Let x_1, x_2, dots, x_n be n observations,bar | vedclass

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(A) Given that the mean of $x_1, x_2, \dots, x_n$ is $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$ and the variance is $\sigma^2 = \frac{1}{n} \sum_{i=1}

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

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(A) Given that the mean of $x_1, x_2, \dots, x_n$ is $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$ and the variance is $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2$.For the observations $2x_1, 2x_2, \dots, 2x_n$:The new mean $\bar{x}' = \frac{1}{n} \sum_{i=1}^{n} (2x_i) = 2 \left( \frac{1}{n} \sum_{i=1}^{n} x_i \right) = 2\bar{x}$.Thus,Statement-$2$ is false because the mean is $2\bar{x}$,not $4\bar{x}$.The new variance $\sigma'^2 = \frac{1}{n} \sum_{i=1}^{n} (2x_i - 2\bar{x})^2 = \frac{1}{n} \sum_{i=1}^{n} 4(x_i - \bar{x})^2 = 4 \left( \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \right) = 4\sigma^2$.Thus,Statement-$1$ is true.Therefore,Statement-$1$ is true and Statement-$2$ is false.

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

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