the means and mean absolute deviations of the individual times of members on two 4×400-meter relay track…

the means and mean absolute deviations of the individual times of members on two 4×400-meter relay track teams are shown in the table below.\nmeans and mean absolute deviations of\nindividual times of members of 4×400-meter relay track teams\n| | team a | team b |\n| ---- | ---- | ---- |\n| mean | 59.32 s | 59.1 s |\n| mean absolute deviation | 1.5 s | 2.4 s |\nwhat is the ratio of the difference in the means of the two teams to the mean absolute deviation of team b?\n○ 0.09\n○ 0.15\n○ 0.25\n○ 0.65

the means and mean absolute deviations of the individual times of members on two 4×400-meter relay track teams are shown in the table below.\nmeans and mean absolute deviations of\nindividual times of members of 4×400-meter relay track teams\n| | team a | team b |\n| ---- | ---- | ---- |\n| mean | 59.32 s | 59.1 s |\n| mean absolute deviation | 1.5 s | 2.4 s |\nwhat is the ratio of the difference in the means of the two teams to the mean absolute deviation of team b?\n○ 0.09\n○ 0.15\n○ 0.25\n○ 0.65

Answer

Explanation:

Step1: Calculate difference in means

$|59.32 - 59.1| = 0.22$

Step2: Divide by Team B's MAD

$\frac{0.22}{2.4} \approx 0.09$

Answer:

0.09