Answer:
0.59375
Step-by-step explanation:
In a uniform distribution the probability that the time t is greater than any given value, X, is:
[tex]P(t\geq X)=1-\frac{X}{b-a}[/tex]
In this problem, the limits of the distribution are a = 0 and b = 8 minutes.
For X =3.25 minutes:
[tex]P(t\geq 3.25)=1-\frac{3.25}{8-0} \\P(t\geq 3.25)=0.59375[/tex]
The probability that a randomly selected passenger has a waiting time greater than 3.25 minutes is 0.59375.
The probability that a randomly selected passenger has a waiting time greater than 3.25 minutes is 0.59375.
Since The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 8 minutes.
So, here the probability is
[tex]= 1 - (3.25 \div (8-0))\\\\[/tex]
= 0.59375
Hence, The probability that a randomly selected passenger has a waiting time greater than 3.25 minutes is 0.59375.
Learn more about probability here: https://brainly.com/question/15241200
