Question

Refer to Column I and Column II and mark the statement(s) that is/are “True” \[ \begin{array}{|c|c|} \hline \textbf{Column I: Class 8, Section A marks} & \textbf{Column II: Class 8, Section B marks} \\ \hline 12 & 18 \\ 12 & 18 \\ 11 & 15 \\ 10 & 15 \\ 10 & 10 \\ 9 & 5 \\ 8 & 2 \\ 8 & 2 \\ \hline \end{array} \]

• A. The mean of Section A is different from that of Section B
• B. The standard deviation of Section A is similar to that of Section B
• C. The mean and the standard deviation of Section A is the same
• D. The mean of Section A and Section B are the same, the standard deviation is different

Hint:

In statistics, the mean is the average of the scores, and the standard deviation measures how spread out the scores are. Comparing these values helps in understanding the distribution and consistency of data.

Answer 1

The Correct Option is D

First, let's calculate the mean and standard deviation for both Section A and Section B: 
- For Section A, the scores are: 12, 12, 11, 10, 10, 9, 8, 8. The mean is calculated as: \[ {Mean of Section A} = \frac{12+12+11+10+10+9+8+8}{8} = \frac{80}{8} = 10. \] 
- For Section B, the scores are: 18, 18, 15, 15, 10, 5, 2, 2. The mean is: \[ {Mean of Section B} = \frac{18+18+15+15+10+5+2+2}{8} = \frac{85}{8} = 10.625. \] Thus, the means of Section A and Section B are slightly different. Therefore, the statement that the means are the same is incorrect. Next, let's calculate the standard deviation: 
- For Section A, the standard deviation is computed using the formula for standard deviation: \( {SD of Section A} = \sqrt{\frac{(12-10)^2 + (12-10)^2 + (11-10)^2 + (10-10)^2 + (10-10)^2 + (9-10)^2 + (8-10)^2 + (8-10)^2}{8}} = \sqrt{2.57} \approx 1.6. \) 
- For Section B, the standard deviation is: \( {SD of Section B} = \sqrt{\frac{(18-10.625)^2 + (18-10.625)^2 + (15-10.625)^2 + (15-10.625)^2 + (10-10.625)^2 + (5-10.625)^2 + (2-10.625)^2 + (2-10.625)^2}{8}} = \sqrt{33.5} \approx 5.79. \) The standard deviation of Section A is much smaller than Section B, confirming that the statement about the standard deviations being different is true. 
Thus, the correct answer is: \[ \boxed{D} \]