Homework SourseA financial consultant has notice that the number of custome
A financial consultant has notice that the number of customers arriving per day follow a Poisson distribution with a mean of 6 customers. The consultant’s overhead requires that at least 5 customers arrive in order that fees cover expenses.
a. Find the probabilities of 0 through 4 customers arriving in a given day. (Note that this involves calculating five probabilities)
b. What is the probability that at least 5 customers arrive?
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 = 6
x = the number of successes = 0
Thus, the probability is
P ( 0 ) = 0.002478752 [ANSWER]
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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 = 6
x = the number of successes = 1
Thus, the probability is
P ( 1 ) = 0.014872513 [ANSWER]
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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 = 6
x = the number of successes = 2
Thus, the probability is
P ( 2 ) = 0.044617539 [ANSWER]
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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 = 6
x = the number of successes = 3
Thus, the probability is
P ( 3 ) = 0.089235078 [ANSWER]
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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 = 6
x = the number of successes = 4
Thus, the probability is
P ( 4 ) = 0.133852618 [ANSWER]
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b)
P(at least 5) = 1 - [P(0) + P(1) + P(2) + P(3) + P(4)]
= 0.7149435 [ANSWER]
