Question

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,$(a ≠ 0).$

Answer 1

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 .