bar{x}\) in (1), we obtain
\(σ^2=\frac{1}{2}\sum_{i=1}^n(\frac{1}{a}y_i-\frac{1}{a}y_i)^2\)
⇒ \(σ^2σ^2=\frac{1}{n}\sum_{i=1}^n(y_i-\bar{y})^2\)
Thus, the variance of the observations, ax1, ax2…..axn, is a2 σ2 .