In a town, 10 accidents take place in a span of 50 days. Assuming that the number of accident follows Poisson distribution, find the probability that there will be 3 or more accidents in a day.
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The probability of the k events in the Poisson distribution is as foolows:
where
λ is the average number of events per interval, in our task: λ = 10 ÷ 50 = 0.2
e = 2.7182... (Euler's number = the base of natural logarithm)
k! = factorial of k, e.g. 5! = 1 × 2 × 3 × 4 × 5
We need to find solutions for k = 3, 4, 5, ... for λ = 0.2 - that is, to calculate the cumulative distribution function for k ≥ 3:
![P(k\ge3)=P(3)+P(4)+P(5)+...=\\1-<span>P(k<3)=1-[P(0)+P(1)+P(2)]=\\1-e^{-0.2}(\frac{0.2^0}{0!}+\frac{0.2^1}{1!}+\frac{0.2^2}{2!})\approx1-0.8187(1+0.2+0.02)=\\1-0.8187\times1.22=0.00115](https://tex.z-dn.net/?f=P%28k%5Cge3%29%3DP%283%29%2BP%284%29%2BP%285%29%2B...%3D%5C%5C1-%3Cspan%3EP%28k%26lt%3B3%29%3D1-%5BP%280%29%2BP%281%29%2BP%282%29%5D%3D%5C%5C1-e%5E%7B-0.2%7D%28%5Cfrac%7B0.2%5E0%7D%7B0%21%7D%2B%5Cfrac%7B0.2%5E1%7D%7B1%21%7D%2B%5Cfrac%7B0.2%5E2%7D%7B2%21%7D%29%5Capprox1-0.8187%281%2B0.2%2B0.02%29%3D%5C%5C1-0.8187%5Ctimes1.22%3D0.00115 "P(k\ge3)=P(3)+P(4)+P(5)+...=\\1-<span>P(k<3)=1-[P(0)+P(1)+P(2)]=\\1-e^{-0.2}(\frac{0.2^0}{0!}+\frac{0.2^1}{1!}+\frac{0.2^2}{2!})\approx1-0.8187(1+0.2+0.02)=\\1-0.8187\times1.22=0.00115")
Answer: 0.00115 = 0.115%
where
λ is the average number of events per interval, in our task: λ = 10 ÷ 50 = 0.2
e = 2.7182... (Euler's number = the base of natural logarithm)
k! = factorial of k, e.g. 5! = 1 × 2 × 3 × 4 × 5
We need to find solutions for k = 3, 4, 5, ... for λ = 0.2 - that is, to calculate the cumulative distribution function for k ≥ 3:
Answer: 0.00115 = 0.115%
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Answer:
Step-by-step explanation:
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