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]
\[
\begin{array}{|c|c|c|}
\hline
\text{Statistic} & \text{Team A} & \text{Team B} \\
\hline
\text{Mean} & 59.32 \, \text{s} & 59.1 \, \text{s} \\
\hline
\text{Mean Absolute Deviation} & 1.5 \, \text{s} & \text{(unknown)} \\
\hline
\end{array}
\]
[/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

Sure! Let's break down the solution to find the ratio of the difference in the means of the two teams to the mean absolute deviation of Team B.

1. Determine the Means of Each Team:
- 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 Means:
- Subtract the mean of Team B from the mean of Team A to find the difference:
[tex]\[
\text{Difference in Means} = 59.32 - 59.1 = 0.22 \text{ seconds}
\][/tex]

3. Mean Absolute Deviation for Team B:
- The mean absolute deviation (MAD) for Team B is given as 1.5 seconds.

4. Calculate the Ratio:
- To find the ratio of the difference in means to the mean absolute deviation of Team B:
[tex]\[
\text{Ratio} = \frac{\text{Difference in Means}}{\text{MAD of Team B}} = \frac{0.22}{1.5}
\][/tex]
- This calculation gives a ratio of approximately 0.15.

Thus, the ratio of the difference in the means of the two teams to the mean absolute deviation of Team B is 0.15.