Respuesta :
Answer:
1) Tom = 82, Abby = 83
2) Tom = 10.7, Abby = 5.7
3) Abby. See below for an explanation.
Step-by-step explanation:
Question 1
To find the mean or average percent for each student's overall unit grade, divide the sum of the individual grades by the number of grades.
[tex]\textsf{Tom: Mean}=\dfrac{74+65+88+90+100+75}{6}=\dfrac{492}{6}=82[/tex]
[tex]\textsf{Abby: Mean}=\dfrac{77+78+80+80+90+93}{6}=\dfrac{498}{6}=83[/tex]
Therefore, the mean of Tom's overall unit grade is 82 and the mean of Abby's overall unit grade is 83.
[tex]\dotfill[/tex]
Question 2
To calculate the MAD (Mean Absolute Deviation), first find the absolute deviation of each individual grade from the mean grade, then take the mean of these absolute deviations.
Tom
Absolute deviations:
|74 - 82| = 8
|65 - 82| = 17
|88 - 82| = 6
|90 - 82| = 8
|100 - 82| = 18
|75 - 82| = 7
Therefore:
[tex]\textsf{Tom: MAD}=\dfrac{8+17+6+8+18+7}{6}=\dfrac{64}{6}=10.7[/tex]
Abby
Absolute deviations:
|77 - 83| = 6
|78 - 83| = 5
|80 - 83| = 3
|80 - 83| = 3
|90 - 83| = 7
|93 - 83| = 10
Therefore:
[tex]\textsf{Abby: MAD}=\dfrac{6+5+3+3+7+10}{6}=\dfrac{34}{6}=5.7[/tex]
So, the MAD of Tom's unit grade is 10.7, and the MAD of Abby's unit grade is 5.7.
[tex]\dotfill[/tex]
Question 3
As the MAD of Abby's unit grade is smaller than Tom's, this indicates that Abby's grades were closer to her overall average.
This suggests that Abby's performance across her unit tests was more consistent, with less deviation from her overall average compared to Tom's grades.