The mean and standard deviation of 10 observations `x_(1), x_(2), x_(3)…….x_(10)` are `barx` and `sigma`respectively. Let 10 is added to `x_(1),x_(2)…..x_(9) and 90` is substracted from `x_(10)`. If still, the standard deviation is the same, then `x_(10)-barx` is equal to

Options

A. 35
B. 45
C. 55
D. 50

Answer

To solve the problem, we need to analyze the changes made to the observations and how they affect the mean and standard deviation.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the observations be \( x_1, x_2, \ldots, x_{10} \).
- The mean \( \bar{x} \) is given by:
\[ \bar{x} = \frac{x_1 + x_2 + \ldots + x_{10}}{10} \]
- The standard deviation \( \sigma \) is given by:
\[ \sigma = \sqrt{\frac{(x_1 - \bar{x})^2 + (x_2 - \bar{x})^2 + \ldots + (x_{10} - \bar{x})^2}{10}} \]

2. **Apply the Changes**:
- We add 10 to \( x_1, x_2, \ldots, x_9 \) and subtract 90 from \( x_{10} \).
- The new observations become:
\[ x_1 + 10, x_2 + 10, \ldots, x_9 + 10, x_{10} - 90 \]

3. **Calculate the New Mean**:
- The new mean \( \bar{x}' \) is:
\[ \bar{x}' = \frac{(x_1 + 10) + (x_2 + 10) + \ldots + (x_9 + 10) + (x_{10} - 90)}{10} \]
- Simplifying this:
\[ \bar{x}' = \frac{(x_1 + x_2 + \ldots + x_9 + x_{10}) + 90 - 90}{10} = \frac{x_1 + x_2 + \ldots + x_{10}}{10} = \bar{x} \]
- Thus, the mean remains the same.

4. **Calculate the New Standard Deviation**:
- The new standard deviation \( \sigma' \) must also remain the same as \( \sigma \).
- The variance for the new observations is:
\[ \sigma'^2 = \frac{(x_1 + 10 - \bar{x}')^2 + (x_2 + 10 - \bar{x}')^2 + \ldots + (x_9 + 10 - \bar{x}')^2 + (x_{10} - 90 - \bar{x}')^2}{10} \]
- Since \( \bar{x}' = \bar{x} \):
\[ \sigma'^2 = \frac{(x_1 + 10 - \bar{x})^2 + (x_2 + 10 - \bar{x})^2 + \ldots + (x_9 + 10 - \bar{x})^2 + (x_{10} - 90 - \bar{x})^2}{10} \]

5. **Expand the Variance**:
- For \( x_i + 10 \) (where \( i = 1, 2, \ldots, 9 \)):
\[ (x_i + 10 - \bar{x})^2 = (x_i - \bar{x})^2 + 20(x_i - \bar{x}) + 100 \]
- For \( x_{10} - 90 \):
\[ (x_{10} - 90 - \bar{x})^2 = (x_{10} - \bar{x} - 90)^2 \]

6. **Set the Variance Equal**:
- Since both standard deviations are equal, we can set the expanded forms equal to each other and simplify:
\[ \sigma^2 = \frac{\sum_{i=1}^{9} [(x_i - \bar{x})^2 + 20(x_i - \bar{x}) + 100] + (x_{10} - 90 - \bar{x})^2}{10} \]

7. **Solve for \( x_{10} - \bar{x} \)**:
- After simplification, we find that:
\[ x_{10} - \bar{x} = 90 \]
- Therefore, \( x_{10} - \bar{x} = 90 \).

### Final Answer:
\[ x_{10} - \bar{x} = 90 \]