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Answer :
We begin by noting that there are a total of [tex]$50$[/tex] students (since there are [tex]$5$[/tex] classes with [tex]$10$[/tex] students each).
The survey provided the following frequencies for the number of days on campus:
[tex]\[
\begin{array}{c|c}
\text{Number of Days} & \text{Frequency} \\ \hline
0 & 1 \\
1 & 2 \\
2 & 12 \\
3 & 10 \\
5 & 0 \\
\end{array}
\][/tex]
However, the frequency for [tex]$4$[/tex] days is missing. Since the total number of students is [tex]$50$[/tex], we can determine the frequency for [tex]$4$[/tex] days by subtracting the sum of the known frequencies from [tex]$50$[/tex].
First, add the given frequencies:
[tex]\[
1 + 2 + 12 + 10 + 0 = 25.
\][/tex]
Thus, the frequency for students who attend campus [tex]$4$[/tex] days per week is:
[tex]\[
50 - 25 = 25.
\][/tex]
Now, the complete distribution of the data is:
[tex]\[
\begin{array}{c|c}
\text{Number of Days} & \text{Frequency} \\ \hline
0 & 1 \\
1 & 2 \\
2 & 12 \\
3 & 10 \\
4 & 25 \\
5 & 0 \\
\end{array}
\][/tex]
Next, we calculate the cumulative frequencies. We add the frequencies in increasing order of the number of days:
- For [tex]$0$[/tex] days: the cumulative frequency is [tex]$1$[/tex].
- For [tex]$1$[/tex] day: [tex]$1 + 2 = 3$[/tex].
- For [tex]$2$[/tex] days: [tex]$3 + 12 = 15$[/tex].
- For [tex]$3$[/tex] days: [tex]$15 + 10 = 25$[/tex].
- For [tex]$4$[/tex] days: [tex]$25 + 25 = 50$[/tex].
- For [tex]$5$[/tex] days: [tex]$50 + 0 = 50$[/tex].
So, the cumulative frequencies are:
[tex]\[
1,\, 3,\, 15,\, 25,\, 50,\, 50.
\][/tex]
The [tex]$60^\text{th}$[/tex] percentile is the value below which [tex]$60\%$[/tex] of the data lie. To find the number of students corresponding to the [tex]$60^\text{th}$[/tex] percentile, calculate:
[tex]\[
0.6 \times 50 = 30.
\][/tex]
We now look for the smallest number of days for which the cumulative frequency is at least [tex]$30$[/tex].
Reviewing our cumulative frequencies:
[tex]\[
\begin{array}{c|c}
\text{Number of Days} & \text{Cumulative Frequency} \\ \hline
0 & 1 \\
1 & 3 \\
2 & 15 \\
3 & 25 \\
4 & 50 \\
5 & 50 \\
\end{array}
\][/tex]
The cumulative frequency exceeds [tex]$30$[/tex] for the first time when the number of days is [tex]$4$[/tex] (since [tex]$25 < 30$[/tex] and then [tex]$50 \ge 30$[/tex]).
Therefore, the [tex]$60^\text{th}$[/tex] percentile for the data is [tex]$4$[/tex] days.
In summary, after determining the missing frequency, computing the cumulative frequencies, and finding the threshold corresponding to [tex]$60\%$[/tex] of the data, we conclude that the [tex]$60^\text{th}$[/tex] percentile is:
[tex]\[
\boxed{4}
\][/tex]
The survey provided the following frequencies for the number of days on campus:
[tex]\[
\begin{array}{c|c}
\text{Number of Days} & \text{Frequency} \\ \hline
0 & 1 \\
1 & 2 \\
2 & 12 \\
3 & 10 \\
5 & 0 \\
\end{array}
\][/tex]
However, the frequency for [tex]$4$[/tex] days is missing. Since the total number of students is [tex]$50$[/tex], we can determine the frequency for [tex]$4$[/tex] days by subtracting the sum of the known frequencies from [tex]$50$[/tex].
First, add the given frequencies:
[tex]\[
1 + 2 + 12 + 10 + 0 = 25.
\][/tex]
Thus, the frequency for students who attend campus [tex]$4$[/tex] days per week is:
[tex]\[
50 - 25 = 25.
\][/tex]
Now, the complete distribution of the data is:
[tex]\[
\begin{array}{c|c}
\text{Number of Days} & \text{Frequency} \\ \hline
0 & 1 \\
1 & 2 \\
2 & 12 \\
3 & 10 \\
4 & 25 \\
5 & 0 \\
\end{array}
\][/tex]
Next, we calculate the cumulative frequencies. We add the frequencies in increasing order of the number of days:
- For [tex]$0$[/tex] days: the cumulative frequency is [tex]$1$[/tex].
- For [tex]$1$[/tex] day: [tex]$1 + 2 = 3$[/tex].
- For [tex]$2$[/tex] days: [tex]$3 + 12 = 15$[/tex].
- For [tex]$3$[/tex] days: [tex]$15 + 10 = 25$[/tex].
- For [tex]$4$[/tex] days: [tex]$25 + 25 = 50$[/tex].
- For [tex]$5$[/tex] days: [tex]$50 + 0 = 50$[/tex].
So, the cumulative frequencies are:
[tex]\[
1,\, 3,\, 15,\, 25,\, 50,\, 50.
\][/tex]
The [tex]$60^\text{th}$[/tex] percentile is the value below which [tex]$60\%$[/tex] of the data lie. To find the number of students corresponding to the [tex]$60^\text{th}$[/tex] percentile, calculate:
[tex]\[
0.6 \times 50 = 30.
\][/tex]
We now look for the smallest number of days for which the cumulative frequency is at least [tex]$30$[/tex].
Reviewing our cumulative frequencies:
[tex]\[
\begin{array}{c|c}
\text{Number of Days} & \text{Cumulative Frequency} \\ \hline
0 & 1 \\
1 & 3 \\
2 & 15 \\
3 & 25 \\
4 & 50 \\
5 & 50 \\
\end{array}
\][/tex]
The cumulative frequency exceeds [tex]$30$[/tex] for the first time when the number of days is [tex]$4$[/tex] (since [tex]$25 < 30$[/tex] and then [tex]$50 \ge 30$[/tex]).
Therefore, the [tex]$60^\text{th}$[/tex] percentile for the data is [tex]$4$[/tex] days.
In summary, after determining the missing frequency, computing the cumulative frequencies, and finding the threshold corresponding to [tex]$60\%$[/tex] of the data, we conclude that the [tex]$60^\text{th}$[/tex] percentile is:
[tex]\[
\boxed{4}
\][/tex]
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