The problem states that the waiting times between subway departures are uniformly distributed between 0 and 7 minutes, and you are asked to find the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes.

Step-by-Step Solution:

1. Define the problem:

   • The waiting time is uniformly distributed between 0 and 7 minutes. This is a uniform distribution problem, where the probability density function (PDF) is constant.
   • The total range is from 0 to 7 minutes.
   • You want the probability that the waiting time is greater than 2.25 minutes.

2. The formula for uniform distribution: The probability that $X$ (waiting time) is greater than a certain value $x$ is: $P(X > x) = \frac{\text{Length of interval where } X > x}{\text{Total length of interval}}$ In this case: $P(X > 2.25) = \frac{7 - 2.25}{7 - 0} = \frac{4.75}{7}$

3. Calculate the probability: $P(X > 2.25) = \frac{4.75}{7} \approx 0.679$

Final Answer:

The probability that a randomly selected passenger has a waiting time greater than 2.25 minutes is approximately 0.679 (rounded to three decimal places).

Do you need more details or have any questions?

Here are 5 related questions to explore further:

1. What is the probability that a passenger has a waiting time less than 2.25 minutes?
2. How do we compute the mean and variance for a uniform distribution?
3. What is the probability that the waiting time is exactly 2.25 minutes?
4. If the waiting time range changed to 0 to 10 minutes, how would the result differ?
5. Can we calculate the median waiting time for this uniform distribution?

Tip: For uniform distributions, probabilities for exact values (like $P(X = 2.25)$) are always zero, since the probability is spread evenly across an interval!