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To solve the problem of determining the standard deviation from the provided frequency distribution, we will follow these steps:

1. **Determine the midpoints of the intervals**: 
   For each class interval (e.g., 5–9, 10–14, etc.), we calculate the midpoint.

2. **Calculate the mean**: 
   Using the midpoints and their respective frequencies, calculate the mean (average) value.

3. **Calculate the variance**:
   Use the formula for variance in grouped dataTo solve the problem of determining the standard deviation from the provided frequency distribution, we will follow these steps:

1. **Determine the midpoints of the intervals**: 
   For each class interval (e.g., 5–9, 10–14, etc.), we calculate the midpoint.

2. **Calculate the mean**: 
   Using the midpoints and their respective frequencies, calculate the mean (average) value.

3. **Calculate the variance**:
   Use the formula for variance in grouped 
   \[
   \text{Variance} = \frac{\sum f(x - \mu)^2}{N}
   \]
   where:
   - \( f \) is the frequency of the class,
   - \( x \) is the midpoint of the class,
   - \( \mu \) is the mean,
   - \( N \) is the total frequency (20 in this case).

4. **Calculate the standard deviation**:
   The standard deviation is the square root of the variance.

Let me begin the calculations.

### Step 1: Midpoints for each class
We calculate the midpoint for each interval:
- For 5–9: \( \frac{5+9}{2} = 7 \)
- For 10–14: \( \frac{10+14}{2} = 12 \)
- For 15–19: \( \frac{15+19}{2} = 17 \)
- For 20–24: \( \frac{20+24}{2} = 22 \)
- For 25–29: \( \frac{25+29}{2} = 27 \)

### Step 2: Frequency and midpoints table
The table becomes:

| Midpoint \( x \) | Frequency \( f \) | \( f \times x \) |
|------------------|-------------------|------------------|
| 7                | 6                 | \( 6 \times 7 = 42 \) |
| 12               | 8                 | \( 8 \times 12 = 96 \) |
| 17               | 3                 | \( 3 \times 17 = 51 \) |
| 22               | 2                 | \( 2 \times 22 = 44 \) |
| 27               | 1                 | \( 1 \times 27 = 27 \) |

Total frequency \( N = 6 + 8 + 3 + 2 + 1 = 20 \).

Let me compute the mean and variance next.The mean of the data is \( 13.0 \), the variance is \( 31.5 \), and the standard deviation is approximately \( 5.61 \).

Thus, the correct answer is **C. 5,61**.

Would you like more details or have any questions? Here are five related questions that expand on the concepts:

1. How is the variance formula for grouped data different from ungrouped data?
2. Can you explain how to find the midpoint of a class interval?
3. What is the significance of using frequency in the variance formula?
4. How does standard deviation help in understanding data spread?
5. Can you explain how the total number of samples (N) affects the standard deviation?

**Tip:** Always double-check the class intervals and midpoints when calculating grouped data to ensure accuracy in your mean and standard deviation.

Terdapat 20 orang yang mengajukan pinjaman di koperasi simpan pinjam pada bulan ini. Berikut data banyaknya pinjaman (dalam juta rupiah). Standar deviasi dari data tabel pinjaman di koperasi simpan pinjam adalah ....

Learn how to calculate the standard deviation for grouped data in a cooperative savings and loan scenario. This problem involves determining the midpoints, calculating the mean, variance, and standard deviation from a frequency distribution table. Suitable for students in Grades 10-12.

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