The probability that a randomly selected passenger has a waiting time greater than 4.25 minutes is 0.5278
The uniform distributed between a and b minutes.
The probability of finding a value above x is:
P(X > x) = [tex]\frac{b-x}{b-a}[/tex]
The time is evenly distributed between 0 and 9 minutes.
This implies that a=0 , b=9
Determine the likelihood that a randomly selected passenger will have to wait longer than 4.25 minutes.
p(X > 4.25) = [tex]\frac{9-4.25}{9-0}[/tex]
= 0.5278
There is a 52.78% chance that a randomly selected passenger will have to wait longer than 4.25 minutes.
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