S.D. of n observation `a_(1),a_(2),a_(3)........a_(n)` is `sigma` then the S.D. of the observations `lambda a_(1), lambda a_(2)...........lambda a_(n)` is

Options

A. `lambda sigma`
B. `-lambda sigma`
C. `|lambda| sigma`
D. `sigma`

Answer

To find the standard deviation of the observations \( \lambda a_1, \lambda a_2, \ldots, \lambda a_n \) given that the standard deviation of \( a_1, a_2, \ldots, a_n \) is \( \sigma \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Standard Deviation**:
The standard deviation (S.D.) of a dataset measures the amount of variation or dispersion from the mean. If the standard deviation of observations \( a_1, a_2, \ldots, a_n \) is \( \sigma \), it is calculated as:
\[ S.D. = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (a_i - \bar{x})^2} \]
where \( \bar{x} \) is the mean of the observations.

2. **Mean of New Observations**:
The mean of the new observations \( \lambda a_1, \lambda a_2, \ldots, \lambda a_n \) can be calculated as:
\[ \bar{x}' = \frac{\lambda a_1 + \lambda a_2 + \ldots + \lambda a_n}{n} = \lambda \cdot \bar{x} \]
where \( \bar{x} \) is the mean of the original observations.

3. **Variance of New Observations**:
The variance of the new observations can be expressed as:
\[ \text{Var}(X') = \frac{1}{n-1} \sum_{i=1}^{n} (\lambda a_i - \bar{x}')^2 \]
Substituting \( \bar{x}' = \lambda \bar{x} \):
\[ \text{Var}(X') = \frac{1}{n-1} \sum_{i=1}^{n} (\lambda a_i - \lambda \bar{x})^2 \]

4. **Factoring Out \( \lambda \)**:
We can factor \( \lambda \) out of the squared term:
\[ \text{Var}(X') = \frac{1}{n-1} \sum_{i=1}^{n} \lambda^2 (a_i - \bar{x})^2 = \lambda^2 \cdot \frac{1}{n-1} \sum_{i=1}^{n} (a_i - \bar{x})^2 \]
This shows that:
\[ \text{Var}(X') = \lambda^2 \cdot \text{Var}(X) \]

5. **Standard Deviation of New Observations**:
The standard deviation is the square root of the variance:
\[ S.D.(X') = \sqrt{\text{Var}(X')} = \sqrt{\lambda^2 \cdot \text{Var}(X)} = |\lambda| \cdot \sqrt{\text{Var}(X)} \]
Since the standard deviation of the original observations is \( \sigma \):
\[ S.D.(X') = |\lambda| \cdot \sigma \]

### Conclusion:
Thus, the standard deviation of the observations \( \lambda a_1, \lambda a_2, \ldots, \lambda a_n \) is:
\[ |\lambda| \cdot \sigma \]