To solve the problem of finding the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes, we need to understand the properties of a uniform distribution.

Given:

• The waiting times are uniformly distributed between 0 and 6 minutes.

For a uniform distribution $U(a, b)$, the probability density function (PDF) is given by:

$f(x) = \frac{1}{b - a} \quad \text{for} \quad a \leq x \leq b$

Here, $a = 0$ and $b = 6$. So the PDF is:

$f(x) = \frac{1}{6 - 0} = \frac{1}{6} \quad \text{for} \quad 0 \leq x \leq 6$

To find the probability that the waiting time $X$ is greater than 2.25 minutes, we calculate:

$P(X > 2.25) = 1 - P(X \leq 2.25)$

First, we find $P(X \leq 2.25)$:

$P(X \leq 2.25) = \int_{0}^{2.25} f(x) \, dx = \int_{0}^{2.25} \frac{1}{6} \, dx$

$P(X \leq 2.25) = \frac{1}{6} \times 2.25 = \frac{2.25}{6} = 0.375$

Now, we calculate $P(X > 2.25)$:

$P(X > 2.25) = 1 - 0.375 = 0.625$

Therefore, the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes is 0.625.

Simplified Answer: 0.625 (rounded to three decimal places).

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:

1. How do you derive the probability density function for a uniform distribution?
2. What is the cumulative distribution function (CDF) for a uniform distribution?
3. How would you calculate the expected value (mean) of a uniformly distributed random variable?
4. What is the variance of a uniformly distributed random variable between $a$ and $b$?
5. How would you calculate the probability that the waiting time is less than 4 minutes?

Tip: For any uniform distribution $U(a, b)$, the area under the curve (PDF) always equals 1, representing the total probability.