Statistics Question - None

Question

Consider the following statements regarding the data set $2x_1, 2x_2, \ldots, 2x_n$ where $\sigma^2$ is the variance of $x_1, x_2, \ldots, x_n$:

Statement-1: The variance of $2x_1, 2x_2, \ldots, 2x_n$ is $4\sigma^2$.

Statement-2: The AM of $2x_1, 2x_2, \ldots, 2x_n$ is $2\bar{x}$.

Which of the following is correct?

Accepted Answer

Step 1: Verify Statement-2 (Mean scaling)

If mean of original data is $\bar{x}$, then mean of $2x_1, 2x_2, \ldots, 2x_n$ is:

$$ ext{AM} = \frac{2x_1 + 2x_2 + \cdots + 2x_n}{n} = 2\left(\frac{x_1 + x_2 + \cdots + x_n}{n} ight) = 2\bar{x}$$

Statement-2 is TRUE ✓

Step 2: Verify Statement-1 (Variance scaling)

Original variance: $\sigma^2 = \frac{1}{n}\sum(x_i - \bar{x})^2$

Variance of transformed data:

$$ ext{Var}(2x_i) = \frac{1}{n}\sum(2x_i - 2\bar{x})^2 = \frac{1}{n}\sum[2(x_i - \bar{x})]^2$$

$$= \frac{1}{n}\sum 4(x_i - \bar{x})^2 = 4 \cdot \frac{1}{n}\sum(x_i - \bar{x})^2 = 4\sigma^2$$

Statement-1 is TRUE ✓

Step 3: Conclusion

Both statements are correct. The key rule: for data scaled by constant k, $ ext{Var}(kX) = k^2\sigma^2$ (quadratic scaling), while $ ext{Mean}(kX) = k\bar{x}$ (linear scaling).

∴ Answer: B (Both statements are correct)

Question:

Step-by-Step Solution

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