g Consider a Poisson distribution with a mean of three occurrences per time period. (a) Write the appropriate Poisson probability function. f(x) = (b) What is the expected number of occurrences in four time periods? (c) Write the appropriate Poisson probability function to determine the probability of x occurrences in four time periods. f(x) = (d) Compute the probability of three occurrences in one time period. (Round your answer to four decimal places.) (e) Compute the probability of twelve occurrences in four time periods. (Round your answer to four decimal places.) (f) Compute the probability of seven occurrences in three time periods. (Round your answer to four decimal places.)

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mtosi17 mtosi17 · Answer 1 · 2020-03-15T05:59:13+01:00

Answer:

a) [tex]P(k,1)=\frac{3^ke^{-3}}{k!}[/tex]

b) E(x)=12

c) [tex]P(x)=\frac{12^xe^{-12}}{x!}[/tex]

d) 0.2240

e) 0.1144

f) 0.1171

Step-by-step explanation:

a) The appropiate Poisson probability function (k events in t interval) is:

[tex]r=3\,ev./t.p.\\\\\\P(k,t)=\frac{(rt)^ke^{-rt}}{k!} \\\\\\P(k,1)=\frac{3^ke^{-3}}{k!}[/tex]

b) In this case, t=4. The expected number of ocurrences is equal to the rate of events multiplied by the time periods:

[tex]E(x)=r\cdot t=3\cdot 4=12[/tex]

c) Probability function for 4 time periods.

[tex]r=3\,,\, t=4\\\\\\P(k,t)=\frac{(rt)^ke^{-rt}}{k!} \\\\\\P(x)=\frac{12^xe^{-12}}{x!}[/tex]

d) Three occurrences in one time period

[tex]P(3)=\frac{3^3e^{-3}}{3!}= \frac{27*0.0498}{6}= 0.2240[/tex]

e) Twelve occurrences in four time periods

[tex]P(12)=\frac{12^{12}e^{-12}}{12!}= \frac{ 54,782,414}{ 479,001,600}= 0.1144[/tex]

f) Seven events in three time periods

[tex]P(7,3)=\frac{9^{7}e^{-9}}{7!}= \frac{ 590.27}{5040}= 0.1171[/tex]