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 113 62 21 4 (a) First, we fit the data to a Poisson distribution. The Poission distribution states P(x) = e−??x x!, where ? ≈ x (sample mean of x values). From our study of weighted averages, we get the following. (b) Now we have P(x) = e−0.58(0.58)x x! for x = 0, 1, 2, 3 . Find P(0), P(1), P(2), and P(3 ≤ x). Round to three places after the decimal. P(0) = P(1) = P(2) = P(3 ≤ x) = (c) The total number of observations is Σf = 200. For a given x, the expected frequency of x deaths is 200P(x). The following table gives the observed frequencies O and the expected frequencies E = 200P(x). x 0 = f E = 200P(x) 0 113 200(0.560) = 112 1 62 200(0.325) = 65 2 21 200(0.094) = 18.8 3 or more 4 200(0.021) = 4.2 Compute ?2 = Σ (O − E)2 E using the values in the table. (Round your answer to two decimal places.)

Name: This cumulative review problem uses material from Chapters 3, 5, and 1
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Description: 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

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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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	113	62	21	4

(a) First, we fit the data to a Poisson distribution.

The Poission distribution states

P(x) =

e^−??^x

where

? ≈ x

(sample mean of x values). From our study of weighted averages, we get the following.

(b) Now we have

P(x) =

e^−0.58(0.58)^x

for

x = 0, 1, 2, 3 .

Find

P(0),

P(1),

P(2),

and

P(3 ≤ x).

Round to three places after the decimal.

P(0)	=
P(1)	=
P(2)	=
P(3 ≤ x)	=

Σf = 200.

For a given x, the expected frequency of x deaths is

200P(x).

The following table gives the observed frequencies O and the expected frequencies

E = 200P(x).

x	0 = f	E = 200P(x)
0	113	200(0.560) = 112
1	62	200(0.325) = 65
2	21	200(0.094) = 18.8
3 or more	4	200(0.021) = 4.2

Compute

?² = Σ

(O − E)²

using the values in the table. (Round your answer to two decimal places.)

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