Data from 14 cities were combined for a 20 - years period, and the total 280 cit
ID: 3153079 • Letter: D
Question
Data from 14 cities were combined for a 20 - years period, and the total 280 city - years included a total of 92 homicides. After finding the mean number of homicides per city - year, find the probability that a randomly selected city - year has the following numbers of homicides, then compare the actual reslts to those expected by using the poison probabilities: Homicides each city - year Actual results 0 201 1 67 2 11 p (0) - (Round to four decimal places as needed P (1) - P (2) - (Round to four decimal places as needed P (3) - (Round to four decimal places as needed.) P (4) - (Round to four decimal places as needed.) The actual results consisted of 201 city - years with 0 homicides; 67 city - years with one homicide; 11 city - years with two homicides; 1 city - year with three homicides; 0 city - years with four homicides. Compare the actual results to those expected by using the poisson probabilities. Does the poison distribution serve as a good tool for predicting the actual results Y es, the results from poison probabilities closely match the actual results. No, the results from the poisson distribution probabilities do not match the actual results.Explanation / Answer
Hence, the mean number of homicides per year is 92/280 = 0.328571429.
a)
Note that the probability of x successes is
P(x) = u^x e^(-u) / x!
where
u = the mean number of successes = 0.328571429
x = the number of successes = 0
Thus, the probability is
P ( 0 ) = 0.719951501 [ANSWER]
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b)
Note that the probability of x successes is
P(x) = u^x e^(-u) / x!
where
u = the mean number of successes = 0.328571429
x = the number of successes = 1
Thus, the probability is
P ( 1 ) = 0.236555493 [ANSWER]
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c)
Note that the probability of x successes is
P(x) = u^x e^(-u) / x!
where
u = the mean number of successes = 0.328571429
x = the number of successes = 2
Thus, the probability is
P ( 2 ) = 0.038862688 [ANSWER]
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d)
Note that the probability of x successes is
P(x) = u^x e^(-u) / x!
where
u = the mean number of successes = 0.328571429
x = the number of successes = 3
Thus, the probability is
P ( 3 ) = 0.00425639 [ANSWER]
***********************
e)
Note that the probability of x successes is
P(x) = u^x e^(-u) / x!
where
u = the mean number of successes = 0.328571429
x = the number of successes = 4
Thus, the probability is
P ( 4 ) = 0.000349632 [ANSWER]
**************************
The expected numbers are thus
E(x=0) = 0.7199*280 = 201.572
E(x=1) = 0.2365*280 = 66.22
E(X=2) = 0.0388*280 = 10.864
E(X=3) = 0.0042*280 = 1.176
E(x=4) = 0.0003*280 = 0.084
As we can see, these are close to 201, 67, 11, 1, 0.
Hence,
OPTION A: YES. [ANSWER]
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