A set of n value `x_(1),x_(2),……..,x_(n)` has mean `barx` and standard deviation `sigma`. The mean and standard deviation of n values `(x_(1))/(k),(x_(2))/(k),.......,(x_(n))/(k)(kne0` respectively are
(i) `kbarx,(sigma)/(k)`
(ii) `(barx)/(k),(sigma)/(k)`
(iii) `kbarx,ksigma`
(iv) `(barx)/(k),ksigma`

Options

A. `kbarx,(sigma)/(k)`
B. `(barx)/(k),(sigma)/(k)`
C. `kbarx,ksigma`
D. `(barx)/(k),ksigma`

Answer

To solve the problem, we need to find the mean and standard deviation of the new set of values \( \frac{x_1}{k}, \frac{x_2}{k}, \ldots, \frac{x_n}{k} \) given that the original set \( x_1, x_2, \ldots, x_n \) has a mean \( \bar{x} \) and standard deviation \( \sigma \).

### Step-by-Step Solution:

1. **Calculate the Mean of the New Set:**
The mean of the original set is given by:
\[ \bar{x} = \frac{x_1 + x_2 + \ldots + x_n}{n} \]
For the new set \( \frac{x_1}{k}, \frac{x_2}{k}, \ldots, \frac{x_n}{k} \), the mean \( \bar{x}' \) can be calculated as:
\[ \bar{x}' = \frac{\frac{x_1}{k} + \frac{x_2}{k} + \ldots + \frac{x_n}{k}}{n} = \frac{1}{k} \cdot \frac{x_1 + x_2 + \ldots + x_n}{n} = \frac{\bar{x}}{k} \]
Thus, the mean of the new set is:
\[ \bar{x}' = \frac{\bar{x}}{k} \]

2. **Calculate the Standard Deviation of the New Set:**
The standard deviation \( \sigma \) of the original set is given by:
\[ \sigma = \sqrt{\frac{\sum_{i=1}^{n} x_i^2}{n} - \left(\frac{\sum_{i=1}^{n} x_i}{n}\right)^2} \]
For the new set, the standard deviation \( \sigma' \) can be calculated as follows:
\[ \sigma' = \sqrt{\frac{\sum_{i=1}^{n} \left(\frac{x_i}{k}\right)^2}{n} - \left(\frac{\sum_{i=1}^{n} \frac{x_i}{k}}{n}\right)^2} \]
Simplifying the first term:
\[ \frac{\sum_{i=1}^{n} \left(\frac{x_i}{k}\right)^2}{n} = \frac{1}{k^2} \cdot \frac{\sum_{i=1}^{n} x_i^2}{n} \]
And the second term:
\[ \left(\frac{\sum_{i=1}^{n} \frac{x_i}{k}}{n}\right)^2 = \left(\frac{1}{k} \cdot \frac{\sum_{i=1}^{n} x_i}{n}\right)^2 = \left(\frac{\bar{x}}{k}\right)^2 \]
Therefore, we can write:
\[ \sigma' = \sqrt{\frac{1}{k^2} \cdot \frac{\sum_{i=1}^{n} x_i^2}{n} - \left(\frac{\bar{x}}{k}\right)^2} \]
Factoring out \( \frac{1}{k^2} \):
\[ \sigma' = \frac{1}{k} \sqrt{\frac{\sum_{i=1}^{n} x_i^2}{n} - \bar{x}^2} = \frac{\sigma}{k} \]

### Final Results:
- The mean of the new set is \( \frac{\bar{x}}{k} \).
- The standard deviation of the new set is \( \frac{\sigma}{k} \).

Thus, the correct answer is:
\[ \text{(ii) } \left(\frac{\bar{x}}{k}, \frac{\sigma}{k}\right) \]