Question

question content area top part 1 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 1.75 minutes.

Answer 1

The probability that a randomly selected passenger has a waiting time of less than 1.75 minutes is 0.35 or 35% when waiting times are uniformly distributed between 0 and 5 minutes.

Step-by-step explanation:

The question is asking to determine the probability that a randomly selected passenger has a waiting time of less than 1.75 minutes when the waiting times are uniformly distributed between 0 and 5 minutes. To calculate this, we use the properties of the uniform distribution. Since the distribution is uniform, the probability is directly proportional to the length of time interval.

The total length of the time interval is 5 minutes. The time interval of interest (less than 1.75 minutes) is a portion of the total interval. So, the probability of a waiting time less than 1.75 minutes is simply the ratio of the time interval of interest (1.75 minutes) to the total interval (5 minutes).

Therefore, the probability is calculated as: Probability = (Time Interval of Interest) / (Total Time Interval) = 1.75 / 5 = 0.35 or 35%.

Answer 2

Step-by-step explanation:

If the waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 5 minutes, then the probability density function is f(x) = 1/5 for 0 ≤ x ≤ 5.

The probability that a randomly selected passenger has a waiting time less than 1.75 minutes is given by the integral of the probability density function from 0 to 1.75:

P(X < 1.75) = ∫[0,1.75] f(x) dx = ∫[0,1.75] (1/5) dx = (1/5) * (1.75 - 0) = 0.35

So the probability that a randomly selected passenger has a waiting time less than 1.75 minutes is 0.35.