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, λ. The original variance is given as σ².

### Step-by-Step Solution:

1. **Understand the Original Variance**:
- The variance of a set of observations \( a_1, a_2, a_3, \ldots, a_n \) is defined as:
\[ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (a_i - \bar{x})^2 \]
where \( \bar{x} \) is the mean of the observations.

2. **Multiply Each Observation by λ**:
- If we multiply each observation by a constant \( \lambda \), the new set of observations becomes \( \lambda a_1, \lambda a_2, \lambda a_3, \ldots, \lambda a_n \).

3. **Calculate the New Mean**:
- The new mean \( \bar{x}_1 \) of the new set is:
\[ \bar{x}_1 = \frac{1}{n} \sum_{i=1}^{n} (\lambda a_i) = \lambda \cdot \frac{1}{n} \sum_{i=1}^{n} a_i = \lambda \bar{x} \]

4. **Calculate the New Variance**:
- The variance of the new set is calculated using the formula:
\[ \sigma_1^2 = \frac{1}{n} \sum_{i=1}^{n} (\lambda a_i - \bar{x}_1)^2 \]
- Substitute \( \bar{x}_1 = \lambda \bar{x} \):
\[ \sigma_1^2 = \frac{1}{n} \sum_{i=1}^{n} (\lambda a_i - \lambda \bar{x})^2 \]
- Factor out \( \lambda \):
\[ \sigma_1^2 = \frac{1}{n} \sum_{i=1}^{n} \lambda^2 (a_i - \bar{x})^2 \]
- This simplifies to:
\[ \sigma_1^2 = \lambda^2 \cdot \frac{1}{n} \sum_{i=1}^{n} (a_i - \bar{x})^2 = \lambda^2 \sigma^2 \]

5. **Conclusion**:
- Therefore, the variance of the new set of observations is:
\[ \sigma_1^2 = \lambda^2 \sigma^2 \]

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