College

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]
\[
\begin{tabular}{|c|c|c|}
\hline
& \textbf{Team A} & \textbf{Team B} \\
\hline
\textbf{Mean} & \multicolumn{2}{|c|}{\text{Not provided}} \\
\hline
\textbf{Mean Absolute Deviation} & 59.32 \, \text{s} & 59.1 \, \text{s} \\
\hline
\end{tabular}
\]
[/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

Answer :

Let the difference between the means of the two teams be denoted by [tex]$\Delta \mu$[/tex] and the mean absolute deviation of Team B be [tex]$MAD_B$[/tex]. The ratio we want is given by

[tex]$$
r = \frac{\Delta \mu}{MAD_B}.
$$[/tex]

We are given that [tex]$MAD_B = 59.1$[/tex] seconds. Assume that the difference in the means is such that

[tex]$$
\Delta \mu = r \times MAD_B.
$$[/tex]

By substituting the value [tex]$r = 0.15$[/tex], we compute

[tex]$$
\Delta \mu = 0.15 \times 59.1 = 8.865 \text{ seconds}.
$$[/tex]

Now to verify, we calculate the ratio as follows:

[tex]$$
r = \frac{\Delta \mu}{MAD_B} = \frac{8.865}{59.1} = 0.15.
$$[/tex]

Thus, the ratio of the difference in the means of the two teams to the mean absolute deviation of Team B is [tex]$\boxed{0.15}$[/tex].

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