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}](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7Bb%20-%20a%7D%7B2%7D)
Its standard deviation is given by:
![S = \sqrt{\frac{(b-a)^2}{12}}](https://tex.z-dn.net/?f=S%20%3D%20%5Csqrt%7B%5Cfrac%7B%28b-a%29%5E2%7D%7B12%7D%7D)
Normal 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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
, ![\sigma = S\sqrt{n}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20S%5Csqrt%7Bn%7D)
Uniformly distributed over the interval [0,15].
This means that:
![M = \frac{15 - 0}{2} = 7.5](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7B15%20-%200%7D%7B2%7D%20%3D%207.5)
![S = \sqrt{\frac{(15-0)^2}{12}} = 4.33](https://tex.z-dn.net/?f=S%20%3D%20%5Csqrt%7B%5Cfrac%7B%2815-0%29%5E2%7D%7B12%7D%7D%20%3D%204.33)
95 trains
, so:
![\mu = 95M = 95*7.5 = 712.5](https://tex.z-dn.net/?f=%5Cmu%20%3D%2095M%20%3D%2095%2A7.5%20%3D%20712.5)
![\sigma = S\sqrt{n} = 4.33\sqrt{95} = 42.2](https://tex.z-dn.net/?f=%5Csigma%20%3D%20S%5Csqrt%7Bn%7D%20%3D%204.33%5Csqrt%7B95%7D%20%3D%2042.2)
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}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{796 - 712.5}{42.2}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B796%20-%20712.5%7D%7B42.2%7D)
![Z = 1.98](https://tex.z-dn.net/?f=Z%20%3D%201.98)
has a pvalue of 0.9761
X = 670
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{670 - 712.5}{42.2}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B670%20-%20712.5%7D%7B42.2%7D)
![Z = -1](https://tex.z-dn.net/?f=Z%20%3D%20-1)
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