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 109 66 21 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 = 0.6.  Hint: For the category 3 or more, use 3. x =  (b) Now we have  P(x) =  e−0.6(0.6)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) =

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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 109 66 21 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 = 0.6.
 Hint: For the category 3 or more, use 3.
x = 


(b) Now we have 
P(x) = 
e−0.6(0.6)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) =  
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