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?(1 point) Responses The mean score for Group B will be equal to the mean score for the entire class. The mean score for Group B will be equal to the mean score for the entire class. The mean score for Group B will be lower than the mean score for the entire class. The mean score for Group B will be lower than the mean score for the entire class. The mean score for Group B will be the same as the mean score for Group A. The mean score for Group B will be the same as the mean score for Group A. The mean score for Group B will be higher than the mean score for the entire class.

1 answer

To find the mean score for Group B, we can use the information given:

  1. The total number of students is 30.
  2. Therefore, each group (Group A and Group B) has 15 students (30 / 2 = 15).
  3. The mean score for Group A is 91%.
  4. The mean score for the entire class of 30 students is 88%.

To find the mean score for Group B, we can set up an equation based on the overall mean:

Let \( \text{Mean score of Group B} = x \).

The contribution to the overall mean from Group A and Group B can be represented as follows:

\[ \frac{(\text{Mean Group A} \times \text{Number of students in Group A}) + (\text{Mean Group B} \times \text{Number of students in Group B})}{\text{Total number of students}} = \text{Overall Mean} \]

Substituting the values:

\[ \frac{(91 \times 15) + (x \times 15)}{30} = 88 \]

Multiplying both sides by 30 to eliminate the denominator:

\[ (91 \times 15) + (x \times 15) = 88 \times 30 \]

Calculating the values:

\[ 1365 + 15x = 2640 \]

Now, solve for \( x \):

\[ 15x = 2640 - 1365 \] \[ 15x = 1275 \] \[ x = \frac{1275}{15} = 85 \]

Thus, the mean score for Group B is 85%.

Since the mean score for Group B (85%) is lower than the overall class mean (88%), we can conclude that:

The mean score for Group B will be lower than the mean score for the entire class.