Q

Given that x bar is the mean and σ^2 is the variance of n observations x1 , x2 , ...,xn . Prove that the mean and variance of the observations ax1 , ax2 , ax3 , ...., axn are a x bar and a^2 σ^2 , respectively, (a ≠ 0).

4.  Given that $\small \bar {x}$ is the mean and  $\small \sigma^2$  is the variance of $\small n$ observations  .Prove that the mean and variance of the observations  $\small ax_1,ax_2,ax_3,....,ax_n$, are  $\small a\bar{x}$ and  $\small a^2 \sigma^2$ respectively, $\small (a \neq 0)$.

Views

Given, Mean = $\small \bar {x}$ and variance = $\small \sigma^2$

Now, Let $y_i$  be the the resulting observations if each observation is multiplied by a:

$\\ \overline y_i = a\overline x_i \\ \implies \overline x_i = \frac{\overline y_i}{a}$

$\dpi{100} \overline y =\frac{1}{n}\sum_{i=1}^ny_i = \frac{1}{n}\sum_{i=1}^nax_i$

$\dpi{100} \overline y = a\left [\frac{1}{n}\sum_{i=1}^nx_i \right] = a\overline x$

Hence the mean of the new observations  $\small ax_1,ax_2,ax_3,....,ax_n$ is $\small a\bar{x}$

We know,

$\dpi{100} \sigma^2=\frac{1}{n}\sum_{i=1}^n(x_i - \overline x)^2$

Now, Substituting the values of $x_i\ and\ \overline x$ :

$\dpi{100} \\ \implies \sigma^2= \frac{1}{n}\sum_{i=1}^n(\frac{y_i}{a} - \frac{\overline y}{a})^2 \\ \implies a^2\sigma^2= \frac{1}{n}\sum_{i=1}^n(y_i - \overline y)^2$

Hence the variance of the new observations  $\small ax_1,ax_2,ax_3,....,ax_n$ is $\dpi{100} a^2\sigma^2$

Hence proved.

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