Question

The relation between standard deviation (SD), mean deviation about mean (MD) and quartile deviation (QD) is
1. 4SD = 6MD = 5QD
2. 4SD = 5MD = 6QD
3. 6SD = 5MD = 4SD
4. 5SD = 4MD = 6QD

• A. 4SD = 6MD = 5QD
• B. 4SD = 5MD = 6QD
• C. 6SD = 5MD = 4SD
• D. 5SD = 4MD = 6QD

Hint:

Remember the approximate fractional relationships for normal distributions:
- \( \text{MD} \approx \frac{4}{5} \text{SD} \implies 5\text{MD} \approx 4\text{SD} \)
- \( \text{QD} \approx \frac{2}{3} \text{SD} \implies 3\text{QD} \approx 2\text{SD} \implies 6\text{QD} \approx 4\text{SD} \)
Combining these gives: \( 4\text{SD} = 5\text{MD} = 6\text{QD} \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question explores the theoretical relationship between standard deviation (SD), mean deviation (MD), and quartile deviation (QD) for a symmetric normal distribution.

Step 2: Detailed Explanation:
For a standard normal distribution \( X \sim N(\mu, \sigma^2) \), these three measures of dispersion have fixed mathematical proportions relative to the standard deviation (\( \sigma \)):
1. Standard Deviation (SD):
\[ \text{SD} = \sigma \] 2. Mean Deviation about Mean (MD):
\[ \text{MD} = \sqrt{\frac{2}{\pi}} \sigma \approx 0.7979 \sigma \approx \frac{4}{5} \sigma \] 3. Quartile Deviation (QD):
\[ \text{QD} \approx 0.6745 \sigma \approx \frac{2}{3} \sigma \] Let us write the approximate ratio of these deviations:
\[ \text{QD} : \text{MD} : \text{SD} \approx \frac{2}{3} \sigma : \frac{4}{5} \sigma : \sigma \] Multiply the entire ratio by 15 (the lowest common multiple of 3 and 5) to convert to integers:
\[ \text{QD} : \text{MD} : \text{SD} \approx 10 : 12 : 15 \] This implies:
\[ \frac{\text{QD}}{10} = \frac{\text{MD}}{12} = \frac{\text{SD}}{15} \] Let us find the common multiple of the numerators (60):
\[ 6 \cdot \text{QD} = 5 \cdot \text{MD} = 4 \cdot \text{SD} \] Rearranging this equality in terms of standard deviation first:
\[ 4 \cdot \text{SD} = 5 \cdot \text{MD} = 6 \cdot \text{QD} \]

Step 3: Final Answer:
The mathematical relationship is \( 4\text{SD} = 5\text{MD} = 6\text{QD} \).
Therefore, the correct choice is Option (B).