To solve this problem, let's use the concept of a uniform distribution.

The waiting time $X$ for a passenger is uniformly distributed between 0 and 8 minutes, which we denote as: $X \sim U(0, 8)$

In a uniform distribution, the probability density function $f(x)$ is constant across the interval: $f(x) = \frac{1}{b - a}$ where $a = 0$ and $b = 8$. Therefore: $f(x) = \frac{1}{8 - 0} = \frac{1}{8}$

We need to find the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes, which is $P(X > 2.25)$.

1. First, find the length of the interval where $X > 2.25$, which is from 2.25 to 8.
2. Calculate this probability by dividing the length of this interval by the total interval length of the uniform distribution.

So, $P(X > 2.25) = \frac{8 - 2.25}{8} = \frac{5.75}{8}$

Now, simplifying: $P(X > 2.25) = 0.71875$

Rounding to three decimal places, we get: $P(X > 2.25) \approx 0.719$

Answer

The probability that a randomly selected passenger has a waiting time greater than 2.25 minutes is approximately 0.719.

Would you like a further breakdown of this solution or have any questions?

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1. What is the probability that a passenger waits between 3 and 7 minutes?
2. How would the probability change if the uniform distribution range was between 0 and 10 minutes?
3. What is the expected waiting time for a passenger under this distribution?
4. How would we calculate the probability of waiting exactly 5 minutes?
5. How would you represent the cumulative distribution function (CDF) for this scenario?

Tip: In a uniform distribution $U(a, b)$, the probability of an event in a subinterval $[c, d]$ is simply the length of $[c, d]$ divided by the length of $[a, b]$.