Answer :
Answer:
0.5357
Step-by-step explanation:
The waiting times are uniformly distributed between 0 and 7 minutes. We need to find the probability that a randomly selected passenger has a waiting time greater than 3.25 minutes.
The formula to use for uniform distribution is:
[tex]P( X > x) = \frac{b-x}{b-a}[/tex]
where,
b is the upper limit of the distribution which is 7 in this case.
a is the lower limit of the distribution which is 0 in this case.
x is the concerned value which is 3.25 in this case.
Using these values, we get:
[tex]P( X > 3.5) = \frac{7-3.5}{7-0}=0.5357[/tex]
This means, the the probability that a randomly selected passenger has a waiting time greater than 3.25 minutes is 0.5357
Answer:
.535714
Step-by-step explanation:
(7-3.25)/7=.535714
Max time in uniform distribution minus the waiting time. divided the sum of this by the waiting time = answer