Consider any set of observations `x_(1),x_(2),x_(3),..,x_(101)`. It is given that `x_(1) lt x_(2) lt x_(3) lt .. Lt x_(100) lt x_(101)`, then the mean deviation of this set of observations about a point k is minimum when k equals

Options

A. `x_(1)`
B. `x_(51)`
C. `(x_(1)+x_(2)+..+x_(101))/(101)`
D. `x_(50)`

Explanation

To find the value of \( k \) for which the mean deviation of the set of observations \( x_1, x_2, x_3, \ldots, x_{101} \) is minimized, we can follow these steps:

### Step 1: Understand Mean Deviation
Mean deviation is defined as the average of the absolute deviations of each observation from a central point \( k \). Mathematically, it is expressed as:
\[ MD(k) = \frac{1}{n} \sum_{i=1}^{n} |x_i - k| \]
where \( n \) is the number of observations.

### Step 2: Identify the Median
For a set of observations sorted in ascending order, the mean deviation is minimized when \( k \) is the median of the dataset. The median is the middle value when the data is arranged in order.

### Step 3: Calculate the Median
Since we have 101 observations (which is an odd number), the median is the value at the position:
\[ \text{Median position} = \frac{n + 1}{2} = \frac{101 + 1}{2} = 51 \]
Thus, the median is \( x_{51} \).

### Step 4: Conclusion
Therefore, the value of \( k \) that minimizes the mean deviation is:
\[ k = x_{51} \]

### Final Answer
The mean deviation of the set of observations about a point \( k \) is minimum when \( k \) equals \( x_{51} \).

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