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This image provides a frequency distribution table showing "Nilai" (values) and "Frekuensi" (frequencies). To solve the question, we likely need to find the mean, median, or another statistical value, and the multiple-choice options at the bottom suggest that we need to calculate something specific from this distribution.

Let's break it down step-by-step, assuming we are tasked with calculating the **mean** (as the provided answer options suggest):

### 1. Midpoint of each class interval:
The midpoint (\( x_i \)) is calculated as the average of the lower and upper limits of each interval:
- \( \text{Midpoint of } 11 - 20 = \frac{11 + 20}{2} = 15.5 \)
- \( \text{Midpoint of } 21 - 30 = \frac{21 + 30}{2} = 25.5 \)
- \( \text{Midpoint of } 31 - 40 = \frac{31 + 40}{2} = 35.5 \)
- \( \text{Midpoint of } 41 - 50 = \frac{41 + 50}{2} = 45.5 \)
- \( \text{Midpoint of } 51 - 60 = \frac{51 + 60}{2} = 55.5 \)
- \( \text{Midpoint of } 61 - 70 = \frac{61 + 70}{2} = 65.5 \)
- \( \text{Midpoint of } 71 - 80 = \frac{71 + 80}{2} = 75.5 \)
- \( \text{Midpoint of } 81 - 90 = \frac{81 + 90}{2} = 85.5 \)
- \( \text{Midpoint of } 91 - 100 = \frac{91 + 100}{2} = 95.5 \)

### 2. Mean Formula:
The formula for the mean (\( \bar{x} \)) in a frequency distribution is:
\[
\bar{x} = \frac{\sum (f_i \cdot x_i)}{\sum f_i}
\]
Where:
- \( f_i \) is the frequency of each class,
- \( x_i \) is the midpoint of each class.

Now, we multiply each midpoint by the corresponding frequency and sum them up.

### 3. Calculations:

| Class Interval | Midpoint \( x_i \) | Frequency \( f_i \) | \( f_i \cdot x_i \) |
|----------------|--------------------|---------------------|---------------------|
| 11 – 20        | 15.5               | 3                   | 46.5                |
| 21 – 30        | 25.5               | 7                   | 178.5               |
| 31 – 40        | 35.5               | 10                  | 355                 |
| 41 – 50        | 45.5               | 16                  | 728                 |
| 51 – 60        | 55.5               | 20                  | 1110                |
| 61 – 70        | 65.5               | 14                  | 917                 |
| 71 – 80        | 75.5               | 10                  | 755                 |
| 81 – 90        | 85.5               | 6                   | 513                 |
| 91 – 100       | 95.5               | 4                   | 382                 |

Now, summing up the products \( f_i \cdot x_i \):

\[
\sum (f_i \cdot x_i) = 46.5 + 178.5 + 355 + 728 + 1110 + 917 + 755 + 513 + 382 = 4985
\]

### 4. Total Frequency \( \sum f_i = 90 \).

### 5. Mean:

\[
\bar{x} = \frac{4985}{90} \approx 55.39
\]

The closest option to this result is **56 (option d)**.

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Would you like a more detailed explanation or to ask about other calculations?

Here are 5 related questions you may want to explore:
1. How do you calculate the median from a frequency distribution?
2. How would you determine the mode in this dataset?
3. What if the class intervals were unequal—how would the calculation change?
4. How can we represent this data graphically (histogram, frequency polygon)?
5. What is the significance of the standard deviation for this dataset?

**Tip:** Always check if the class intervals are equal, as unequal intervals require additional adjustments when calculating the mean, median, or mode.

Calculate a specific statistical value (likely the mean) from the frequency distribution table provided.

Learn how to calculate the mean from a frequency distribution table, with a step-by-step breakdown using midpoints and frequencies. This guide is ideal for students in Grades 8-10, covering key statistical concepts like the mean and frequency distribution.

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