Question

The scores for an online exam for a class of 30 students were randomly split into two groups of equal size, Group A and Group B. Group A had a mean score of 91%. It was then found that the entire class had a mean score of 88%. What can be said about the mean score for Group B?
A. The mean score for Group B will be equal to the mean score for the entire class.
B. The mean score for Group B will be lower than the mean score for the entire class.
C. The mean score for Group B will be the same as the mean score for Group A.
D. The mean score for Group B will be higher than the mean score for the entire class.

Answer 1

Let \( n = 30 \) be the total number of students in the class. The class is split into two equal groups: Group A and Group B, each with \( n/2 = 15 \) students.

Given:
- Mean score for Group A (\( M_A \)) = 91%
- Mean score for the entire class (\( M \)) = 88%

First, we can find the total scores for both groups using the mean:

1. Total score of Group A:
\[
\text{Total score for Group A} = M_A \times \text{Number of students in Group A} = 91 \times 15 = 1365
\]

2. Total score of the entire class:
\[
\text{Total score for the entire class} = M \times n = 88 \times 30 = 2640
\]

Now, let's find the total score for Group B. Since the total score for the entire class is the sum of the scores of Group A and Group B:
\[
\text{Total score for Group B} = \text{Total score for the entire class} - \text{Total score for Group A} = 2640 - 1365 = 1275
\]

Next, we can calculate the mean score for Group B (\( M_B \)):
\[
M_B = \frac{\text{Total score for Group B}}{\text{Number of students in Group B}} = \frac{1275}{15} = 85%
\]

Now we compare the mean score for Group B with the mean score of the entire class. We have:
- \( M_B = 85\% \)
- \( M = 88\% \)

Since 85% is less than 88%, we conclude that the mean score for Group B will be lower than the mean score for the entire class.

Therefore, the correct answer is:
**B. The mean score for Group B will be lower than the mean score for the entire class.**