The following dot plots represent the scores on the Chapter 5 quiz for Mrs. Chin's 4th and 6th period classes.

Calculate the mean and mean absolute deviation (rounded to the nearest tenth) for both classes.
Use your answer calculations from part A to answer all of the following questions: Which class period, on average, scored better on the quiz? By how much did they score better? How does the difference between the mean scores compare to the mean absolute deviation? Is there much overlap in the data? Write your answers in complete sentences.

(for set A: Chapter 5 math quiz 4th period)

1) Mean:  

1 x 10 = 10

3 x 12 = 36  

4 x 13 = 52

2 x 14 = 28

6 x 15 = 90

2 x 16 = 32

1 x 19 = 19

Mean: 10 + 36 + 52 + 28 + 90 + 32 + 19 = 267/19 = 14.05

(MAD) Mean Absolute Deviation:

14.05 - 10 = 4.05

14.05 - 12 = 2.05 x 3 = 6.15

14.05 - 13 = 1.05 x 4 = 4.2

14.05 - 14 = 0.05 x 2 = 0.1

14.05 - 15 = 0.95 x 6 = 5.7

14.05 - 16 = 1.95 x 2 = 3.9  

14.05 - 19 = 4.95

Mean Absolute Deviation: 4.05 + 6.15 + 4.2 + 0.1 + 5.7 + 3.9 + 4.95 = 29.05/ 19 = 1.5

(for set B: chapter 5 math quiz 6th period)

2) Mean:  

13 x 1 = 13  

14 x 1 = 14  

19 x 1 = 19  

20 x 1 = 20  

3 x 15 = 45  

2 x 16 = 32  

4 x 17 = 68  

5 x 18 = 90

Mean: 13 + 14 + 19 + 20 + 45 + 32 + 68 + 90 = 301/18 = 16.7

16.7 - 13 = 3.7  

16.7 - 14 = 2.7  

19 - 16.7 = 2.3

20 - 16.7 = 3.3

16.7 - 15 = 1.7 x 3 = 5.1

16.7 - 16 = 0.7 x 2 = 1.4

17 - 16.7 = 0.3 x 4 = 1.2

18 - 16.7 = 1.3 x 5 = 6.5

Mean Absolute Deviation: 3.7 + 2.7 + 2.3 + 3.3 + 5.1 + 1.4 + 1.2 + 6.5 = 26.2/18 = 1.45

answer: the anser is 3

step-by-step explanation:

c- 2o is equal to 9

asdsaddsa

step-by-step explanation: