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

Let the original data set be $$x_1,x_2,\dots ,x_n$$ with arithmetic mean $$\bar{x}$$ and variance $$\sigma^2$$.

Step 1: Recall the formulae
Arithmetic mean: $$\bar{x}=\frac{x_1+x_2+\dots+x_n}{n}$$
Variance: $$\sigma^2=\frac{1}{n}\sum_{i=1}^{n}(x_i-\bar{x})^{2}$$

Step 2: Effect of multiplying each observation by a constant
If every observation is multiplied by a constant $$k$$, the new observations become $$kx_1,kx_2,\dots ,kx_n$$.

• New mean: $$\overline{kx}=k\bar{x}$$ (because each term in the numerator gets multiplied by $$k$$).
• New variance: $$\sigma_{k}^{2}=k^{2}\sigma^{2}$$ (because each squared deviation $$\bigl(kx_i-k\bar{x}\bigr)^2=k^{2}(x_i-\bar{x})^{2}$$).

Step 3: Apply $$k=2$$ to the given statements

• New mean of $$2x_1,2x_2,\dots ,2x_n$$ is $$2\bar{x}$$, not $$4\bar{x}$$.
• New variance is $$2^{2}\sigma^{2}=4\sigma^{2}$$.

Step 4: Verify the statements
Statement 1: “Variance of $$2x_1,2x_2,\dots ,2x_n$$ is $$4\sigma^{2}$$” — TRUE.
Statement 2: “Arithmetic mean of $$2x_1,2x_2,\dots ,2x_n$$ is $$4\bar{x}$$” — FALSE (it is $$2\bar{x}$$).

Hence Statement 1 is true and Statement 2 is false.

Option D which is: Statement 1 is true, statement 2 is false