Jumlah peserta ujian yang akan diterima sebagai mahasiswa adalah 60% dari jumlah peserta ujian. Nilai terendah dari peserta ujian yang lolos tes adalah ...

Sum the frequencies from the table:

So, the total number of participants is 300.

The problem states that 60% of the total participants will be accepted.

These 180 students are those with the highest scores.

If 180 students are accepted (the top 180 scores), then $300 - 180 = 120$ students are not accepted (the bottom 120 scores). The lowest score among the accepted students is the score that separates the bottom 120 students from the rest. This means we are looking for the score that corresponds to the 120th data point from the bottom when scores are arranged in ascending order. This is equivalent to finding the 40th percentile ($P_{40}$), since $\frac{120}{300} \times 100\% = 40\%$.

The formula for the $k$-th percentile ($P_k$) is:

Where:

• $L$ = lower boundary of the percentile class
• $N$ = total number of data points (300)
• $F$ = cumulative frequency of the class before the percentile class
• $f$ = frequency of the percentile class
• $c$ = class width

First, find the position of the 40th percentile:

Now, let's look at the cumulative frequencies:

• Class 36-45: Cumulative frequency = 12
• Class 46-55: Cumulative frequency = $12 + 48 = 60$
• Class 56-65: Cumulative frequency = $60 + 60 = 120$

The 120th data point falls exactly at the end of the class 56-65.
Therefore, the percentile class is 56-65.

• $L = 56 - 0.5 = 55.5$ (lower boundary of the class 56-65)
• $F = 60$ (cumulative frequency of the class before 56-65, which is 46-55)
• $f = 60$ (frequency of the class 56-65)
• $c = 65 - 56 + 1 = 10$ (class width)

Substitute these values into the formula:

The minimum score for a participant to be accepted is 65.5.

The final answer is $\text{D. 65,5}$.