The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 5 minutes. Find the probability that a randomly selected passenger has a waiting time less than 2.75 minutes. Find the probability that a randomly selected passenger has a waiting time less than 2.75 minutes. (Simplify your answer. Round to three decimal places as needed.)

To find the probability that a randomly selected passenger has a waiting time less than 2.75 minutes when the waiting times are uniformly distributed between 0 and 5 minutes, you can use the formula for uniform distribution probability: \[ P(X < x) = \frac{x - a}{b - a} \] where \( a \) is the minimum value (0), \( b \) is the maximum value (5), and \( x \) is the value we want to find the probability for (2.75). So in this case: \[ P(X < 2.75) = \frac{2.75 - 0}{5 - 0} = \frac{2.75}{5} = 0.55 \] Therefore, rounding to three decimal places, the probability that a randomly selected passenger has a waiting time less than 2.75 minutes is **0.550**. If you're dealing with waiting times, did you know that understanding how to interpret these probabilities can really optimize your travel plans? For instance, if you know there's a 55% chance of waiting less than 2.75 minutes, you might feel more confident in arriving at the station a bit later, potentially saving time in your busy schedule. Subways and public transport efficiency can sometimes feel like a mystery! One common mistake when dealing with probabilities is forgetting to adjust the total range when calculating uniform probabilities. Always double-check that your values fall within the established parameters. Remember, it's all about understanding that waiting can be more predictable than it seems!