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Let ${x_1}\;,\;{x_2}\;,\;.\;.\;.\;,{x_n}$ be $n$ observations, and let $\bar x$ be their arithmaetic mean and ${\sigma ^2}$ be the variance
Statement $-1$ :Variance of $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4{\sigma ^2}$ .
Statement $-2$: Arithmetic mean $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4\bar x$.
Statement $-1$ is false, Statement $-2$ is true;
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not acorrect explanation for Statement $-1$
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$
Statement $-1$ is true, Statement $-2$ is false
Solution
$x_{1}, x_{2}, x_{3}, \ldots . x_{n}, \mathrm{A.M} .=\bar{x}, \text { Variance }=\sigma^{2}$
Statement $2 : A.M.$ of $2 x_{1}, 2 x_{2}, \ldots ., 2 x_{n}$
$=\frac{2\left(x_{1}+x_{2}+\ldots . .+x_{n}\right)}{n}=2 \bar{x}$
Given $A . M .=4 \bar{x} $
$ \therefore$ Statement $2$ is false.
