In the average there are 2 suicides.if 1 month equal to 30 days find the probability that:at least 5 suicides
Poison distribution
P(X≥5)=1−P(4)−P(3)−P(2)−P(1)−P(0)P(x)=e−λλxx!P(X\ge5)=1-P(4)-P(3)-P(2)-P(1)-P(0)\\P(x)=\frac{e^{-\lambda}\lambda^x}{x!}P(X≥5)=1−P(4)−P(3)−P(2)−P(1)−P(0)P(x)=x!e−λλx
λ=2\lambda=2λ=2
P(X=4)=e−2244!=0.09P(X=4)=\frac{e^{-2}2^4}{4!}=0.09P(X=4)=4!e−224=0.09
P(X=3)=e−2233!=0.18P(X=3)=\frac{e^{-2}2^3}{3!}=0.18P(X=3)=3!e−223=0.18
P(X=2)=e−2222!=0.27P(X=2)=\frac{e^{-2}2^2}{2!}=0.27P(X=2)=2!e−222=0.27
P(X=1)=e−2211!=0.27P(X=1)=\frac{e^{-2}2^1}{1!}=0.27P(X=1)=1!e−221=0.27
P(X=0)=e−2200!=0.135P(X=0)=\frac{e^{-2}2^0}{0!}=0.135P(X=0)=0!e−220=0.135
P(X≥5)=1−0.09−0.18−0.27−0.27−0.135=0.055P(X\ge5)=1-0.09-0.18-0.27-0.27-0.135=0.055P(X≥5)=1−0.09−0.18−0.27−0.27−0.135=0.055
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