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?

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.

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~

lol 2

step-by-step explanation: