Question

which team has the higher mean absolute deviation? team a 8 5 4 3 6 3 team b 7 4 5 6 3 4

Explanation:

Step1: Calculate the mean of Team A

The data for Team A is \(8,5,4,3,6,3\). The mean \(\bar{x}_A=\frac{8 + 5+4+3+6+3}{6}=\frac{29}{6}\approx4.83\).

Step2: Calculate the absolute - deviations of Team A

\(\vert8 - 4.83\vert=3.17\), \(\vert5 - 4.83\vert = 0.17\), \(\vert4 - 4.83\vert=0.83\), \(\vert3 - 4.83\vert = 1.83\), \(\vert6 - 4.83\vert=1.17\), \(\vert3 - 4.83\vert = 1.83\). The mean absolute deviation \(MAD_A=\frac{3.17+0.17 + 0.83+1.83+1.17+1.83}{6}=\frac{8.9}{6}\approx1.48\).

Step3: Calculate the mean of Team B

The data for Team B is \(7,4,5,6,3,4\). The mean \(\bar{x}_B=\frac{7 + 4+5+6+3+4}{6}=\frac{29}{6}\approx4.83\).

Step4: Calculate the absolute - deviations of Team B

\(\vert7 - 4.83\vert=2.17\), \(\vert4 - 4.83\vert = 0.83\), \(\vert5 - 4.83\vert=0.17\), \(\vert6 - 4.83\vert=1.17\), \(\vert3 - 4.83\vert = 1.83\), \(\vert4 - 4.83\vert = 0.83\). The mean absolute deviation \(MAD_B=\frac{2.17+0.83+0.17+1.17+1.83+0.83}{6}=\frac{7}{6}\approx1.17\).

Answer:

Team A