Question

the means and mean absolute deviations of the individual times of members of two relay swim teams are shown in the table. the difference of the means is found and then compared to each of the mean absolute deviations. which is true?

Explanation:

Step1: Calculate the difference of the means

The mean of Team A is \(127.9\) s and the mean of Team B is \(127.4\) s.
The difference of the means is \(127.9 - 127.4=0.5\) s.

Step2: Compare the difference of the means with the mean absolute deviations

The mean absolute deviation of Team A is \(0.26\) s and of Team B is \(0.23\) s.
Let's check the ratios:

• For Team A: \(\frac{0.5}{0.26}\approx1.92\)
• For Team B: \(\frac{0.5}{0.23}\approx2.17\)

The difference of the means (\(0.5\) s) is about \(2\) times the mean absolute deviation of the data sets.

Answer:

The difference between the mean times is about 2 times the mean absolute deviation of the data sets.