Answer:
0.625.
Explanation:
To find the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes, we can use the properties of a uniform distribution. In a uniform distribution, the probability density function (PDF) is constant over the range of possible values.
Given that the waiting times are uniformly distributed between 0 and 6 minutes, the probability density function is given by:
f(x) = 1 / (b - a) for a ≤ x ≤ b
where 'a' is the lower bound (0 minutes) and 'b' is the upper bound (6 minutes).
Now, we can find the probability of the waiting time being greater than 2.25 minutes:
P(X > 2.25) = ∫[2.25, 6] f(x) dx
P(X > 2.25) = ∫[2.25, 6] (1 / (6 - 0)) dx
P(X > 2.25) = (1 / 6) ∫[2.25, 6] dx
P(X > 2.25) = (1 / 6) * [x] from 2.25 to 6
P(X > 2.25) = (1 / 6) * (6 - 2.25)
P(X > 2.25) = 3.75 / 6
P(X > 2.25) = 0.625
Rounded to three decimal places, the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes is approximately 0.625.