Answer:
0.67% probability he will have to shut down after this month
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/2021/formulas/mathematics/college/frjienvs346ki5axyreyxszxd4zhu8xxhm.png)
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given time interval.
On average sells 8.9 machines per month.
So
![\mu = 8.9](https://img.qammunity.org/2021/formulas/mathematics/college/pclgmthjfn9mxpbisl13lenxvxk2ehojba.png)
Using the Poisson distribution, what is the probability he will have to shut down after this month
If he sells less than 3 machines.
![P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)](https://img.qammunity.org/2021/formulas/mathematics/college/qy48506au00yifb1sgaflt85xqwsjgfmxr.png)
![P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)](https://img.qammunity.org/2021/formulas/mathematics/college/frjienvs346ki5axyreyxszxd4zhu8xxhm.png)
![P(X = 0) = (e^(-8.9)*8.9^(0))/((0)!) = 0.0001](https://img.qammunity.org/2021/formulas/mathematics/college/mbp12fpx4flfllvln312si1afc68jo2u7b.png)
![P(X = 1) = (e^(-8.9)*8.9^(1))/((1)!) = 0.0012](https://img.qammunity.org/2021/formulas/mathematics/college/439hjv2yjymzcl86n8lpih0ohxpbh8ihms.png)
![P(X = 2) = (e^(-8.9)*8.9^(2))/((2)!) = 0.0054](https://img.qammunity.org/2021/formulas/mathematics/college/krjavrzx44r7u69xfh44594pufsb5phuas.png)
![P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0001 + 0.0012 + 0.0054 = 0.0067](https://img.qammunity.org/2021/formulas/mathematics/college/9y5xgex34ekewfbnhdta7xh9wez50ju47y.png)
0.67% probability he will have to shut down after this month