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

Statement 2: Arithmetic mean of $2x_1,2x_2,\dots \dots ,2x_n$ is $4\overline{x}$

Given that  $x_1,x_2,\dots \dots ,x_n$ be n observations, and let $\overline{x}$ be their arithematic mean and ${\sigma }^2$ be their variance.
 

Then ${\sigma }^2=\sum \frac{x_i^2}{n}-{\left(\sum \frac{x_i}{n}\right)}^2$

Variance of $2x_1,2x_2,\dots .,2x_n=\sum \frac{{\left(2x_i\right)}^2}{n}-{\left(\sum \frac{2x_i}{n}\right)}^2=4\left[\sum \frac{x_i^2}{n}-{\left(\sum \frac{x_i}{n}\right)}^2\right]=4{\sigma }^2$

Statement 1 is true.
A.M. of $2x_1,2x_2,\dots \dots ,2x_n=\frac{2x_1+2x_2+\cdots +2x_n}{n}=2\left(\frac{x_1+x_2+\cdots \cdot +x_n}{n}\right)=2\overline{x}$

Statement 2 is false.