If the mean of n observations `x_1,x_2,x_3...x_n` is `barx` then the sum of deviations of observations from mean is

Options

A. 0
B. `(bar(X))/(n)`
C. `n bar(X)`
D. none of these

Answer

To solve the problem, we need to find the sum of deviations of observations from the mean. Let's break it down step by step.

### Step-by-Step Solution:

1. **Understand the Mean**:
The mean (average) of n observations \( x_1, x_2, x_3, \ldots, x_n \) is given by the formula:
\[ \bar{x} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} \]

2. **Define the Sum of Deviations**:
The sum of deviations from the mean is defined as:
\[ S = \sum_{i=1}^{n} (x_i - \bar{x}) \]
This means we will sum up the differences between each observation and the mean.

3. **Expand the Sum**:
We can expand the sum:
\[ S = (x_1 - \bar{x}) + (x_2 - \bar{x}) + (x_3 - \bar{x}) + \ldots + (x_n - \bar{x}) \]

4. **Rearranging the Terms**:
We can rearrange this expression:
\[ S = (x_1 + x_2 + x_3 + \ldots + x_n) - n\bar{x} \]
Here, \( n\bar{x} \) represents the sum of the mean added n times.

5. **Substituting the Mean**:
From the definition of the mean, we know that:
\[ n\bar{x} = x_1 + x_2 + x_3 + \ldots + x_n \]
Therefore, we can substitute this back into our equation for S:
\[ S = (x_1 + x_2 + x_3 + \ldots + x_n) - (x_1 + x_2 + x_3 + \ldots + x_n) \]

6. **Final Calculation**:
This simplifies to:
\[ S = 0 \]
Thus, the sum of deviations of observations from the mean is always 0.

### Conclusion:
The sum of deviations of observations from the mean \( \bar{x} \) is 0.

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