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 66 minutes?

Part c) Find the probability (to 2 decimal places) that 92 or more of the 95 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 5656 or more of the 9595 wait times recorded exceed 55 minutes.

Answer 1

Mean wait time for a train = (12 + 0)/2 = 6 minutes

Variance of the wait time for a train = = 12

The sum of the wait time for next 95 trains, Y would follow Normal distribution (Using Central Limit Theorem)

Thus, the distribution of Y has mean = 6*95 = 570 minutes

and Variance = 12*95 = 1140

-> standard deviation = = 33.76 minutes

Thus,

We know Z =

(a) The required probability = P(536 < Y < 637)

= = P(-1.007 < Z < 1.985)

= 0.8194

(b) The sampling distribution of the sample mean of the 95 wait times will have

Mean = 6 minutes

and Standard deviation = = 0.3554

To find

The corresponding z value = 0

Thus, the required probability = P(Z > 0) = 0.5

(c) The probability that the wait time for a train exceeds 11 minutes = (12 - 11)/(12 - 0) = 1/12

Let N be a binomial random variable which denotes the number of wait times which exceeds 11 minutes

Thus, the required probability = P(N >= 92)

= P(N = 92) + P(N = 93) + P(N = 94) + P(N = 95)

=

=

(d) The probability that the wait time for a train exceeds 11 minutes = (12 - 5)/(12 - 0) = 7/12

Let M be a binomial random variable which denotes the number of wait times which exceeds 5 minutes

Mean of M = 95*7/12 = 55.417

Standard deviation of M = = 4.8

Since both (95*7/12) and (95 * 5/12) are greater than 10, M can be approximated to normal distribution

where M ~

To find P(M >= 56)

Using correction of continuity, the required probability = P(M >= 56) = P(M > 55.5)

The corresponding z score = (55.5 - 55.417)/4.8 = 0.0173

Thus, the required probability = P(Z > 0.0173) = 0.493