SOLUTION: A bus comes by every 7 minutes. The times from when a person arives at the bus stop until the bus arrives follows a Uniform distribution from 0 to 7 minutes. A person arrives at th

Question 1173094: A bus comes by every 7 minutes. The times from when a person arives at the bus stop until the bus arrives follows a Uniform distribution from 0 to 7 minutes. A person arrives at the bus stop at a randomly selected time. Round to 4 decimal places where possible.
a. The mean of this distribution is
b. The standard deviation is
c. The probability that the person will wait more than 3 minutes is
d. Suppose that the person has already been waiting for 1 minutes. Find the probability that the person's total waiting time will be between 2.1 and 3 minutes
e. 56% of all customers wait at least how long for the train?
minutes.
Found 2 solutions by Boreal, ewatrrr:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The mean is the midpoint between the ends or 3.5 min
variance is (b-a)^2/12 or 49/12 or 4.0833 min
sd is sqrt(V)=2.02 min
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wait more than 3 min is (4/7), which may be converted to the decimal.
waiting 1 minute. The total waiting time must be between 2.1 and 3 minutes. That means the person will wait between 2.1 and 3 minutes 0.9 minute range, with 6 minutes to go, since the clock has started.
That is 0.1500 probability.
-
56% wait 3.92 minutes, 7*0.56

Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!

Hi
The times from when a person arrives at the busstop until the bus arrives follows a Uniform distribution from 0 to 7
Uniform Distribution a - 0 and b= 7
, and
a) μ = (0 + 7)/2 = 7/2
b) σ = sqrt((7-0)^2/12) = 2.02
c) P(x > 3mim) = (7-3)/7 = 4/7
d) Already waited 1min f(x) =
P ( 2.1 < x < 3) = (3 - 2.1)(1/6) = .15
e.)P(x < k) = (k-0)(1/7) = .56
k = .56/7 = .08
Wish You the Best in your Studies.
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