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

Probability of Vehicles Exceeding the Speed Limit

Step 1: Define Given Data

For trucks:

• Speed range: \(30 \leq x \leq 60\) km/h
• Probability density function (PDF):

\[ f(x) = \frac{1}{60 - 30} = \frac{1}{30} \]

For cars:

• Speed range: \(40 \leq y \leq 80\) km/h
• Probability density function (PDF):

\[ f(y) = \frac{1}{80 - 40} = \frac{1}{40} \]

Step 2: Compute Probability of Speed Exceeding 50 km/h

For trucks:

\[ P(50 < x < 60) = \int_{50}^{60} \frac{1}{30} dx = \frac{60 - 50}{30} = \frac{10}{30} = \frac{1}{3} \]

For cars:

\[ P(50 < y < 80) = \int_{50}^{80} \frac{1}{40} dy = \frac{80 - 50}{40} = \frac{30}{40} = \frac{3}{4} \]

Step 3: Compute Weighted Probability

Given proportions:

• Trucks: 30%
• Cars: 70%

\[ \text{Total percentage} = \left(\frac{1}{3} \times 30\%\right) + \left(\frac{3}{4} \times 70\%\right) \]

\[ = 10\% + 52.5\% = 62.5\%. \]

Final Answer:

Correct Answer: \( \mathbf{62.5\%} \).