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{tabular}{|l|l|l|}

\hline

\multicolumn{3}{|c|}{Means and Mean Absolute Deviations of Individual Times of Members of $4 \times 400$-meter Relay Track Teams} \\

\hline

& Team A & Team B \\

\hline

Mean & 59.32 s & 59.1 s \\

\hline

Mean Absolute Deviation & 1.5 s & 2.45 s \\

\hline

\end{tabular}

\][/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

We are given the following information for the two teams:

- Team A has a mean time of [tex]$59.32$[/tex] seconds.
- Team B has a mean time of [tex]$59.1$[/tex] seconds and a mean absolute deviation (MAD) of [tex]$2.45$[/tex] seconds.

We want to find the ratio of the absolute difference in the means of the two teams to the MAD of Team B.

Step 1. Calculate the difference in the means.

The difference is given by:
[tex]$$
\text{Difference} = \lvert 59.32 - 59.1 \rvert.
$$[/tex]
Subtracting, we get:
[tex]$$
59.32 - 59.1 = 0.22 \text{ seconds (approximately)}.
$$[/tex]

Step 2. Calculate the ratio.

Now, divide the absolute difference by the MAD of Team B:
[tex]$$
\text{Ratio} = \frac{0.22}{2.45}.
$$[/tex]

Evaluating this gives:
[tex]$$
\text{Ratio} \approx 0.09.
$$[/tex]

Thus, the ratio of the difference in the means of the two teams to the mean absolute deviation of Team B is approximately [tex]$0.09$[/tex].