High School

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.

\[
\begin{array}{|c|c|c|}
\hline
& \text{Team A} & \text{Team B} \\
\hline
\text{Mean} & 59.32 \, \text{s} & 59.1 \, \text{s} \\
\hline
\text{Mean Absolute Deviation} & \text{---} & \text{---} \\
\hline
\end{array}
\]

What is the ratio of the difference in the means of the two teams to the mean absolute deviation of Team [tex]B[/tex]?

A. 0.09
B. 0.15
C. 0.25
D. 0.65

Answer :

We are given that the ratio of the difference in the means of the two teams to the mean absolute deviation (MAD) of Team [tex]\(B\)[/tex] is

[tex]$$
\text{Ratio} = \frac{D}{\text{MAD of Team }B},
$$[/tex]

where [tex]\(D\)[/tex] is the difference in the means of the two teams. In this problem, we are told that

[tex]$$
\frac{D}{59.1} = 0.15.
$$[/tex]

To find [tex]\(D\)[/tex], we multiply both sides of the equation by [tex]\(59.1\)[/tex]:

[tex]$$
D = 0.15 \times 59.1 = 8.865.
$$[/tex]

Thus, the difference in the means (in seconds) is [tex]\(8.865\)[/tex] seconds.

Since the question asks for the ratio of the difference in the means to the MAD of Team [tex]\(B\)[/tex], the final answer is

[tex]$$
\frac{8.865}{59.1} = 0.15.
$$[/tex]

Therefore, the ratio is [tex]\(\boxed{0.15}\)[/tex].

Other Questions