If each observation of a raw data, whose variance is `sigma^(2)`, is multiplied by `lambda`, then the variance of the new set is

Options

A. `sigma^(2)`
B. `lambda^(2) sigma^(2)`
C. `lambda+sigma^(2)`
D. `lambda^(2)+sigma^(2)`

Answer

To solve the problem, we need to determine the variance of a new set of observations obtained by multiplying each observation of a raw data set by a constant factor, \( \lambda \). The original variance of the data set is given as \( \sigma^2 \).

### Step-by-Step Solution:

1. **Understanding Variance**:
The variance of a set of observations measures how much the observations deviate from their mean. If the original observations are \( x_1, x_2, \ldots, x_n \), the variance \( \sigma^2 \) is calculated as:
\[ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \]
where \( \bar{x} \) is the mean of the observations.

2. **New Observations**:
When each observation \( x_i \) is multiplied by \( \lambda \), the new observations become \( \lambda x_1, \lambda x_2, \ldots, \lambda x_n \).

3. **Calculating the New Mean**:
The mean of the new observations \( \bar{x}' \) is given by:
\[ \bar{x}' = \frac{1}{n} \sum_{i=1}^{n} (\lambda x_i) = \lambda \left(\frac{1}{n} \sum_{i=1}^{n} x_i\right) = \lambda \bar{x} \]

4. **Calculating the New Variance**:
The variance of the new observations \( \sigma'^2 \) can be calculated as:
\[ \sigma'^2 = \frac{1}{n} \sum_{i=1}^{n} (\lambda x_i - \bar{x}')^2 \]
Substituting \( \bar{x}' = \lambda \bar{x} \):
\[ \sigma'^2 = \frac{1}{n} \sum_{i=1}^{n} (\lambda x_i - \lambda \bar{x})^2 \]
Factoring out \( \lambda \):
\[ \sigma'^2 = \frac{1}{n} \sum_{i=1}^{n} \lambda^2 (x_i - \bar{x})^2 = \lambda^2 \left(\frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2\right) \]
Thus, we have:
\[ \sigma'^2 = \lambda^2 \sigma^2 \]

5. **Conclusion**:
The variance of the new set of observations, after multiplying each observation by \( \lambda \), is given by:
\[ \sigma'^2 = \lambda^2 \sigma^2 \]

### Final Answer:
The variance of the new set is \( \lambda^2 \sigma^2 \).