Consider any set of 201 observations x1, x2, ....x200, x201. It is given that x1

Q: Wait, isn't the mean the value that minimizes the sum of deviations? Why are we using the median?

That is a classic trap! The mean minimizes the sum of *squared* deviations, i.e., $\sum (x_i - k)^2$. However, when we are dealing with the sum of *absolute* deviations, i.e., $\sum |x_i - k|$, the median is the mathematical champion that minimizes the total distance.

Q: What happens if $n$ were even, say $n=200$? Would there still be a unique median?

Excellent question! If $n$ is even, the sum of absolute deviations is minimized by *any* value between the two middle terms (inclusive). In that case, the median is not unique. But for an odd number of observations like $n=201$, the median is perfectly unique.

Q: Why do we ignore the $\frac{1}{n}$ factor in the minimization?

Because $n$ is a positive constant ($201$). Minimizing $\frac{1}{n} \sum |x_i - k|$ is mathematically equivalent to minimizing just the sum $\sum |x_i - k|$. We focus on the sum to simplify the problem without changing the result.

Question

Consider any set of 201 observations $x_1, x_2, \dots, x_{200}, x_{201}$. It is given that $x_1 < x_2 < \dots < x_{200} < x_{201}$. Then the mean deviation of this set of observations about a point $k$ is minimum when $k$ equals

Options:

(A) $(x_1 + x_2 + \dots + x_{200} + x_{201})/201$

(B) $x_1$

(C) $x_{101}$

(D) $x_{201}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a long, straight road. Scattered along this road are 201 milestones, labeled $x_1, x_2, \dots, x_{201}$, arranged in perfect, strictly increasing order.

You are tasked with choosing a single point $k$ on this road such that the average distance from $k$ to all these milestones is as small as possible. This is the essence of minimizing the mean deviation.

### The Tug-of-War Analogy

Let us define the mean deviation about a point $k$ as:

$$MD(k) = \frac{1}{n} \sum_{i=1}^{n} |x_i - k|$$

Since the number of observations $n=201$ is a constant, our mission is simply to minimize the sum:

$$S = \sum_{i=1}^{201} |x_i - k|$$

Picture yourself standing at point $k$. If you decide to take a step to the right, you move closer to every milestone located to your right, effectively reducing the distance to those points. However, you simultaneously move further away from every milestone located to your left, increasing those distances.

This is a classic tug-of-war. If there are more milestones to your left than to your right, moving to the right will increase the total distance. If there are more to your right, moving to the right will decrease it.

The only point where this tug-of-war reaches a stalemate—where the total distance is minimized—is the point that perfectly balances the number of milestones on either side. This point is the **median**.

### The Mathematical Proof of the Median

Why does the median win? Consider the derivative of the sum $S$ with respect to $k$.

The derivative of $|x_i - k|$ is $-1$ if $k < x_i$ and $+1$ if $k > x_i$. Thus, the derivative of the total sum $S$ is the number of points to the left of $k$ minus the number of points to the right of $k$.

For the sum to be at a minimum, this derivative must be zero. This happens precisely when the number of points to the left equals the number of points to the right. This is the definition of the median.

### The Final Calculation

Now that we have established that $k$ must be the median, we simply need to locate it. For a set of $n$ ordered observations where $n$ is odd, the median is the term at the position:

$$	ext{Position} = \frac{n+1}{2}$$

With $n=201$, we calculate the position as:

$$\frac{201+1}{2} = \frac{202}{2} = 101$$

Therefore, the median is the $101^{st}$ observation in our sequence, which is **$x_{101}$**.

By choosing $k = x_{101}$, we ensure that there are exactly 100 points to the left and 100 points to the right, creating the perfect balance that minimizes the mean deviation. This problem is a beautiful reminder that in statistics, the 'best' measure of central tendency depends entirely on what you are trying to minimize.

Animated Solution for Mathematics - Statistics: Consider any set of 201 observations $x_{1}, x_{2}, \dots, x_{200}, x_{201}$ . It is given that $x_{1} < x_{2} < \dots < x_{200} < x_{201}$ . Then the mean deviation of this set of observations about a point $k$ is minimum when $k$ equals

Select Answer:

Visualized Solution

Visualizing the Observations

Defining Mean Deviation

The Minimization Goal

Visualizing Distances

The Median Property

Why the Median?

Finding the Median

Position of the Median

Calculating the Index

Final Conclusion

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

The Tug-of-War Analogy

The Mathematical Proof of the Median

The Final Calculation

Similar Questions

Let the median and the mean deviation about the median of 7 observation 170, 125, 230, 190, 210, a, b be 170 and $\frac{205}{7}$ respectively. Then the mean deviation about the mean of these 7 observations is :

Let the mean and the variance of 20 observations $x_{1}, x_{2}, \dots, x_{20}$ be 15 and 9, respectively. For $α \in R$ , if the mean of $(x_{1} + α)^{2}, (x_{2} + α)^{2}, \dots, (x_{20} + α)^{2}$ is 178, then the square of the maximum value of $α$ is equal to

Let the mean of 6 observations 1, 2, 4, 5, $x$ and $y$ be 5 and their variance be 10. Then their mean deviation about the mean is equal to

If the mean deviation about the median of the numbers $k, 2 k, 3 k, \dots, 1000 k$ is 500, then $k^{2}$ is equal to :

If the mean deviation about the mean of the numbers $1, 2, 3, \dots, n$ , where $n$ is odd, is $\frac{5 ( n + 1 )}{n}$ , then $n$ is equal to ______.

The variance $σ^{2}$ of the data $x_{i} f_{i} 03125362106123173$ is ________.

The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations are $2, 3, 5, 10, 11, 13, 15, 21$ , then the mean deviation about the median of all the 10 observations is

Let $S$ be the set of all values of $a_{1}$ for which the mean deviation about the mean of 100 consecutive positive integers $a_{1}, a_{2}, a_{3}, \dots, a_{100}$ is 25. Then $S$ is

Let the mean of the data $x Frequency (f) 14324528 7 α 98$ be 5. If $m$ and $σ^{2}$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3 α}{m + σ ^{2}}$ is equal to _______.

For observations $x_{i}$ given $\sum_{i = 1}^{10} (x_{i} - 5) = 10$ and $\sum_{i = 1}^{10} (x_{i} - 5)^{2} = 40$ . If mean and variance of observations $(x_{1} - 3), (x_{2} - 3) \dots (x_{10} - 3)$ is $λ$ & $μ$ respectively then ordered pair $(λ, μ)$ is