Respuesta :
Answer:
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:
[tex]M = \frac{b - a}{2}[/tex]
The standard deviation is given by:
[tex]S = \sqrt{\frac{(b-a)^2}{12}}[/tex]
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = nM[/tex] and standard deviation [tex]\sigma = S\sqrt{n}[/tex]
Uniformly distributed over the interval [0,15].
This means that [tex]a = 0, b = 15[/tex]. So
[tex]M = \frac{b - a}{2} = \frac{15}{2} = 7.5[/tex]
[tex]S = \sqrt{\frac{(b-a)^2}{12}} = \sqrt{\frac{(15-0)^2}{12}} = 4.33[/tex]
You observe the wait time for the next 95 trains to arrive.
This means that [tex]n = 95[/tex], so [tex]\mu = 95*7.5 = 712.5[/tex], and [tex]\sigma = 4.33\sqrt{95} = 42.2[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{796 - 712.5}{42.2}[/tex]
[tex]Z = 1.98[/tex]
[tex]Z = 1.98[/tex] has a p-value of 0.976.
X = 670
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{670 - 712.5}{42.2}[/tex]
[tex]Z = -1.01[/tex]
[tex]Z = -1.01[/tex] 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
