To find the mean score for Group B, we can use the information provided about the mean scores of both groups and the overall class.
- The class consists of 30 students, split into two equal groups, A and B, so each group has 15 students.
- The mean score for Group A is 91%.
- The overall mean score for the entire class is 88%.
Let's denote:
- Sum of scores for Group A = Mean score of Group A × Number of students in Group A = 91% × 15 = 1365% (relative to a score out of 100).
- Let the sum of scores for Group B be \( S_B \).
- The total sum of scores for the class is equal to the mean score of the entire class multiplied by the number of students, which gives us 88% × 30 = 2640%.
From this, we can set up the equation for the total scores:
\[ \text{Total scores} = \text{Scores from Group A} + \text{Scores from Group B} \]
So,
\[ 2640 = 1365 + S_B \]
Now, solve for \( S_B \):
\[ S_B = 2640 - 1365 = 1275% \]
Next, we find the mean score for Group B:
\[ \text{Mean score for Group B} = \frac{S_B}{\text{Number of students in Group B}} = \frac{1275}{15} = 85% \]
Comparing the mean score of Group B (85%) to the overall class mean (88%):
- Group B's mean score (85%) is lower than the mean score for the entire class (88%).
So, the correct response is:
The mean score for Group B will be lower than the mean score for the entire class.