Let's breakdown the problem step-by-step to find the required solutions for parts (a) and (b).
### Part (a): Estimate the Mean Time
To estimate the mean time, we use the formula for the mean of a grouped frequency distribution.
1. Identify the midpoints of each class interval:
The midpoints ([tex]\(x\)[/tex]) are calculated by taking the average of the lower and upper bounds of each interval.
- For [tex]\(5 < t < 10\)[/tex]: midpoint = [tex]\(\frac{5 + 10}{2} = 7.5\)[/tex]
- For [tex]\(10 < t < 15\)[/tex]: midpoint = [tex]\(\frac{10 + 15}{2} = 12.5\)[/tex]
- For [tex]\(15 < t < 20\)[/tex]: midpoint = [tex]\(\frac{15 + 20}{2} = 17.5\)[/tex]
- For [tex]\(20 < t < 25\)[/tex]: midpoint = [tex]\(\frac{20 + 25}{2} = 22.5\)[/tex]
- For [tex]\(25 < t < 30\)[/tex]: midpoint = [tex]\(\frac{25 + 30}{2} = 27.5\)[/tex]
Therefore, the midpoints are [tex]\(7.5, 12.5, 17.5, 22.5\)[/tex], and [tex]\(27.5\)[/tex].
2. Calculate the total number of students ([tex]\(N\)[/tex]):
Sum of the frequencies:
[tex]\[
N = 2 + 9 + 5 + 5 + 3 = 24
\][/tex]
3. Calculate the sum of the products of each midpoint and its respective frequency:
[tex]\[
\text{Sum of } (x_i \times f_i) = (7.5 \times 2) + (12.5 \times 9) + (17.5 \times 5) + (22.5 \times 5) + (27.5 \times 3)
\][/tex]
[tex]\[
= 15 + 112.5 + 87.5 + 112.5 + 82.5 = 410
\][/tex]
4. Estimate the mean time:
[tex]\[
\text{Mean} = \frac{\text{Sum of } (x_i \times f_i)}{N} = \frac{410}{24} \approx 17.08 \text{ minutes}
\][/tex]
Rounded to 2 significant figures, the mean time is [tex]\(\boxed{17.08}\)[/tex] minutes.
### Part (b): Fraction of Students who finished in less than 20 minutes
1. Sum of the frequencies of students who finished in less than 20 minutes:
These are the students in the first three intervals:
[tex]\[
2 + 9 + 5 = 16
\][/tex]
2. Calculate the fraction of students who finished in less than 20 minutes:
[tex]\[
\frac{16}{24} = \frac{2}{3}
\][/tex]
So, the fraction of students who finished in less than 20 minutes is [tex]\(\boxed{\frac{2}{3}}\)[/tex].
