If the standard deviation of n observation `x_(1), x_(2),…….,x_(n)` is 5 and for another set of n observation `y_(1), y_(2),………., y_(n)` is 4, then the standard deviation of n observation
`x_(1)-y_(1), x_(2)-y_(2),………….,x_(n)-y_(n)` is

Options

A. 1
B. `(sqrt5)/(2)`
C. 5
D. Data insufficient

Answer

To find the standard deviation of the observations \( z_i = x_i - y_i \) where \( i = 1, 2, \ldots, n \), we can use the properties of variance and standard deviation.

### Step-by-Step Solution

1. **Understanding Standard Deviation and Variance**:
The standard deviation (SD) of a set of observations is the square root of the variance. If the standard deviation of \( x_i \) is given as \( \sigma_x = 5 \), then the variance \( \sigma_x^2 = 25 \). Similarly, for \( y_i \), where the standard deviation \( \sigma_y = 4 \), the variance \( \sigma_y^2 = 16 \).

\[ \sigma_x^2 = 25, \quad \sigma_y^2 = 16 \]

2. **Using the Formula for Variance of Differences**:
The variance of the difference of two independent random variables \( x \) and \( y \) is given by:
\[ \sigma_{z}^2 = \sigma_x^2 + \sigma_y^2 \]
where \( z_i = x_i - y_i \).

3. **Substituting the Values**:
Now substituting the variances we calculated:
\[ \sigma_{z}^2 = 25 + 16 = 41 \]

4. **Calculating the Standard Deviation**:
To find the standard deviation \( \sigma_z \), we take the square root of the variance:
\[ \sigma_z = \sqrt{41} \]

5. **Conclusion**:
Thus, the standard deviation of the observations \( z_i = x_i - y_i \) is \( \sqrt{41} \).

### Final Answer
The standard deviation of the observations \( x_i - y_i \) is \( \sqrt{41} \).