the wait time (after a scheduled arrival time) in minutes for a train to arrive is Uniformly distributed over the interval [0,15]. 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 670 and 796

0.82 = 82% probability that the sum of the 95 wait times you observed is between 670 and 796

Step-by-step explanation:

Uniform distribution:

The uniform distribution has two bounds, a and b.

The mean is given by:

The standard deviation is given by:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

n values from an uniform distribution can be approximated to a normal with mean and standard deviation

Uniformly distributed over the interval [0,15].

This means that . So

You observe the wait time for the next 95 trains to arrive.

This means that , so , and

a) What is the approximate probability (to 2 decimal places) that the sum of the 95 wait times you observed is between 670 and 796

This is the p-value of Z when X = 796 subtracted by the p-value of Z when X = 670. So

X = 796

has a p-value of 0.976.

X = 670

has a p-value of 0.156.

0.976 - 0.156 = 0.82

0.82 = 82% probability that the sum of the 95 wait times you observed is between 670 and 796

true

step-by-step explanation: