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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).

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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}

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

\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 :

\\ \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 a^2\sigma^2

Hence proved. 

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