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Let’s work through each part of the question.

### Given Data Table

| Distance \(d\) (km) | Frequency |
|---------------------|-----------|
| \(0 < d \leq 2\)   | 22        |
| \(2 < d \leq 5\)   | 13        |
| \(5 < d \leq 10\)  | 9         |
| \(10 < d \leq 20\) | 12        |
| \(20 < d \leq 40\) | 4         |

### Solution

#### (a) Modal Class
The **modal class** is the class interval with the highest frequency. Here, the highest frequency is 22, which corresponds to the interval:
\[
0 < d \leq 2
\]
So, the **modal class** is \(0 < d \leq 2\).

#### (b) Median Class
To find the median class, we need to determine where the middle value falls within the cumulative frequency distribution. Since there are 60 people, the median position is the \(\frac{60 + 1}{2} = 30.5\)th value.

Let's compute the cumulative frequency:
- \(0 < d \leq 2\): 22
- \(2 < d \leq 5\): 22 + 13 = 35
- \(5 < d \leq 10\): 35 + 9 = 44
- \(10 < d \leq 20\): 44 + 12 = 56
- \(20 < d \leq 40\): 56 + 4 = 60

The 30.5th value lies in the cumulative frequency range between 22 and 35, which corresponds to the interval:
\[
2 < d \leq 5
\]
So, the **median class** is \(2 < d \leq 5\).

#### (c) Mean Distance Estimate
To estimate the mean, we use the midpoint of each class interval and multiply by the frequency, then divide by the total frequency.

1. **Calculate midpoints** for each class:
   - \(0 < d \leq 2\): midpoint = \( \frac{0 + 2}{2} = 1\)
   - \(2 < d \leq 5\): midpoint = \( \frac{2 + 5}{2} = 3.5\)
   - \(5 < d \leq 10\): midpoint = \( \frac{5 + 10}{2} = 7.5\)
   - \(10 < d \leq 20\): midpoint = \( \frac{10 + 20}{2} = 15\)
   - \(20 < d \leq 40\): midpoint = \( \frac{20 + 40}{2} = 30\)

2. **Calculate the weighted sum** of distances:
   \[
   (1 \times 22) + (3.5 \times 13) + (7.5 \times 9) + (15 \times 12) + (30 \times 4)
   \]
   Calculating each term:
   - \(1 \times 22 = 22\)
   - \(3.5 \times 13 = 45.5\)
   - \(7.5 \times 9 = 67.5\)
   - \(15 \times 12 = 180\)
   - \(30 \times 4 = 120\)

   Total sum:
   \[
   22 + 45.5 + 67.5 + 180 + 120 = 435
   \]

3. **Calculate the mean**:
   \[
   \text{Mean} = \frac{435}{60} \approx 7.25 \, \text{km}
   \]

### Final Answers
- (a) **Modal class**: \(0 < d \leq 2\)
- (b) **Median class**: \(2 < d \leq 5\)
- (c) **Mean estimate**: \(7.25\) km

Would you like more details on any part of the solution?

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Here are 5 related questions to expand your understanding:
1. What is the difference between the mean, median, and mode in data analysis?
2. How is the median position determined in grouped data?
3. Why is an estimate used for the mean in grouped data, rather than an exact calculation?
4. How would the mean change if one of the frequencies were increased significantly?
5. What other statistical measures could be calculated from this data?

**Tip:** In grouped data, always use the midpoints of intervals for estimating the mean, as it provides a reasonable average for each range.

The table below gives some information about the distance travelled to work, in km, each day by 60 people.

(a) Write down the modal class.
(b) Find the class that contains the median distance travelled to work.
(c) Calculate an estimate, in km, for the mean distance travelled to work.

Learn how to determine the modal, median, and mean class intervals for a set of distances travelled to work by 60 people. This problem covers statistical concepts in grouped data analysis, suitable for students in Grades 8-10.

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