Answer: 0.625
Step-by-step explanation:
Given : The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed with interval= [0 minutes, 6 minutes]
The probability density function :-
[tex]f(x)=\dfrac{1}{6-0}=\dfrac{1}{6}[/tex]
The interval of required event = [2.25 seconds, 6 seconds]
Now, the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes is given by :-
[tex]\dfrac{6-2.25}{6}=\dfrac{3.75}{6}=0.625[/tex]
Hence, the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes = 0.625
Using the uniform distribution, it is found that there is a 0.625 = 62.5% probability that a randomly selected passenger has a waiting time greater than 2.25 minutes.
An uniform distribution has two bounds, a and b.
The probability of finding a value above x is:
[tex]P(X > x) = \frac{b - x}{b - a}[/tex]
In this problem, uniformly distributed between 0 and 6 minutes, thus [tex]a = 0, b = 6[/tex].
The probability of a time greater than 2.25 minutes is P(X > 2.25), thus:
[tex]P(X > x) = \frac{6 - 2.25}{6 - 0} = 0.625[/tex]
0.625 = 62.5% probability that a randomly selected passenger has a waiting time greater than 2.25 minutes.
A similar problem is given at https://brainly.com/question/13547683
