Question

The wait time (after a scheduled arrival time) in minutes for a train to arrive is Uniformly distributed over the interval [0,12]. You observe the wait time for the next 95 trains to arrive. Assume wait times are independent.
Part a) What is the approximate probability (to 2 decimal places) that the sum of the 95 wait times you observed is between 536 and 637?

Part b) What is the approximate probability (to 2 decimal places) that the average of the 95 wait times exceeds 6 minutes?

Answer 1

The question relates to a continuous uniform distribution which is constant over a range of values and zero elsewhere.
The mean is
\[\mu=\frac{a+b}{2}\]
The given values are a = 0 and b = 12 therefore the mean is (0 + 12)/2 = 6
Also
\[\frac{1}{b-a}=\frac{1}{12}\]
When the sum of 95 wait times is 536 the average time is 536/95 = 5.642 minutes.
When the sum of 95 wait times is 637 the average time is 637/95 = 6.705 minutes.
The probability that the sum of the 95 wait times that were observed is between 536 and 637 is found from the area under the continuous uniform distribution between 5.642 and 6.705 minutes:\[P=\frac{6.705-5.642}{12}=you\ can\ calculate\]

Answer 2

ohh i see. I had forgotten to divide by 12. no wonder, thank you :)

Answer 3

You're welcome :)