Question

Please help me on this one..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.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 above graph represents the situation. The graph shows a continuous uniform distribution with the area under the graph equal to 1.
The mean wait time is given by:
\[Mean\ \mu=\frac{0+12}{2}=6\ minutes\]
When the sum of the 95 wait times is 536 minutes the average wait time is:
\[\bar{x _{1}}=\frac{536}{95}\]
When the sum of the 95 wait times is 637 minutes the average wait time is:
\[\bar{x _{2}}=\frac{637}{95}\]
The required probability is given by the area of the curve between x-bar 2 and x-bar 1 as follows:
\[\frac{1}{12}(\bar{x _{2}}-\bar{x _{1}})=\frac{1}{12}(\frac{637}{95}-\frac{536}{95})=you\ can\ calculate\]

Answer 2

is this an answer for part a or b?

Answer 3

My post deals with the first part of the question.