Question

17#1

a)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 100100 trains to arrive. Assume wait times are independent.

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

Part c) Find the probability (to 2 decimal places) that 97 or more of the 100 wait times exceed 11 minute. Please carry answers to at least 6 decimal places in intermediate steps.

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 55 minutes

Answer 1

For each train we are given here that

Therefore, the mean and standard deviation here are computed as:

Therefore using Central limit theorem for 100 sample size, the distribution of 100 trains mean time is given here as:

Therefore the probability here would be given as:

as normal distribution is symmetric about its mean. Therefore 0.5 is the required probability here.

c) P(X > 11) = 1/12

Therefore the probability that 97 or more than 100 wait time exceeds 11 minutes is computed using the binomial probability distribution here as:

P(Y >= 97) = 1 - P(Y <= 96)

This is computed in EXCEL here as:
=1-binom.dist(96,100,1/12,TRUE)

0 is the required output here.

Therefore 0 is the required probability here.

d) Note that as the waiting time ranges from 0 to 12 minutes, therefore the time cannot exceed 55 minutes for any train here. Therefore 0 is the required probability here.