A motoring magazine collected data on cars on a particular stretch of road - Leaving Cert Mathematics - Question 8 - 2019

To find the proportion of cars traveling over 95 km/h, we need to calculate the z-score:

1. First, find the z-score for 95 km/h: $z = \frac{95 - 87.3}{12} = 0.5617$
2. Now, calculate the probability: $P(Z > 0.5617) = 1 - P(Z \leq 0.5617) \approx 1 - 0.7133 = 0.2867$
3. Therefore, we would expect approximately 28.67% of cars to be traveling over 95 km/h.

We need to find the speed corresponding to the 70th percentile, where 70% of speeds were higher:

1. To find this, determine the z-score for 70%: $z = 0.524$
2. Now, use the z-score formula to find the speed: $0.524 = \frac{x - 87.3}{12}$ leading to:
   $x = z \cdot 12 + 87.3 \approx 81.06 \text{ km/h}$
3. Therefore, the speed is approximately 81 km/h when rounded to the nearest whole number.

Given the p-value of 0.024, which is less than 0.05, we reject the null hypothesis. This suggests there is significant evidence to conclude that the average speed had changed.

To find the average speed given that it is lower than 87.3 km/h:

1. Use the z-score formula: $z = \frac{x - 87.3}{12/\sqrt{100}} = -2.26 \text{ (using normal distribution)}$
2. Re-arranging gives: $-2.26 = \frac{x - 87.3}{1.2} \implies x = 87.3 - 2.26 \times 1.2 \approx 84.6$
3. Thus, the average speed is approximately 84.6 km/h, rounded to 1 decimal place.