Question

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

• A. Statement-1 is false, Statement-2 is true
• B. Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1
• C. Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1
• D. Statement-1 is true, statement-2 is false

Answer 1

The Correct Option is D

To solve this problem, we need to analyze both statements given in the question.

Statement 1: Variance of \(2x_1, 2x_2, \ldots, 2x_n\) is \(4\sigma^2\).

The variance of a set of numbers \(x_1, x_2, \ldots, x_n\) with variance \(\sigma^2\) is given by:

\[ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \]

If each observation is multiplied by a constant \(k\), the new variance becomes \(k^2\) times the original variance. In this case, each observation is multiplied by 2.

\[ \text{New Variance} = (2^2) \sigma^2 = 4\sigma^2 \]

Therefore, Statement 1 is true.

Statement 2: Arithmetic mean of \(2x_1, 2x_2, \ldots, 2x_n\) is \(4\bar{x}\).

The arithmetic mean of a set of numbers \(x_1, x_2, \ldots, x_n\) is:

\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \]

For the sequence \(2x_1, 2x_2, \ldots, 2x_n\), the new arithmetic mean is:

\[ \text{New Mean} = \frac{1}{n} \sum_{i=1}^{n} 2x_i = 2 \times \frac{1}{n} \sum_{i=1}^{n} x_i = 2\bar{x} \]

This is clearly \(2\bar{x}\), not \(4\bar{x}\). Therefore, Statement 2 is false.

Conclusion

Statement 1 is true, and Statement 2 is false. The correct answer is: Statement-1 is true, statement-2 is false.