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 data obtained by multiplying each observation of a raw data set by a constant factor \( \lambda \). The original variance of the raw data is given as \( \sigma^2 \).

### Step-by-Step Solution:

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

2. **Transforming the Data**:
If we multiply each observation by \( \lambda \), the new observations become \( \lambda x_1, \lambda x_2, \ldots, \lambda x_n \).

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

4. **Calculating the New Variance**:
The new variance \( \sigma'^2 \) of the transformed data can be calculated as follows:
\[ \sigma'^2 = \frac{1}{n-1} \sum_{i=1}^{n} (\lambda x_i - \bar{x}')^2 \]
Substituting \( \bar{x}' = \lambda \bar{x} \):
\[ \sigma'^2 = \frac{1}{n-1} \sum_{i=1}^{n} (\lambda x_i - \lambda \bar{x})^2 \]

5. **Factoring Out \( \lambda \)**:
We can factor out \( \lambda \) from the expression:
\[ \sigma'^2 = \frac{1}{n-1} \sum_{i=1}^{n} \lambda^2 (x_i - \bar{x})^2 = \lambda^2 \cdot \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2 \]
This simplifies to:
\[ \sigma'^2 = \lambda^2 \sigma^2 \]

6. **Final Result**:
Thus, the variance of the new set after multiplying each observation by \( \lambda \) is:
\[ \sigma'^2 = \lambda^2 \sigma^2 \]

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

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