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 greater than 3.25 minutes.
Find the probability that a randomly selected passenger has a waiting time greater than 3.25 minutes.
(Simplify your answer. Round to three decimal places as needed.)
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The probability that a randomly selected passenger has a waiting time greater than 3.25 minutes is 0.594.

What is probability?

It is defined as the ratio of the number of favorable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

We have:

The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 5 minutes.

a = 0
b = 8

Let's suppose the x is the waiting time:

Density function f(x)= 1/(b-a) = 1/8

P(X > 3.25) = 1 - P(x < 3.25)

[tex]\rm 1 - [\int\limits^8_0 {\dfrac{1}{8}} \, dx ][/tex]

After solving and plugging the upper and lower limit:

= 1 - (3.25/8)

= 0.59375 ≈ 0.594

P(X > 3.25) = 0.594

Thus, the probability that a randomly selected passenger has a waiting time greater than 3.25 minutes is 0.594.

Learn more about the probability here:

brainly.com/question/11234923

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