5 On average a factory has 17 accidents per month a What is
5. On average, a factory has 1.7 accidents per month. a. What is the probability that they have no accidents in this month? b. What is the probability that they have zero accidents in a 3-month period? c. What is the probability that they have two accidents or fewer in a month? d. What is the probability that they have more than four accidents in a month?
Solution
a)
Note that the probability of x successes out of n trials is
P(x) = u^x e^(-u) / x!
where
u = the mean number of successes = 1.7
x = the number of successes = 0
Thus, the probability is
P ( 0 ) = 0.182683524 [ANSWER]
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b)
Note that the probability of x successes out of n trials is
P(x) = u^x e^(-u) / x!
where
u = the mean number of successes = 3 mo*1.7 accidents/mo = 5.1
x = the number of successes = 0
Thus, the probability is
P ( 0 ) = 0.006096747 [ANSWER]
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c)
Using a cumulative poisson distribution table or technology, matching
u = the mean number of successes = 1.7
x = the maximum number of successes = 2
Then the cumulative probability is
P(at most 2 ) = 0.757223207 [ANSWER]
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d)
Note that P(more than x) = 1 - P(at most x).
Using a cumulative poisson distribution table or technology, matching
u = the mean number of successes = 1.7
x = our critical value of successes = 4
Then the cumulative probability of P(at most x) from a table/technology is
P(at most 4 ) = 0.970385194
Thus, the probability of at least 5 successes is
P(more than 4 ) = 0.029614806 [ANSWER]

