Question

The mean deviation from the mean $m$ in a normal distribution with variance $v^2$ is approximately :

• A. (4/5)v
• B. (3/5)v
• C. (4/9)v
• D. (1/5)v

Hint:

For any normal distribution, the relationship between measures of dispersion is approximately:
\[ \text{QD} : \text{MD} : \text{SD} \approx 10 : 12 : 15 \]
where QD is Quartile Deviation, MD is Mean Deviation, and SD is Standard Deviation.
This gives \( \text{MD} \approx \frac{12}{15} \text{SD} = \frac{4}{5} \text{SD} \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In a normal distribution \( N(\mu, \sigma^2) \), where \( \mu \) is the mean and \( \sigma \) is the standard deviation, the mean deviation about the mean is a measure of dispersion.
Key Formula or Approach:
For a normal distribution, the Mean Deviation (MD) about the mean is related to the standard deviation \( \sigma \) by the following exact relation:
\[ \text{MD} = \sqrt{\frac{2}{\pi}} \sigma \]

Step 2: Detailed Explanation:
We are given that the normal distribution has mean \( m \) (which corresponds to \( \mu \)) and variance \( v^2 \).
Therefore, the standard deviation of this distribution is \( \sigma = \sqrt{v^2} = v \).
We substitute \( \sigma = v \) into our expression for the mean deviation:
\[ \text{MD} = \sqrt{\frac{2}{\pi}} v \]
Let us evaluate the numerical value of the constant factor \( \sqrt{\frac{2}{\pi}} \):
\[ \pi \approx 3.14159 \]
\[ \frac{2}{\pi} \approx \frac{2}{3.14159} \approx 0.63662 \]
Taking the square root:
\[ \sqrt{\frac{2}{\pi}} \approx \sqrt{0.63662} \approx 0.79788 \]
Let us check the fractions given in the options to see which is the closest approximation to \( 0.79788 \):
- Option (A): \( \frac{4}{5} = 0.8 \)
- Option (B): \( \frac{3}{5} = 0.6 \)
- Option (C): \( \frac{4}{9} \approx 0.444 \)
- Option (D): \( \frac{1}{5} = 0.2 \)
Since \( 0.8 \) is extremely close to \( 0.79788 \), we approximate the mean deviation as:
\[ \text{MD} \approx 0.8 v = \frac{4}{5} v \]

Step 3: Final Answer:
The mean deviation from the mean is approximately (4/5)v.