If S.D of `x_(1), x_(2), x_(3),….x_(n),…` is `sigma`, then find S.D of `-x_(1), -x_(2), -x_(3),….-x_(n)` ?

Answer

To find the standard deviation of the values `-x1, -x2, -x3, ..., -xn` given that the standard deviation of `x1, x2, x3, ..., xn` is `σ`, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Standard Deviation**:
The standard deviation (S.D.) of a set of values measures the amount of variation or dispersion from the mean. It is defined as:
\[ S.D. = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}} \]
where \( \bar{x} \) is the mean of the values \( x_1, x_2, ..., x_n \).

2. **Mean of the Original Set**:
Let the mean of the original set \( x_1, x_2, ..., x_n \) be \( \bar{x} \). Thus, we have:
\[ \bar{x} = \frac{x_1 + x_2 + ... + x_n}{n} \]

3. **Mean of the New Set**:
The mean of the new set \( -x_1, -x_2, ..., -x_n \) can be calculated as:
\[ \bar{-x} = \frac{-x_1 + -x_2 + ... + -x_n}{n} = -\bar{x} \]

4. **Calculating the S.D. of the New Set**:
The standard deviation of the new set \( -x_1, -x_2, ..., -x_n \) is given by:
\[ S.D. = \sqrt{\frac{\sum (-x_i - \bar{-x})^2}{n}} \]
Substituting \( \bar{-x} = -\bar{x} \):
\[ S.D. = \sqrt{\frac{\sum (-x_i + \bar{x})^2}{n}} \]

5. **Simplifying the Expression**:
Notice that:
\[ -x_i + \bar{x} = -(x_i - \bar{x}) \]
Therefore, we have:
\[ S.D. = \sqrt{\frac{\sum (-(x_i - \bar{x}))^2}{n}} = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}} \]
Since squaring a negative value results in a positive value, we can simplify this to:
\[ S.D. = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}} = S.D. \text{ of the original set} \]

6. **Conclusion**:
Thus, the standard deviation of the set \( -x_1, -x_2, ..., -x_n \) is the same as that of the original set:
\[ S.D. = \sigma \]

### Final Answer:
The standard deviation of `-x1, -x2, -x3, ..., -xn` is `σ`.