Refer to Column I and Column II and mark the statement(s) that is/are “True”
\[
\begin{array}{|c|c|}
\hline
\textbf{Column I: Class 8, Section A marks} & \textbf{Column II: Class 8, Section B marks} \\
\hline
12 & 18 \\
12 & 18 \\
11 & 15 \\
10 & 15 \\
10 & 10 \\
9 & 5 \\
8 & 2 \\
8 & 2 \\
\hline
\end{array}
\]
Show Hint
- The mean of Section A is different from that of Section B
- The standard deviation of Section A is similar to that of Section B
- The mean and the standard deviation of Section A is the same
- The mean of Section A and Section B are the same, the standard deviation is different
The Correct Option is D
Solution and Explanation
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: \[ \boxed{D} \]
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