Question

Help.

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

whats the question

Answer 2

Have you considered determining the Mean and Standard Deviation of the Uniform Distribution on [0,12]?

Answer 3

Yes, so the mean is (0+12)/2 which is 6. Variance is (12-0)^2/12 which is 12.

Answer 4

Sum of 95 wait times you observed is between 536 and 637?

Observed Mean is between 536/95 = 5.64 and 637/95 = 6.71?

Do we have enough samples to suggest a Normal Approximation?

Answer 5

Oh okay. So
536/95 = 5.642
637/95 = 6.705

Therefore the answer for part a is,
the pobability
= (6.705 - 5.642)/12
= 0.089 (2dp)

Answer 6

Is 12 the Standard Deviation or the Variance?