Question

A one-way, single lane road has traffic that consists of 30% trucks and 70% cars. The speed of trucks (in km/h) is a uniform random variable on the interval (30, 60), and the speed of cars (in km/h) is a uniform random variable on the interval (40, 80). The speed limit on the road is 50 km/h. The percentage of vehicles that exceed the speed limit is ........ (rounded off to 1 decimal place). 

Hint:

When dealing with uniform random variables, the probability of exceeding a given value is found by integrating the probability density function over the appropriate interval.

Answer 1

Correct Answer: 61

For trucks: \[ x = {Truck} = 30%, \quad x \in (30, 60) \] \[ f(x) = \frac{1}{\beta - \alpha} = \frac{1}{60 - 30} = \frac{1}{30} \] For cars: \[ y = {Car} = 70%, \quad y \in (40, 80) \] \[ f(y) = \frac{1}{80 - 40} = \frac{1}{40} \] For the truck, the probability of exceeding the limit (50 km/h) is: \[ P(50<x<60) = \int_{50}^{60} f(x) dx = \int_{50}^{60} \frac{1}{30} dx = \frac{60 - 50}{30} = \frac{10}{30} = \frac{1}{3} \] For the car, the probability of exceeding the limit (50 km/h) is: \[ P(50<y<80) = \int_{50}^{80} f(y) dy = \int_{50}^{80} \frac{1}{40} dy = \frac{80 - 50}{40} = \frac{30}{40} = \frac{3}{4} \] Now, the percentage of vehicle time that exceeds the speed limit is: \[ \frac{1}{3} \times 30% + \frac{3}{4} \times 70% = 10% + 52.5% = 62.5% \] Thus, the percentage of vehicles that exceed the speed limit is 62.5%