Question #3: A group of high school students were asked to report the number of times they missed
school in the past year. Their responses are shown in the grouped frequency distribution below.
 # of Days f
__________________________
 0-4 70
 5-9 52
 10-14 38
 15-19 21
 20-24 10
 25-29 8
_________________________
According to the above data, the mean number of days the students missed school in the past year is
___________ (5 points).

To calculate the **mean number of days** missed by students, we'll need to use a weighted average formula. Since the data is grouped into intervals, we'll follow these steps:

### Steps to Calculate the Mean:
1. **Find the midpoint of each class interval**: The midpoint (\( x_i \)) is the average of the lower and upper bounds of each class interval.
2. **Multiply the midpoint by the frequency for each class**: This will give us the total number of missed days contributed by each group.
3. **Find the total frequency**: Add up all the frequencies.
4. **Find the sum of the products of midpoints and frequencies**.
5. **Divide the sum of the products by the total frequency**: This gives the mean.

### 1. Midpoints for each interval:
\[
\text{Midpoint of } 0-4 = \frac{0 + 4}{2} = 2
\]
\[
\text{Midpoint of } 5-9 = \frac{5 + 9}{2} = 7
\]
\[
\text{Midpoint of } 10-14 = \frac{10 + 14}{2} = 12
\]
\[
\text{Midpoint of } 15-19 = \frac{15 + 19}{2} = 17
\]
\[
\text{Midpoint of } 20-24 = \frac{20 + 24}{2} = 22
\]
\[
\text{Midpoint of } 25-29 = \frac{25 + 29}{2} = 27
\]

### 2. Multiply midpoints by their frequencies:

| Interval | Midpoint (\( x_i \)) | Frequency (\( f_i \)) | \( f_i \times x_i \) |
|----------|----------------------|-----------------------|----------------------|
| 0-4      | 2                    | 70                    | \( 70 \times 2 = 140 \) |
| 5-9      | 7                    | 52                    | \( 52 \times 7 = 364 \) |
| 10-14    | 12                   | 38                    | \( 38 \times 12 = 456 \) |
| 15-19    | 17                   | 21                    | \( 21 \times 17 = 357 \) |
| 20-24    | 22                   | 10                    | \( 10 \times 22 = 220 \) |
| 25-29    | 27                   | 8                     | \( 8 \times 27 = 216 \) |

### 3. Total frequency:
\[
70 + 52 + 38 + 21 + 10 + 8 = 199
\]

### 4. Sum of the products of midpoints and frequencies:
\[
140 + 364 + 456 + 357 + 220 + 216 = 1753
\]

### 5. Mean:
\[
\text{Mean} = \frac{\text{Sum of } f_i \times x_i}{\text{Total frequency}} = \frac{1753}{199} \approx 8.81
\]

### Final Answer:
The mean number of days the students missed school in the past year is approximately **8.81 days**.

---

Would you like more details on this process or have any questions?

### Related Questions:
1. How would the mean change if the frequency of the 25-29 interval doubled?
2. What is the median number of days missed based on this grouped frequency distribution?
3. How can we estimate the mode from a grouped frequency distribution?
4. What would the standard deviation of this data be?
5. How do changes in the upper intervals affect the mean?

### Tip:  
When calculating the mean from grouped data, always ensure that the midpoints represent the center of each class interval.

A group of high school students were asked to report the number of times they missed school in the past year. Their responses are shown in the grouped frequency distribution below. # of Days f: 0-4, 70; 5-9, 52; 10-14, 38; 15-19, 21; 20-24, 10; 25-29, 8. According to the above data, the mean number of days the students missed school in the past year is ___________ (5 points).

This problem involves calculating the mean number of days students missed school based on a grouped frequency distribution. It covers key statistical concepts such as calculating midpoints and using a weighted average to find the mean. Suitable for high school students in Grades 10-12.

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