Question

Match the following List-I with List-II.

\[ \begin{array}{|c|l|c|l|} \hline \text{List-I} & & \text{List-II} & \\ \hline (a) & \text{Mean} & (i) & \text{List of individual values} \\ \hline (b) & \text{Median} & (ii) & \text{Value appears most often} \\ \hline (c) & \text{Mode} & (iii) & \text{Average for complete data} \\ \hline (d) & \text{Frequency Distribution} & (iv) & \text{Location average} \\ \hline \end{array} \]

• A. a-i, b-iii, c-ii, d-iv
• B. a-iii, b-iv, c-ii, d-i
• C. a-i, b-iv, c-ii, d-iii
• D. a-iii, b-iv, c-i, d-ii

Hint:

Measures of Central Tendency: \[ \text{Mean}=\text{Arithmetic Average} \] \[ \text{Median}=\text{Positional Average} \] \[ \text{Mode}=\text{Most Frequent Value} \]

Answer 1

The Correct Option is B

Step 1: Match Mean.
Mean is the arithmetic average of all observations. \[ (a)\rightarrow(iii) \]

Step 2: Match Median.
Median represents the central or positional value. It is often called a location average. \[ (b)\rightarrow(iv) \]

Step 3: Match Mode.
Mode is the value occurring with maximum frequency. \[ (c)\rightarrow(ii) \]

Step 4: Match Frequency Distribution.
Frequency distribution is formed from a list of observations and their frequencies. \[ (d)\rightarrow(i) \]

Step 5: Write the final matching.
\[ a-iii,\quad b-iv,\quad c-ii,\quad d-i \] Therefore, \[ {\text{Option (B)}} \]