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The number of social interactions of ten minutes or longer over a one-week period for a group of college students is given in the following frequency table (These interactions excluded family and work situations.) Find the mean, median, mode, and midrange for these data
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline \begin{tabular}{c}
Social interactions in \\
a week: $x$
\end{tabular} & 6 & 7 & 13 & 16 & 25 & 26 & 32 & 38 & 45 & 46 \\
\hline \begin{tabular}{c}
Number of \\
college students: $\mathrm{f}$
\end{tabular} & 14 & 17 & 17 & 17 & 12 & 13 & 3 & 2 & 2 & 2 \\
\hline
\end{tabular}

The mean is approximately $\square$
(Round to the nearest tenth as needed.)
The median is $\square$.
(Round to the nearest tenth as needed)
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The mode of the data is
(Use a comma to separate answers as needed.)
B. There is no mode for the given data.
(1) Time Remaining: 01:19:39
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Step 1 ：To find the mean, we need to multiply each social interaction by the number of students who had that many interactions, add those products together, and then divide by the total number of students. The formula is \(\frac{(6 \times 14) + (7 \times 17) + (13 \times 17) + (16 \times 17) + (25 \times 12) + (26 \times 13) + (32 \times 3) + (38 \times 2) + (45 \times 2) + (46 \times 2)}{14 + 17 + 17 + 17 + 12 + 13 + 3 + 2 + 2 + 2}\)

Step 2 ：After calculating the above expression, we get \(\frac{1788}{99}\)

Step 3 ：Solving the above expression, we get the mean as \(\boxed{18.06}\) (rounded to the nearest tenth)

Step 4 ：To find the median, we need to find the middle value. Since there are 99 students, the median will be the 50th value. Looking at the cumulative frequency, we can see that the 50th value falls in the 16 social interactions category. So, the median is \(\boxed{16}\)

Step 5 ：The mode is the value that appears most frequently. In this case, 7, 13, and 16 all appear 17 times, which is the highest frequency. So, the modes are \(\boxed{7, 13, 16}\)

Step 6 ：The midrange is the average of the highest and lowest values. In this case, the highest value is 46 and the lowest value is 6. The formula is \(\frac{46 + 6}{2}\)

Step 7 ：Solving the above expression, we get the midrange as \(\boxed{26}\)