1. **Problem statement:** We have a frequency table of distances travelled by 200 cars, divided into intervals with frequencies. We need to estimate the mean distance, find the difference in means with a changed interval, calculate histogram block heights, and find probabilities related to the distances. 2. **Estimate the mean (part a(i))** - Use the midpoint of each interval as the representative value. - Midpoints: - $80 < d \leq 100$: midpoint $= \frac{80+100}{2} = 90$ - $100 < d \leq 150$: midpoint $= \frac{100+150}{2} = 125$ - $150 < d \leq 200$: midpoint $= \frac{150+200}{2} = 175$ - $200 < d \leq 300$: midpoint $= \frac{200+300}{2} = 250$ - $300 < d \leq 400$: midpoint $= \frac{300+400}{2} = 350$ - Multiply each midpoint by its frequency and sum: $$\text{Sum} = 7 \times 90 + 33 \times 125 + 76 \times 175 + 52 \times 250 + 32 \times 350$$ $$= 630 + 4125 + 13300 + 13000 + 11200 = 42255$$ - Total frequency $= 200$ - Estimate of mean: $$\bar{x} = \frac{42255}{200} = 211.275$$ - Rounded to 3 decimal places: $211.275$ km 3. **Difference in mean estimates (part a(ii))** - Ann changes the last interval to $300 < d \leq 360$. - Midpoint for last interval changes to $\frac{300+360}{2} = 330$. - Calculate new sum with last midpoint 330: $$\text{New sum} = 7 \times 90 + 33 \times 125 + 76 \times 175 + 52 \times 250 + 32 \times 330$$ $$= 630 + 4125 + 13300 + 13000 + 10560 = 41615$$ - New mean estimate: $$\bar{x}_{new} = \frac{41615}{200} = 208.075$$ - Difference: $$211.275 - 208.075 = 3.2$$ - So, Kai's estimate is 3.2 km higher than Ann's. 4. **Histogram block heights (part a(iii))** - Height of block $= \frac{\text{frequency}}{\text{class width}}$ (frequency density) - Given height for $200 < d \leq 300$ is 2.6 cm. - Frequency density for this interval: $$\frac{52}{300-200} = \frac{52}{100} = 0.52$$ - Scale factor: $$\text{height} = 2.6 \text{ cm} \Rightarrow \text{density} = 0.52$$ $$\Rightarrow \text{scale} = \frac{2.6}{0.52} = 5$$ - Calculate heights for other intervals: - $80 < d \leq 100$: $$\text{density} = \frac{7}{100-80} = \frac{7}{20} = 0.35$$ $$\text{height} = 0.35 \times 5 = 1.75 \text{ cm}$$ - $150 < d \leq 200$: $$\text{density} = \frac{76}{200-150} = \frac{76}{50} = 1.52$$ $$\text{height} = 1.52 \times 5 = 7.6 \text{ cm}$$ - $300 < d \leq 400$: $$\text{density} = \frac{32}{400-300} = \frac{32}{100} = 0.32$$ $$\text{height} = 0.32 \times 5 = 1.6 \text{ cm}$$ 5. **Probability car travelled more than 300 km (part b)** - Frequency for $300 < d \leq 400$ is 32. - Total cars = 200. - Probability: $$P = \frac{32}{200} = 0.16$$ 6. **Probability both cars travelled 150 km or less (part c(i))** - Intervals $80 < d \leq 100$ and $100 < d \leq 150$ have frequencies 7 and 33. - Total cars with $d \leq 150$: $$7 + 33 = 40$$ - Probability first car $\leq 150$ km: $$\frac{40}{200} = 0.2$$ - After picking one car, remaining cars $= 199$, remaining cars $\leq 150$ km $= 39$. - Probability second car $\leq 150$ km: $$\frac{39}{199}$$ - Combined probability: $$0.2 \times \frac{39}{199} = \frac{40}{200} \times \frac{39}{199} = \frac{1560}{39800} \approx 0.0392$$ 7. **Probability one car travelled more than 200 km and the other 100 km or less (part c(ii))** - Cars travelled more than 200 km: intervals $200 < d \leq 300$ and $300 < d \leq 400$ with frequencies 52 and 32. - Total cars $> 200$ km: $$52 + 32 = 84$$ - Cars travelled 100 km or less: frequency 7. - Probability first car $> 200$ km and second car $\leq 100$ km: $$\frac{84}{200} \times \frac{7}{199} = \frac{588}{39800}$$ - Probability first car $\leq 100$ km and second car $> 200$ km: $$\frac{7}{200} \times \frac{84}{199} = \frac{588}{39800}$$ - Total probability: $$\frac{588}{39800} + \frac{588}{39800} = \frac{1176}{39800} \approx 0.0295$$ **Final answers:** - (a)(i) Mean estimate $= 211.275$ km - (a)(ii) Difference in means $= 3.2$ km - (a)(iii) Heights: $1.75$ cm, $7.6$ cm, $1.6$ cm - (b) Probability $= 0.16$ - (c)(i) Probability $\approx 0.0392$ - (c)(ii) Probability $\approx 0.0295$