The table shows the traveling time to work of 50 people.

| Traveling time (t) in minutes | Frequency |
|-------------------------------|-----------|
| 0 < t ≤ 10 | 5 |
| 10 < t ≤ 20 | 15 |
| 20 < t ≤ 30 | 13 |
| 30 < t ≤ 40 | 10 |
| 40 < t ≤ 50 | 7 |

Calculate an estimate of the mean traveling time.



Answer :

To estimate the mean traveling time, we will use the method of midpoints for each interval and then the frequencies given for those intervals. Here's a detailed step-by-step solution:

1. Identify the midpoints of each interval:
- For [tex]\(0 < t \leq 10\)[/tex], the midpoint is [tex]\( \frac{0 + 10}{2} = 5 \)[/tex].
- For [tex]\(10 < t \leq 20\)[/tex], the midpoint is [tex]\( \frac{10 + 20}{2} = 15 \)[/tex].
- For [tex]\(20 < t \leq 30\)[/tex], the midpoint is [tex]\( \frac{20 + 30}{2} = 25 \)[/tex].
- For [tex]\(30 < t \leq 40\)[/tex], the midpoint is [tex]\( \frac{30 + 40}{2} = 35 \)[/tex].
- For [tex]\(40 < t \leq 50\)[/tex], the midpoint is [tex]\( \frac{40 + 50}{2} = 45 \)[/tex].

So, the midpoints are [tex]\( [5, 15, 25, 35, 45] \)[/tex].

2. Multiply each midpoint by the corresponding frequency:
- For the interval [tex]\(0 < t \leq 10\)[/tex], [tex]\( 5 \times 5 = 25 \)[/tex].
- For the interval [tex]\(10 < t \leq 20\)[/tex], [tex]\( 15 \times 15 = 225 \)[/tex].
- For the interval [tex]\(20 < t \leq 30\)[/tex], [tex]\( 25 \times 13 = 325 \)[/tex].
- For the interval [tex]\(30 < t \leq 40\)[/tex], [tex]\( 35 \times 10 = 350 \)[/tex].
- For the interval [tex]\(40 < t \leq 50\)[/tex], [tex]\( 45 \times 7 = 315 \)[/tex].

3. Sum these results to get the total of the products of midpoints and frequencies:
- [tex]\( 25 + 225 + 325 + 350 + 315 = 1240 \)[/tex].

4. Calculate the total frequency: The total frequency is the sum of all frequencies.
- [tex]\( 5 + 15 + 13 + 10 + 7 = 50 \)[/tex].

5. Estimate the mean traveling time: Divide the total of the products by the total frequency.
- Mean traveling time [tex]\( = \frac{1240}{50} = 24.8 \)[/tex].

Thus, the estimated mean traveling time is [tex]\( 24.8 \)[/tex] minutes.