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Let $\mathrm{x}_1, \mathrm{x}_2, \ldots \ldots, \mathrm{x}_{\mathrm{n}}$ be $\mathrm{n}$ observations, and let $\overline{\mathrm{x}}$ be their arithematic mean and $\sigma^2$ be their variance.
Statement $1$: Variance of $2 x_1, 2 x_2, \ldots \ldots, 2 x_n$ is $4 \sigma^2$.
Statement $2$: Arithmetic mean of $2 \mathrm{x}_1, 2 \mathrm{x}_2, \ldots . ., 2 \mathrm{x}_{\mathrm{n}}$ is $4 \overline{\mathrm{x}}$.
Options:
Statement $1$: Variance of $2 x_1, 2 x_2, \ldots \ldots, 2 x_n$ is $4 \sigma^2$.
Statement $2$: Arithmetic mean of $2 \mathrm{x}_1, 2 \mathrm{x}_2, \ldots . ., 2 \mathrm{x}_{\mathrm{n}}$ is $4 \overline{\mathrm{x}}$.
Solution:
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Verified Answer
The correct answer is:
Statement $1$ is true, statement $2$ is false
Statement $1$ is true, statement $2$ is false
$\sigma^2=\sum \frac{x_i^2}{n}-\left(\sum \frac{x_i}{n}\right)^2$
Variance of $2 x_1, 2 x_2, \ldots . ., 2 x_n=\sum \frac{\left(2 x_i\right)^2}{n}-\left(\sum \frac{2 x_i}{n}\right)^2=4\left[\sum \frac{x_i^2}{n}-\left(\sum \frac{x_i}{n}\right)^2\right]=4 \sigma^2$ Statement $1$ is true.
A.M. of $2 x_1, 2 x_2, \ldots \ldots, 2 x_n=\frac{2 x_1+2 x_2+\cdots+2 x_n}{n}=2\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)=2 \bar{x}$
Statement $2$ is false.
Variance of $2 x_1, 2 x_2, \ldots . ., 2 x_n=\sum \frac{\left(2 x_i\right)^2}{n}-\left(\sum \frac{2 x_i}{n}\right)^2=4\left[\sum \frac{x_i^2}{n}-\left(\sum \frac{x_i}{n}\right)^2\right]=4 \sigma^2$ Statement $1$ is true.
A.M. of $2 x_1, 2 x_2, \ldots \ldots, 2 x_n=\frac{2 x_1+2 x_2+\cdots+2 x_n}{n}=2\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)=2 \bar{x}$
Statement $2$ is false.
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