For (2n+1) observations `x_(1), x_(2),-x_(2),..,x_(n),-x_(n)` and 0, where all x's are distinct, let SD and MD denote the standard deviation and median, respectively. Then which of the following is always true ?

Options

A. `SD lt MD`
B. `SD gt MD`
C. SD=MD
D. Nothing can be said in general about the relationship between SD and MD

Explanation

To solve the problem, we need to analyze the given observations and calculate the median (MD) and standard deviation (SD) for the set of observations: \( x_1, x_2, -x_2, \ldots, x_n, -x_n, 0 \).

### Step 1: Arrange the Observations
The observations given are:
- \( x_1, x_2, -x_2, -x_1, \ldots, x_n, -x_n, 0 \)

To find the median, we need to arrange these observations in ascending order. The arrangement will be:
- \( -x_n, -x_{n-1}, \ldots, -x_2, -x_1, 0, x_1, x_2, \ldots, x_n \)

### Step 2: Determine the Median (MD)
Since there are \( 2n + 1 \) observations (an odd number), the median is the middle value. In our arranged list, the middle value is the \( (n + 1)^{th} \) term, which is \( 0 \).

Thus,
\[ \text{MD} = 0 \]

### Step 3: Calculate the Mean
Next, we calculate the mean of the observations. The mean is given by:
\[ \text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} \]

The sum of the observations is:
\[ x_1 + x_2 + (-x_2) + (-x_1) + x_3 + (-x_3) + \ldots + x_n + (-x_n) + 0 = 0 \]

Thus,
\[ \text{Mean} = \frac{0}{2n + 1} = 0 \]

### Step 4: Calculate the Standard Deviation (SD)
The standard deviation is calculated using the formula:
\[ \text{SD} = \sqrt{\frac{\sum (x_i - \bar{x})^2}{N}} \]
where \( \bar{x} \) is the mean and \( N \) is the number of observations.

Since the mean \( \bar{x} = 0 \), we have:
\[ \text{SD} = \sqrt{\frac{\sum x_i^2}{2n + 1}} \]

The individual terms \( x_i \) are distinct, and thus \( \sum x_i^2 \) will be a positive value (since all \( x_i \) are distinct and non-zero). Therefore, the standard deviation will be greater than zero.

### Conclusion
From our calculations, we have:
- Median (MD) = 0
- Standard Deviation (SD) > 0

Thus, we conclude that:
\[ \text{SD} > \text{MD} \]

### Final Answer
The correct option is that the standard deviation is always greater than the median.

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