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

Answer :

To solve this problem, we need to find the ratio of the difference in the means of two teams to the mean absolute deviation (MAD) of Team B.

Here's how we do it step-by-step:

1. Identify the Means of the Two Teams:
- The mean time for Team A: 59.32 seconds.
- The mean time for Team B: 59.1 seconds.

2. Calculate the Difference in Means:
- Subtract the mean of Team B from the mean of Team A:
[tex]\[
\text{Difference in means} = 59.32 - 59.1 = 0.22 \text{ seconds}
\][/tex]

3. Identify the Mean Absolute Deviation (MAD) of Team B:
- The MAD for Team B is given as 59.1 seconds.

4. Calculate the Ratio:
- Divide the difference in means by the MAD of Team B:
[tex]\[
\text{Ratio} = \frac{0.22}{59.1}
\][/tex]

5. Find the Result:
- After performing the division, the ratio is approximately 0.0037.

Given the choices provided, it appears that there might be a misunderstanding or misalignment due to rounding or interpretation errors. The expected answers are a bit off from this calculation. It's common for multiple-choice questions to use rounding or to check for specific value interpretations. Unfortunately, without further context, none of the given options (0.09, 0.15, 0.25, 0.65) match the calculated ratio closely.

If there is additional context or if the question has a different interpretation of the provided values, please double-check the parameters or verify any assumptions.