greg teaches an art class. the table below shows how many drawings his students had submitted by last…

greg teaches an art class. the table below shows how many drawings his students had submitted by last friday. greg calculates the mean absolute deviation (mad) of the data. then, one student submits 25 additional drawings. greg cannot remember whether the drawings are amy’s or emily’s, but he thinks the mad will increase no matter who submitted the drawings. is greg correct? use the drop - down menus to explain your reasoning.\n\ngreg’s students’ drawings\n| student | number of drawings submitted |\n| ---- | ---- |\n| amy | 6 |\n| bob | 34 |\n| christa | 35 |\n| diego | 37 |\n| emily | 43 |\n\nclick the arrows to choose an answer from each menu.\nthe mad of the data in the table is choose... . if the additional drawings are amy’s, the mad of the data set will choose... . if they are emily’s, the mad will choose... . the mad of the new data set choose... depend on whether it was amy or emily who turned in the additional drawings. so, greg is choose... .

greg teaches an art class. the table below shows how many drawings his students had submitted by last friday. greg calculates the mean absolute deviation (mad) of the data. then, one student submits 25 additional drawings. greg cannot remember whether the drawings are amy’s or emily’s, but he thinks the mad will increase no matter who submitted the drawings. is greg correct? use the drop - down menus to explain your reasoning.\n\ngreg’s students’ drawings\n| student | number of drawings submitted |\n| ---- | ---- |\n| amy | 6 |\n| bob | 34 |\n| christa | 35 |\n| diego | 37 |\n| emily | 43 |\n\nclick the arrows to choose an answer from each menu.\nthe mad of the data in the table is choose... . if the additional drawings are amy’s, the mad of the data set will choose... . if they are emily’s, the mad will choose... . the mad of the new data set choose... depend on whether it was amy or emily who turned in the additional drawings. so, greg is choose... .

Answer

Explanation:

Step1: Calculate the mean of the original data - set

The original data set is (6,34,35,37,43). The mean (\bar{x}=\frac{6 + 34+35+37+43}{5}=\frac{155}{5}=31).

Step2: Calculate the absolute - deviations and the MAD of the original data - set

The absolute deviations are (|6 - 31|=25), (|34 - 31| = 3), (|35 - 31|=4), (|37 - 31|=6), (|43 - 31| = 12). The MAD of the original data set is (\frac{25+3+4+6+12}{5}=\frac{50}{5}=10).

Step3: Case 1: If the additional drawings are Amy's

The new data set is (6 + 25=31,34,35,37,43). The new mean (\bar{x}_{1}=\frac{31+34+35+37+43}{5}=\frac{180}{5}=36). The absolute deviations are (|31 - 36|=5), (|34 - 36| = 2), (|35 - 36|=1), (|37 - 36|=1), (|43 - 36| = 7). The new MAD is (\frac{5+2+1+1+7}{5}=\frac{16}{5}=3.2), which is less than the original MAD.

Step4: Case 2: If the additional drawings are Emily's

The new data set is (6,34,35,37,43 + 25=68). The new mean (\bar{x}_{2}=\frac{6+34+35+37+68}{5}=\frac{180}{5}=36). The absolute deviations are (|6 - 36|=30), (|34 - 36| = 2), (|35 - 36|=1), (|37 - 36|=1), (|68 - 36| = 32). The new MAD is (\frac{30+2+1+1+32}{5}=\frac{66}{5}=13.2), which is greater than the original MAD.

Answer:

The MAD of the data in the table is (10). If the additional drawings are Amy's, the MAD of the data set will decrease. If they are Emily's, the MAD will increase. The MAD of the new data set does depend on whether it was Amy or Emily who turned in the additional drawings. So, Greg is incorrect.