Answer:
0.8743 = 87.43% probability that more than one accident occurs per year
Explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
![P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)](https://img.qammunity.org/2022/formulas/mathematics/college/fc9bfg9bauetugxxr4o8egdqz83cs0jk74.png)
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
Buchtal, a manufacturer of ceramic tiles, reports on average 3.1 job-related accidents per year.
This means that
![\mu = 3.1](https://img.qammunity.org/2022/formulas/mathematics/college/48uleadvf71hx876fdjeaot87osxhhe5hz.png)
What is the probability that more than one accident occurs per year?
This is:
![P(X > 1) = 1 - P(X \leq 1)](https://img.qammunity.org/2022/formulas/mathematics/college/jl9o2pp824g5j1sqyvfmx7urxf4lu1qr9y.png)
In which
![P(X \leq 1) = P(X = 0) + P(X = 1)](https://img.qammunity.org/2022/formulas/mathematics/college/tf7uvvdo0e1p0zojpev4l2uifzrtqvot1r.png)
Then
![P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)](https://img.qammunity.org/2022/formulas/mathematics/college/fc9bfg9bauetugxxr4o8egdqz83cs0jk74.png)
![P(X = 0) = (e^(-3.6)*(3.6)^(0))/((0)!) = 0.0273](https://img.qammunity.org/2022/formulas/mathematics/college/80d7bmruwebqy8mwniuu54p3md8huipgwh.png)
![P(X = 1) = (e^(-3.6)*(3.6)^(1))/((1)!) = 0.0984](https://img.qammunity.org/2022/formulas/mathematics/college/tueo91jkys6y741wk81l5axptl1sf08qbq.png)
![P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0273 + 0.0984 = 0.1257](https://img.qammunity.org/2022/formulas/mathematics/college/bqfi9v09hovgpmf9z4vbk2wa8336a9n7jb.png)
![P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1257 = 0.8743](https://img.qammunity.org/2022/formulas/mathematics/college/x5jgvo1syj3r0r3w59y5ygq02r6bby2wzr.png)
0.8743 = 87.43% probability that more than one accident occurs per year