Question

The mean square deviation of the variate X from ------ is minimum.

• A. C (constant)
• B. Mode
• C. Mean
• D. Median

Hint:

Remember these two critical minimization properties for competitive exams:
- Sum of Squared Deviations \(\sum(X_i - A)^2\) is minimum about the Mean.
- Sum of Absolute Deviations \(\sum|X_i - A|\) is minimum about the Median.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The mean square deviation is a measure of the dispersion or spread of data points around a central value.
In statistical mathematics, the sum of squared deviations from a variable has a unique minimization property.

Step 2: Key Formula or Approach:
Let \(X_1, X_2, \dots, X_n\) be a set of \(n\) observations.
The sum of squared deviations about an arbitrary constant \(A\) is:
\[ S(A) = \sum_{i=1}^n (X_i - A)^2 \] To find the value of \(A\) that minimizes this function, we take the derivative of \(S(A)\) with respect to \(A\) and set it to zero:
\[ \frac{dS}{dA} = 0 \]

Step 3: Detailed Explanation:
Let us perform the mathematical minimization:
1. Taking the first derivative:
\[ \frac{d}{dA} \left[ \sum_{i=1}^n (X_i - A)^2 \right] = -2 \sum_{i=1}^n (X_i - A) = 0 \] 2. Simplifying the summation:
\[ \sum_{i=1}^n (X_i - A) = 0 \] \[ \sum_{i=1}^n X_i - \sum_{i=1}^n A = 0 \] 3. Since \(A\) is a constant, \(\sum_{i=1}^n A = nA\):
\[ \sum_{i=1}^n X_i - nA = 0 \] \[ nA = \sum_{i=1}^n X_i \] 4. Solving for \(A\):
\[ A = \frac{\sum_{i=1}^n X_i}{n} = \bar{X} \text{ (Arithmetic Mean)} \] This mathematical proof shows that the sum of squared deviations is at its absolute minimum when the deviations are taken about the arithmetic mean.
- Let us compare this with the median:
The sum of absolute (non-squared) deviations is minimized when taken about the median:
\[ \sum |X_i - A| \text{ is minimized when } A = \text{Median} \] Therefore, the mean square deviation is minimized when taken about the mean.

Step 4: Final Answer:
The mean square deviation is minimum when taken about the arithmetic mean, matching Option (C).