Question

A bus comes by every 12 minutes. The times from when a person arrives at the bus stop until the bus arrives follows a uniform distribution from 0 to 12 minutes. A person arrives at the bus stop at a randomly selected time. Round to 4 decimal places where possible.
a) What is the mean of this distribution?
b) What is the standard deviation?
c) What is the probability that the person will wait more than 5 minutes?
d) Suppose that the person has already been waiting for 2.8 minutes. Find the probability that the person's total waiting time will be between 3.4 and 5.7 minutes.
e) 10% of all customers wait at least how long for the train?

Answer 1

Answer:

a. The mean of uniform distribution is (a+b)/2, where a and b are limits of the uniform distribution.  So, mean is (0+12)/2 = 6 minutes

b. Standard deviation is (a-b)^2/12 = (0-12)^2/12 = (-12^2)/12 = 144/12 = 12

c. Prob (Wait for more than 5 min) = (12-5)/(12-0) = 7/12 = 0.5833

d. So, given he has waited for 2.8 minutes, what is prob that he will wait between 3.4 and 5.7 minutes

= (5.7 - 3.4) / ((12-2.8) - 0)  

= (5.7 - 3.4) / (9.2 - 0)

= 2.3/9.2

= 0.25

e. 10% of all customers wait at least how long for the train?

10% of all customers wait for?

Since its a uniform distribution, people are uniformly spread. So, the 10% of people will be in the top 10% percentile group of waiting time which essentially means:

0.10 = (12-b)/(12-0)

0.10 = (12-b)/12

1.2 = 12 - b

b = 12 - 1.2

b = 10.8 minutes

So, 10% of all customers wait for at least 10.8 minutes for the train.