Question

Tom and Abby wanted to see whose grades were the closest to their final grade after their last unit test. For Tom's 5 lesson/test grades he has: 74, 65, 88, 90, 100, 75 For Abby's 5 lesson/test grades she has: 77, 78, 80, 80, 90, 93. Feel free to use the worksheet to show your work, or complete all work on your own paper. Either way, all work must be shown. If needed, round to the nearest tenths place. Question 1: What is the mean or average percent for both Tom and Abby's overall unit grade? Question 2: What is the MAD of Tom and Abby's unit grade? Question 3: Which student's (Tom or Abby) grades were the closest to their overall average? For this student, what could this mean about their grades? Make sure to answer both parts of this question.​

Answer 1

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.

$\textsf{Tom: Mean}=\dfrac{74+65+88+90+100+75}{6}=\dfrac{492}{6}=82$

$\textsf{Abby: Mean}=\dfrac{77+78+80+80+90+93}{6}=\dfrac{498}{6}=83$

Therefore, the mean of Tom's overall unit grade is 82 and the mean of Abby's overall unit grade is 83.

$\dotfill$

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:

$\textsf{Tom: MAD}=\dfrac{8+17+6+8+18+7}{6}=\dfrac{64}{6}=10.7$

Abby

Absolute deviations:

|77 - 83| = 6

|78 - 83| = 5

|80 - 83| = 3

|80 - 83| = 3

|90 - 83| = 7

|93 - 83| = 10

Therefore:

$\textsf{Abby: MAD}=\dfrac{6+5+3+3+7+10}{6}=\dfrac{34}{6}=5.7$

So, the MAD of Tom's unit grade is 10.7, and the MAD of Abby's unit grade is 5.7.

$\dotfill$

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.