Answers

2016-04-20T04:42:54+05:30
The probability of the k events in the Poisson distribution is as foolows:
P(k)=\frac{\lambda^ke^{-\lambda}}{k!}
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

Answer: 0.00115 = 0.115%
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