Question

For a discrete data set when all the values are not same then the relationship between mean deviation about mean (MD) and standard deviation (SD) is

• A. MD = SD
• B. MD $<$ SD
• C. MD $>$ SD
• D. MD $\leq$ SD

Hint:

For any non-constant dataset, the Standard Deviation is always strictly greater than the Mean Deviation about the mean because squaring larger deviations gives them disproportionately more weight.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Mean Deviation (MD) about the mean and Standard Deviation (SD) are both measures of dispersion that quantify the spread of data points around the arithmetic mean.
Key Formula or Approach:
For a discrete dataset $\{x_1, x_2, \dots, x_n\}$ with mean $\bar{x}$: \[ \text{MD} = \frac{1}{n} \sum_{i=1}^n |x_i - \bar{x}| \] \[ \text{SD} = \sqrt{\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2} \]

Step 2: Detailed Explanation:
Let $d_i = |x_i - \bar{x}|$.
By definition, $d_i \geq 0$ for all $i$.
According to the Cauchy-Schwarz inequality (or the properties of variances of non-negative numbers): \[ \left( \frac{1}{n} \sum_{i=1}^n d_i \right)^2 \leq \frac{1}{n} \sum_{i=1}^n d_i^2 \]
Taking the square root on both sides: \[ \frac{1}{n} \sum_{i=1}^n d_i \leq \sqrt{\frac{1}{n} \sum_{i=1}^n d_i^2} \implies \text{MD} \leq \text{SD} \]
Equality holds if and only if all $d_i$ values are identical, which occurs either when all observations are equal, or in specific symmetric two-point configurations.
Since the problem states that all values are not the same, the deviations are not uniform, making the inequality strict: \[ \text{MD} < \text{SD} \]

Step 3: Final Answer:
The strict relationship is MD $<$ SD.