Consider any set of 201 observations...

MathematicsMeasures of DispersionJEE Advanced 1981Easy

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

(A)

(x_{1} + x_{2} + \dots + x_{200} + x_{201}) /201

(B)

x_{1}

(C)

x_{101}

(D)

x_{201}

Answer:C

Visualized Solution (Hindi)

Visualizing the Observations

Given observations: $x_{1} < x_{2} < \dots < x_{201}$
Total number of observations $n = 201$
Objective: Find $k$ that minimizes Mean Deviation.

Defining Mean Deviation

Mean Deviation about $k$ : $M D (k) = \frac{1}{n} \sum_{i = 1}^{n} ∣ x_{i} - k ∣$
To minimize $M D (k)$ , we must minimize the sum of absolute differences: $\sum ∣ x_{i} - k ∣$

The Minimization Property

Key Property: Mean deviation is minimum when $k$ equals the median of the data set.
This is because the median is the central point that balances the number of observations on either side.

Finding the Median Position

Number of observations $n = 201$ (Odd)
Position of median $= (\frac{n + 1}{2})^{t h}$ term

Calculating the Index

Median position $= \frac{201 + 1}{2} = 10 1^{s t}$ term
Median $= x_{101}$
Therefore, $k = x_{101}$

Final Result & Takeaway

Final Answer: $k = x_{101}$
Key Takeaway: Median minimizes absolute deviations ( $∣ x_{i} - k ∣$ ).
Next Challenge: What if $n$ was even? In that case, any value in the interval $[x_{n /2}, x_{n /2 + 1}]$ would minimize the mean deviation.

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Visualized Solution (Hindi)

Visualizing the Observations

Defining Mean Deviation

The Minimization Property

Finding the Median Position

Calculating the Index

Final Result & Takeaway

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