To solve this, we need to calculate the probability that a randomly selected passenger's waiting time is less than 1.75 minutes, given that the waiting times are uniformly distributed between 0 and 99 minutes.

For a uniform distribution, the probability density function is given by: $f(x) = \frac{1}{b - a}$ where $a$ and $b$ are the lower and upper bounds of the distribution. Here, $a = 0$ and $b = 99$.

Step 1: Calculate the total range of the distribution

The total range is: $b - a = 99 - 0 = 99 \text{ minutes}$

Step 2: Find the probability

Since the distribution is uniform, the probability of waiting less than 1.75 minutes is the length of the interval from 0 to 1.75 divided by the total range (99 minutes): $P(X < 1.75) = \frac{1.75}{99}$

Step 3: Perform the calculation

$P(X < 1.75) = \frac{1.75}{99} \approx 0.0177$

Thus, the probability that a randomly selected passenger has a waiting time of less than 1.75 minutes is approximately 0.018 (rounded to three decimal places).

Would you like more details about this method? Here's related questions that may help or let me know:

Would you like to answer related questions: