Consider a Poisson distribution with a mean of two occurrences per time period.
ID: 3128067 • Letter: C
Question
Consider a Poisson distribution with a mean of two occurrences per time period.
a. Write the appropriate Poisson probability function.
b. What is the expected number of occurrences in three time periods?
c. Write the appropriate Poisson probability function to determine the probability of x occurrences in three time periods.
d. Compute the probability of two occurrences in one time period.
e. Compute the probability of six occurrences in three time periods.
f. Compute the probability of five occurrences in two time periods.
Explanation / Answer
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 = 2
x = the number of successes
hence,
P(x) = 2^x e^(-2) / x! [ANSWER]
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b)
For 1 period it is 2 occurrences, so for 3 period,
E(3x) = 3E(x) = 3*2 = 6 [ANSWER]
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c)
As the new mean is now u = 6,
P(x) = u^x e^(-u) / x!
P(x) = 6^x e^(-6) / x! [ANSWER]
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d)
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 = 2
x = the number of successes = 2
Thus, the probability is
P ( 2 ) = 0.270670566 [ANSWER]
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e)
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 = 6
Thus, the probability is
P ( 6 ) = 0.160623141 [ANSWER]
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f)
There are 2*2 = 4 mean occurrences in 2 periods.
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
x = the number of successes = 5
Thus, the probability is
P ( 5 ) = 0.156293452 [ANSWER]
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