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Mathematics 20 Online
OpenStudy (anonymous):

Poisson Distribution: A small life insurance company has determined that on the average it receives 3 death claims per day. Find the probability that the company receives at least four death claims on a randomly selected day.

OpenStudy (anonymous):

can you change the 3 to 6 and the 4 to 7

OpenStudy (anonymous):

you have to compute several numbers

OpenStudy (anonymous):

at least 4 means not 0, 1, 2, or 3, so you need to compute \[1-\left(P(x=0)+P(x=1)+P(x=2)+P(x=3)\right))\]

OpenStudy (anonymous):

\(P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!}\)

OpenStudy (anonymous):

in your case \(\lambda = 3\)

OpenStudy (anonymous):

\[P(x=0)=e^{-3}\] \[P(x=1)=3e^{-3}\] \[P(x=2)=\frac{9e^{-3}}{2}\] \[P(x=3)=\frac{3^3e^{-3}}{6}=\frac{9e^{-\lambda}}{2}\]

OpenStudy (anonymous):

thanks so much for the explanation Satellite 73

OpenStudy (anonymous):

yw

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