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 100 trains to arrive. Assume wait times are independent.

could someone guide me on how to do these?

Part c) Find the probability (to 2 decimal places) that 97 or more of the 100 wait times exceed 1 minute.

Part d) Use the Normal approximation to the Binomial distribution (with continuity correction) to find the probability (to 2 decimal places) that 56 or more of the 100 wait times recorded exceed 5 minutes.

Answer 1

did you use uniform distribution

Answer 2

what is part a and part b

Answer 3

@perl
yes i used uniform distribution.

Part a) What is the approximate probability (to 2 decimal places) that the sum of the 100 wait times you observed is between 535 and 665?

Part b) What is the approximate probability (to 2 decimal places) that the average of the 100 wait times exceeds 6 minutes?

Answer 4

i got part a and b, but i really dont know how to start c and d. what approach should i take?

Answer 5

c) The probability that a wait time is I minute or less is 1/12. If we find the probability that 3 or fewer wait times out of the 100 are one minute or less, and subtract that probability from 1, we will have the probability that 97 or more of the 100 wait times exceed 1 minute.
\[P(0\ WaitTime\ is\ \le1\ \min.)=\left(\begin{matrix}100 \\ 0\end{matrix}\right)(\frac{1}{12})^{0}(\frac{11}{12})^{100}=0.000014\]
\[P(1\ WaitTime\ is\ \le1\ \min.)=\left(\begin{matrix}100 \\ 1\end{matrix}\right)(\frac{1}{12})^{1}(\frac{11}{12})^{99}=0.0015\]
\[P(2\ WaitTime\ are\ \le1\ \min.)=\left(\begin{matrix}100 \\ 2\end{matrix}\right)(\frac{1}{12})^{2}\frac{11}{12})^{98}=0.0068\]

\[P(3\ WaitTime\ are\ \le1\ \min.)=\left(\begin{matrix}100 \\ 3\end{matrix}\right)(\frac{1}{12}^{3}(\frac{11}{12})^{97}=0.02\]
The probability that 97 or more of the 100 wait times exceed 1 minute is given by:
1.000 - (0.02 + 0.0068 + 0.0015 + 0.000014) = 0.972