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

[tex]
\[
\begin{array}{|c|c|c|}
\hline
\multicolumn{3}{|c|}{\text{Means and Mean Absolute Deviations of Individual Times of Members of } 4 \times 400\text{-meter Relay Track Teams} } \\
\hline
& \text{Team A} & \text{Team B} \\
\hline
\text{Mean} & 59.32 \, \text{s} & 59.1 \, \text{s} \\
\hline
\text{Mean Absolute Deviation} & 0.22 \, \text{s} & 0.15 \, \text{s} \\
\hline
\end{array}
\]
[/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 asked to find the ratio of the difference in the means of two teams' relay times to the mean absolute deviation of Team B.

Here are the steps to solve it:

1. Identify the means of the two teams:
- The mean time for Team A is 59.32 seconds.
- The mean time for Team B is 59.1 seconds.

2. Calculate the difference in the means of the two teams:
- Subtract the mean of Team B from the mean of Team A:
[tex]\[
\text{Difference in means} = 59.32 - 59.1 = 0.22
\][/tex]

3. Identify the mean absolute deviation of Team B:
- This value is given as 0.09.

4. Calculate the ratio of the difference in the means to the mean absolute deviation of Team B:
- Divide the difference in means by the mean absolute deviation of Team B:
[tex]\[
\text{Ratio} = \frac{0.22}{0.09} \approx 2.44
\][/tex]

So, the ratio of the difference in the means to the mean absolute deviation of Team B is approximately [tex]$2.44$[/tex].