The means and mean absolute deviations of the individual times of members of two relay swim teams are shown in the table below.

\begin{tabular}{|c|c|c|}
\hline \multicolumn{2}{|c|}{\begin{tabular}{c}
Means and Mean Absolute Deviations of \\
Individual Times of Members of [tex]$4 \times 200$[/tex]-meter Relay Swim Teams \\
\end{tabular}} & \\
\hline Team & A & B \\
\hline Mean & [tex]$127.9 s$[/tex] & [tex]$127.4 s$[/tex] \\
\hline Mean Absolute Deviation & [tex]$0.26 s$[/tex] & [tex]$0.23 s$[/tex] \\
\hline
\end{tabular}

The difference of the means is found and then compared to each of the mean absolute deviations. Which is true?

A. The difference between the mean times is about equal to the mean absolute deviation of the data sets.
B. The difference between the mean times is about 2 times the mean absolute deviation of the data sets.
C. The difference between the mean times is about 5 times the mean absolute deviation of the data sets.
D. The difference between the mean times is about 16 times the mean absolute deviation of the data sets.



Answer :

To determine which statement is true about the difference between the mean times, we start by finding the difference between the means of the two relay swim teams.

1. Calculate the difference between the means:
- Mean of Team A: [tex]\( 127.9 \)[/tex] seconds
- Mean of Team B: [tex]\( 127.4 \)[/tex] seconds
- Difference between the means: [tex]\( |127.9 - 127.4| = 0.5 \)[/tex] seconds

2. Mean Absolute Deviation (MAD):
- MAD of Team A: [tex]\( 0.26 \)[/tex] seconds
- MAD of Team B: [tex]\( 0.23 \)[/tex] seconds

3. Compare the difference between the means to each of the MADs:
- The MADs for the two teams are: [tex]\( 0.26 \)[/tex] seconds and [tex]\( 0.23 \)[/tex] seconds.
- The maximum of these MADs is [tex]\( \text{max}(0.26, 0.23) = 0.26 \)[/tex] seconds.

4. Evaluate relative comparisons:
- The difference [tex]\( 0.5 \)[/tex] seconds is about:
- [tex]\( \dfrac{0.5 \text{ seconds}}{0.26 \text{ seconds}} \approx 1.92 \)[/tex]

5. Interpret the relative comparison:
- The difference between the mean times [tex]\( 0.5 \)[/tex] seconds is approximately [tex]\( 2 \)[/tex] times the mean absolute deviation [tex]\( (2 \times 0.26 \approx 0.52) \)[/tex].

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

Therefore, the correct statement is:
"The difference between the mean times is about 2 times the mean absolute deviation of the data sets."