Question

he 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

Answer 1

Answer:

81.74% probability that the sum of the 95 wait times you observed is between 670 and 796

Step-by-step explanation:

To solve this question, the uniform probability distribution and the normal probability distribution must be understood.

Uniform distribution:

A distribution is called uniform if each outcome has the same probability of happening.

The distribution has two bounds, a and b.

Its mean is given by:

$M = \frac{b - a}{2}$

Its standard deviation is given by:

$S = \sqrt{\frac{(b-a)^2}{12}}$

Normal distribution:

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

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

$Z = \frac{X - \mu}{\sigma}$

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 instances of the uniform distribution can be approximated to the normal with $\mu = nM$, $\sigma = S\sqrt{n}$

Uniformly distributed over the interval [0,15].

This means that:

$M = \frac{15 - 0}{2} = 7.5$

$S = \sqrt{\frac{(15-0)^2}{12}} = 4.33$

95 trains

$n = 95$, so:

$\mu = 95M = 95*7.5 = 712.5$

$\sigma = S\sqrt{n} = 4.33\sqrt{95} = 42.2$

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 pvalue of Z when X = 796 subtracted by the pvalue of Z when X = 670. So

X = 796

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{796 - 712.5}{42.2}$

$Z = 1.98$

$Z = 1.98$ has a pvalue of 0.9761

X = 670

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{670 - 712.5}{42.2}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

0.9761 - 0.1587 = 0.8174

81.74% probability that the sum of the 95 wait times you observed is between 670 and 796