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Sagot :
Let's solve each part of the question step-by-step.
### Part a) Estimate for the Mean Time
1. Identify Midpoints of Each Interval:
- For the interval [tex]\( 5 \leqslant t < 10 \)[/tex]: midpoint [tex]\( = \frac{5 + 10}{2} = 7.5 \)[/tex]
- For the interval [tex]\( 10 \leqslant t < 15 \)[/tex]: midpoint [tex]\( = \frac{10 + 15}{2} = 12.5 \)[/tex]
- For the interval [tex]\( 15 \leqslant t < 20 \)[/tex]: midpoint [tex]\( = \frac{15 + 20}{2} = 17.5 \)[/tex]
- For the interval [tex]\( 20 \leqslant t < 25 \)[/tex]: midpoint [tex]\( = \frac{20 + 25}{2} = 22.5 \)[/tex]
- For the interval [tex]\( 25 \leqslant t < 30 \)[/tex]: midpoint [tex]\( = \frac{25 + 30}{2} = 27.5 \)[/tex]
2. Multiply Each Midpoint by its Frequency to Get the Total Sum:
- [tex]\( 7.5 \times 2 = 15 \)[/tex]
- [tex]\( 12.5 \times 9 = 112.5 \)[/tex]
- [tex]\( 17.5 \times 5 = 87.5 \)[/tex]
- [tex]\( 22.5 \times 5 = 112.5 \)[/tex]
- [tex]\( 27.5 \times 3 = 82.5 \)[/tex]
3. Sum These Products:
- Total sum = [tex]\( 15 + 112.5 + 87.5 + 112.5 + 82.5 = 410 \)[/tex]
4. Calculate the Total Number of Students:
- Total frequency = [tex]\( 2 + 9 + 5 + 5 + 3 = 24 \)[/tex]
5. Estimate the Mean Time:
- Mean time [tex]\( = \frac{\text{Total sum}}{\text{Total frequency}} = \frac{410}{24} \approx 17.08 \)[/tex]
Answer for Part a): The estimated mean time it took the students is [tex]\( 17.08 \)[/tex] minutes.
### Part b) Fraction of Students Who Finished in Less Than 20 Minutes
1. Calculate the Number of Students Who Finished in Less Than 20 Minutes:
- Combine the frequencies of intervals less than 20 minutes:
- [tex]\( 5 \leqslant t < 10 \)[/tex]: 2 students
- [tex]\( 10 \leqslant t < 15 \)[/tex]: 9 students
- [tex]\( 15 \leqslant t < 20 \)[/tex]: 5 students
Total = [tex]\( 2 + 9 + 5 = 16 \)[/tex]
2. Fraction of Students:
- Total number of students = 24
- Fraction [tex]\( = \frac{16}{24} = \frac{2}{3} \approx 0.6667 \)[/tex]
Answer for Part b): The fraction of the students who finished in less than 20 minutes is [tex]\( \frac{2}{3} \)[/tex] or approximately [tex]\( 0.6667 \)[/tex].
### Part a) Estimate for the Mean Time
1. Identify Midpoints of Each Interval:
- For the interval [tex]\( 5 \leqslant t < 10 \)[/tex]: midpoint [tex]\( = \frac{5 + 10}{2} = 7.5 \)[/tex]
- For the interval [tex]\( 10 \leqslant t < 15 \)[/tex]: midpoint [tex]\( = \frac{10 + 15}{2} = 12.5 \)[/tex]
- For the interval [tex]\( 15 \leqslant t < 20 \)[/tex]: midpoint [tex]\( = \frac{15 + 20}{2} = 17.5 \)[/tex]
- For the interval [tex]\( 20 \leqslant t < 25 \)[/tex]: midpoint [tex]\( = \frac{20 + 25}{2} = 22.5 \)[/tex]
- For the interval [tex]\( 25 \leqslant t < 30 \)[/tex]: midpoint [tex]\( = \frac{25 + 30}{2} = 27.5 \)[/tex]
2. Multiply Each Midpoint by its Frequency to Get the Total Sum:
- [tex]\( 7.5 \times 2 = 15 \)[/tex]
- [tex]\( 12.5 \times 9 = 112.5 \)[/tex]
- [tex]\( 17.5 \times 5 = 87.5 \)[/tex]
- [tex]\( 22.5 \times 5 = 112.5 \)[/tex]
- [tex]\( 27.5 \times 3 = 82.5 \)[/tex]
3. Sum These Products:
- Total sum = [tex]\( 15 + 112.5 + 87.5 + 112.5 + 82.5 = 410 \)[/tex]
4. Calculate the Total Number of Students:
- Total frequency = [tex]\( 2 + 9 + 5 + 5 + 3 = 24 \)[/tex]
5. Estimate the Mean Time:
- Mean time [tex]\( = \frac{\text{Total sum}}{\text{Total frequency}} = \frac{410}{24} \approx 17.08 \)[/tex]
Answer for Part a): The estimated mean time it took the students is [tex]\( 17.08 \)[/tex] minutes.
### Part b) Fraction of Students Who Finished in Less Than 20 Minutes
1. Calculate the Number of Students Who Finished in Less Than 20 Minutes:
- Combine the frequencies of intervals less than 20 minutes:
- [tex]\( 5 \leqslant t < 10 \)[/tex]: 2 students
- [tex]\( 10 \leqslant t < 15 \)[/tex]: 9 students
- [tex]\( 15 \leqslant t < 20 \)[/tex]: 5 students
Total = [tex]\( 2 + 9 + 5 = 16 \)[/tex]
2. Fraction of Students:
- Total number of students = 24
- Fraction [tex]\( = \frac{16}{24} = \frac{2}{3} \approx 0.6667 \)[/tex]
Answer for Part b): The fraction of the students who finished in less than 20 minutes is [tex]\( \frac{2}{3} \)[/tex] or approximately [tex]\( 0.6667 \)[/tex].
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