A bus comes by every 11 minutes. The times from when a person arives at the busstop until the bus arrives follows a Uniform distribution from 0 to 11 minutes. A person arrives at the bus stop at a randomly selected time. Round to 4 decimal places where possible. The mean of this distribution is 5.50 Correct The standard deviation is 3.1754 Correct The probability that the person will wait more than 4 minutes is 0.6364 Correct Suppose that the person has already been waiting for 2.4 minutes. Find the probability that the person's total waiting time will be between 3.8 and 4.5 minutes ________ 29% of all customers wait at least how long for the train?______minutes.
just need help with the bottom 2. if possible can you show me if there is way to to on the ti-84 calculator
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a. The mean of uniform distribution is (a+b)/2 , where a and b are limits of the uniform distribution.
So, mean is (0+11)/2 =5.5000 minutes
b. Standard deviation is (a-b)^2/12 =sqrt (0-11)^2/12 = 3.1754
c. Prob( wait for more than 7 min) = (11-4)/(11-0) = 0.6364
d. So, given he has waited for ,2.4 minutes what is prob that he will wait between 3,8 and 4.5 minutes
e. 29% of all customers wait atleast how long for the train ?
29% of all customers wait for ?
Since its a unfiorm distribution people are uniformly spread.
So, the 29% of people will be inthe top 29% percentile group of waiting time.
which essentially means
.29 = (11-b)/(11-0)
b = 7.81 minutes
So, 29% of all customers wait atleast 7.81 min for the train
These are just normal calculations and don't necessitate the use of ti-84 calculator
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