Question

Jason and Jill or two students in mr. White's math class. On the last 5 quizzes Jason scored an 80 90 95 85 and 70. Jill scored a 70 75 90 195 find the mean and me in mean absolute deviation for each student?

Answer 1

Jason's scores : 80 90 95 85 70 and

Jill's score : 70 75 90 100 95.

Mean of Jason's scores = $\frac{80 + 90 + 95 + 85 + 70}{5}=\frac{420}{5}=84.$

Mean of Jill's scores = $\frac{70+75+90+100+95}{5}=\frac{430}{5}=86$

Now, in order to find the mean absolute deviation, need to find the difference of each score from means.

Mean absolute deviation for Jason's scores.        

|84-80| = 4

|84-90| = 6

|84-95| = 9

|84-85| = 1

|84-70|= 14

$\frac{4+6+9+1+14}{5}=\frac{34}{5}=6.8$

Mean absolute deviation for Jill's scores

|86-70| = 16

|86-75| = 11

|86-90| = 4

|86-100| = 14

|86-95|= 9

$\frac{16+11+4+14+9}{5}=\frac{54}{5}=10.8$

Jill got average quiz score 86 and Jason got 84.

Therefore, Jill got better quiz average.

Also, the mean absolute deviation for Jason scores is less that is 6.8 than 10.8.

Therefore, Jason got more consistent grades.