8. Lincoln County EMS is used on average 500 times over the course of a year. The number of times Lincoln County EMS is used on a given day can be modeled with a Poisson Distribution.D. Compute P(6 or more times used) (hint: use its complement); round to the nearest hundred thousandth.E. While working for 500 days on Lincoln County EMS a person has had both a day where they were used 4 times and another where they were used 5 times. Is that to be expected or not? Explain.

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Answered: Lincoln County EMS is used on average…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Do it using each one formula seperatly.
Consider that we are interested in computing the time weighted average number of cars parked in the parking garage on a Sunday in December. We start at midnight and collect the entry and exit times of cars in the garage until the following midnight as follows. Entry 12:45:00 AM Entry 1:00:00 AM Entry 3:30:00 AM Exit 5:15:00 AM Entry 7:00:00 AM Entry 8:15:00 AM Exit 8:45:00 AM Entry 10:00:00 AM Exit 10:30:00 AM Entry 2:30:00 PM Exit 4:15:00 PM Entry 6:00:00 PM Exit 7:45:00 PM Entry 9:30:00 PM Exit 11:15:00 PM What was the time weight average number of cars parked in the parking garage on that Sunday?
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This cumulative review problem uses material from Chapters 3, 5, and 10. Recall that the Poisson distribution deals with rare events. Death from the kick of a horse is a rare event, even in the Prussian army. The following data are a classic example of a Poisson application to rare events. The data represent the number of deaths from the kick of a horse per army corps per year for 10 Prussian army corps over a period of time. Let x represent the number of deaths and f the frequency of x deaths. x 0 1 2 3 or more f 111 61 24 4 (a) First, we fit the data to a Poisson distribution.The Poission distribution states P(x) = e−λλx x! , where λ ≈ x (sample mean of x values). From our study of weighted averages, we get the following. x = Σxf Σf Verify that x = 0.605. Hint: For the category 3 or more, use 3. x = (b) Now we have P(x) = e−0.605(0.605)x x! for x = 0, 1, 2, 3 . Find P(0), P(1), P(2), and P(3 ≤ x). Round to three places…

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