If the variance of the observations `x_(1),x_(2),……,x_(n)` is `sigma^(2)`, then the variance of `alphax_(1),alphax_(2),….,alphax_(n),alphane0` is

Options

A. `sigma^(2)`
B. `alphasigma^(2)`
C. `alpha^(2)sigma^(2)`
D. `(sigma^(2))/(alpha^(2))`

Answer

To solve the problem, we need to find the variance of the observations \( \alpha x_1, \alpha x_2, \ldots, \alpha x_n \) given that the variance of the observations \( x_1, x_2, \ldots, x_n \) is \( \sigma^2 \).

### Step-by-Step Solution:

1. **Understanding Variance**:
The variance of a set of observations \( x_1, x_2, \ldots, x_n \) is given by:
\[ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \]
where \( \bar{x} \) is the mean of the observations.

2. **Mean of New Observations**:
The new observations are \( y_i = \alpha x_i \) for \( i = 1, 2, \ldots, n \). The mean of the new observations \( y \) is:
\[ \bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = \frac{1}{n} \sum_{i=1}^{n} \alpha x_i = \alpha \left( \frac{1}{n} \sum_{i=1}^{n} x_i \right) = \alpha \bar{x} \]

3. **Variance of New Observations**:
The variance of the new observations \( y_1, y_2, \ldots, y_n \) is given by:
\[ \text{Var}(Y) = \frac{1}{n} \sum_{i=1}^{n} (y_i - \bar{y})^2 \]
Substituting \( y_i = \alpha x_i \) and \( \bar{y} = \alpha \bar{x} \):
\[ \text{Var}(Y) = \frac{1}{n} \sum_{i=1}^{n} (\alpha x_i - \alpha \bar{x})^2 \]

4. **Factoring Out \( \alpha \)**:
We can factor out \( \alpha \) from the expression:
\[ \text{Var}(Y) = \frac{1}{n} \sum_{i=1}^{n} \alpha^2 (x_i - \bar{x})^2 = \alpha^2 \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \]

5. **Final Expression**:
Since \( \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 = \sigma^2 \), we can substitute this into our equation:
\[ \text{Var}(Y) = \alpha^2 \sigma^2 \]

Thus, the variance of the observations \( \alpha x_1, \alpha x_2, \ldots, \alpha x_n \) is \( \alpha^2 \sigma^2 \).

### Final Answer:
\[ \text{Variance} = \alpha^2 \sigma^2 \]