The means and mean absolute deviations of the individual times of members on two [tex]4 \times 400[/tex]-meter relay track teams are shown in the table below.

[tex]
\[
\begin{array}{|c|c|c|}
\hline
\multicolumn{3}{|c|}{\text{Means and Mean Absolute Deviations of Individual Times of Members of } 4 \times 400 \text{-meter Relay Track Teams}} \\
\hline
& \text{Team A} & \text{Team B} \\
\hline
\text{Mean} & 59.32 \, \text{s} & 59.1 \, \text{s} \\
\hline
\text{Mean Absolute Deviation} & 1.5 \, \text{s} & 2.4 \, \text{s} \\
\hline
\end{array}
\]
[/tex]

What is the ratio of the difference in the means of the two teams to the mean absolute deviation of Team B?

A. 0.09
B. 0.15
C. 0.25
D. 0.65

To solve the problem, we're going to find the ratio of the difference in the means of the two teams' times to the mean absolute deviation (MAD) of Team B.

1. Identify the means of both teams:
- Mean time for Team A: 59.32 seconds
- Mean time for Team B: 59.1 seconds

2. Calculate the difference in the means:
- Subtract the mean time of Team B from the mean time of Team A:
- Difference in means = 59.32 - 59.1 = 0.22 seconds

3. Identify the Mean Absolute Deviation (MAD) of Team B:
- MAD for Team B = 2.4 seconds

4. Calculate the ratio:
- Divide the difference in means by the MAD of Team B:
- Ratio = 0.22 / 2.4 ≈ 0.0917

Considering the options provided, the closest to this ratio is 0.09. So, the answer to the question is:

0.09