I'm really stuck with this question can someone please help me.

A bus is scheduled to stop at a certain bus stop every 45 minutes. At the end of the day, buses still stop after every 45 minutes, but because delays often occur earlier in the day, the bus is never earlyand likely to be late. The director of the bus line claims that the length of time a bus is late is uniformly distributed over the interval 0 to 35 minutes. If the director's claim is true, what is theprobability that the last bus on a given day will be more than 22 minutes late?

Christina I'm stuck in this same question well actually similar to this can anyone be kind enough to help us out? thanks

IDK ask someone DUH

To find the probability that the last bus on a given day will be more than 22 minutes late, we need to determine the proportion of the interval [0, 35] that represents being more than 22 minutes late.

First, let's visualize the problem. Imagine a number line representing the interval [0, 35], where 0 represents being exactly on time and 35 represents being exactly 35 minutes late.

We know that the length of time a bus is late is uniformly distributed over this interval, which means that every point within the interval has an equal probability of being the actual delay time.

Since delays are likely to be late and never early, we are only interested in the region of the number line that represents being late. In this case, that would be the portion of the interval [0, 35] that is greater than 22.

To find the length of this region, we subtract the starting point of the region (22) from the endpoint of the interval (35):

35 - 22 = 13

Therefore, the length of the region representing being more than 22 minutes late is 13 minutes.

To find the probability, we need to divide this length by the total length of the interval:

13 / 35 ≈ 0.371

So, the probability that the last bus on a given day will be more than 22 minutes late is approximately 0.371 or 37.1%.