Homework SourseA health clinic gets an average of 45 pediatric appointments
A health clinic gets an average of 4.5 pediatric appointments per day. The appointments are unrelated and so we can assume that they are random, independent events. Let X be the count of pediatric appointments per day. What distribution does X follow, approximately? Give the mean and standard deviation.
A. What is the probability that the health clinic gets no pediaric appointments on a given day?
B. What is the probability that the health clinic gets exactly 1 such appointment on a given day?
C. What is the probability that the health clinic gets 2 or more such appointments on a given day?
Solution
It follors a Poisson Distribution, with
mean = 4.5, and
standard deviation = sqrt(mean) = sqrt(4.5) = 2.121320344
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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 = 4.5
x = the number of successes = 0
Thus, the probability is
P ( 0 ) = 0.011108997 [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 = 4.5
x = the number of successes = 1
Thus, the probability is
P ( 1 ) = 0.049990484 [ANSWER]
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c)
Note that P(at least x) = 1 - P(at most x - 1).
Using a cumulative poisson distribution table or technology, matching
u = the mean number of successes = 4.5
x = our critical value of successes = 2
Then the cumulative probability of P(at most x - 1) from a table/technology is
P(at most 1 ) = 0.061099481
Thus, the probability of at least 2 successes is
P(at least 2 ) = 0.938900519 [ANSWER]
