Question

Consider the data of scores obtained by students in an examination. If the score of every student is increased by 2 marks, then which of the following statements is TRUE?

• A. The mean deviation about the mean does not change.
• B. The mean deviation about the mean is increased by 2.
• C. The mean deviation about the median is increased by 2.
• D. The variance is increased by 2.

Hint:

Measures of dispersion such as Range, Mean Deviation, Standard Deviation, and Variance are independent of the change of origin (i.e., adding or subtracting a constant to/from each observation).
They only change when there is a change of scale (i.e., multiplying or dividing by a constant).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are analyzing how statistical measures of dispersion (specifically mean deviation and variance) behave when a constant value (2 marks) is added to every data point in a set of exam scores.

Step 2: Key Formula or Approach: Let the original scores be $x_1, x_2, \dots, x_N$.
The mean of the original data is $\bar{x}$.
The new scores are $y_i = x_i + 2$ for all $i$.
The mean of the new data is $\bar{y} = \bar{x} + 2$.
We will calculate the mean deviation of the new dataset and compare it to that of the original dataset.

Step 3: Detailed Explanation:

• The mean deviation about the mean for the original data is defined as:
\[ \text{MD}_x = \frac{1}{N} \sum_{i=1}^{N} |x_i - \bar{x}| \]
• For the new dataset, the mean deviation about the mean is:
\[ \text{MD}_y = \frac{1}{N} \sum_{i=1}^{N} |y_i - \bar{y}| \]
• Substituting $y_i = x_i + 2$ and $\bar{y} = \bar{x} + 2$:
\[ |y_i - \bar{y}| = |(x_i + 2) - (\bar{x} + 2)| = |x_i - \bar{x}| \]
• Therefore:
\[ \text{MD}_y = \frac{1}{N} \sum_{i=1}^{N} |x_i - \bar{x}| = \text{MD}_x \] This proves that the mean deviation about the mean remains completely unchanged.

• Similarly, since the median also increases by 2, the deviations from the median remain unchanged, making option (C) false.

• Let us check the variance:
The variance of $x$ is $\sigma_x^2 = \frac{1}{N} \sum (x_i - \bar{x})^2$.
The variance of $y$ is $\sigma_y^2 = \frac{1}{N} \sum (y_i - \bar{y})^2 = \frac{1}{N} \sum (x_i - \bar{x})^2 = \sigma_x^2$.
Thus, the variance does not change, making option (D) false.

Step 4: Final Answer:
The mean deviation about the mean does not change when a constant is added to all data values.