Let x_1, x_2, x_3, dots, x_n be n observation | vedclass

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(B) Let the original observations be $x_i$ with mean $\bar{x} = \frac{1}{n} \sum x_i$ and variance $\sigma^2 = \frac{1}{n} \sum (x_i - \bar{x})^2$.For

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Statement $-1$ is true,Statement $-2$ is false.

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(B) Let the original observations be $x_i$ with mean $\bar{x} = \frac{1}{n} \sum x_i$ and variance $\sigma^2 = \frac{1}{n} \sum (x_i - \bar{x})^2$.For the new observations $y_i = 2x_i$,the new mean $\bar{y}$ is:$\bar{y} = \frac{1}{n} \sum (2x_i) = 2 \left( \frac{1}{n} \sum x_i \right) = 2\bar{x}$.Thus,Statement $-2$ is false because the mean is $2\bar{x}$,not $4\bar{x}$.The new variance $\sigma_y^2$ is:$\sigma_y^2 = \frac{1}{n} \sum (y_i - \bar{y})^2 = \frac{1}{n} \sum (2x_i - 2\bar{x})^2 = \frac{1}{n} \sum 4(x_i - \bar{x})^2 = 4 \sigma^2$.Thus,Statement $-1$ is true.

Let $x_1, x_2, x_3, \dots, x_n$ be $n$ observations,$\bar{x}$ be their arithmetic mean,and $\sigma^2$ be their variance.
Statement $-1$: The variance of observations $2x_1, 2x_2, 2x_3, \dots, 2x_n$ is $4\sigma^2$.
Statement $-2$: The arithmetic mean of $2x_1, 2x_2, 2x_3, \dots, 2x_n$ is $4\bar{x}$.

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Let $x_1, x_2, x_3, \dots, x_n$ be $n$ observations,$\bar{x}$ be their arithmetic mean,and $\sigma^2$ be their variance.Statement $-1$: The variance of observations $2x_1, 2x_2, 2x_3, \dots, 2x_n$ is $4\sigma^2$.Statement $-2$: The arithmetic mean of $2x_1, 2x_2, 2x_3, \dots, 2x_n$ is $4\bar{x}$.