Answer: 0.375
Step-by-step explanation:
The given interval : (0,6) [in minutes]
Let X represents the waiting time of a passenger.
We know that the cumulative uniform distribution function for interval (a,b) is given by :_
[tex]F(x)=\begin{cases}0,&\text{ for } x<a\\\frac{x-a}{b-a},& \text{for } a\leq x\leq 1\\1,& \text{for }x>b\end{cases}[/tex]
Then , the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes. is given by :_
[tex]P(2.25<x)=\dfrac{2.25-0}{6-0}=0.375[/tex]
Hence, the required probability : 0.375
