Answer:
75.8% probability that a 31 square foot metal sheet has at least 4 defects.
Step-by-step 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) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given time interval.
In this problem, we have that:
3 defects per 18 square feet.
So for 31 square feet, we have to solve a rule of three
3 defects - 18 square feet
x defects - 31 square feet
![18x = 3*31](https://tex.z-dn.net/?f=18x%20%3D%203%2A31)
![x = \frac{31*3}{18}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B31%2A3%7D%7B18%7D)
![x = 5.17](https://tex.z-dn.net/?f=x%20%3D%205.17)
So ![\mu = 5.17](https://tex.z-dn.net/?f=%5Cmu%20%3D%205.17)
Assuming a Poisson distribution, find the probability that a 31 square foot metal sheet has at least 4 defects.
Either it has three or less defects, or it has at least 4 defects. The sum of the probabilities of these events is decimal 1.
So
![P(X \leq 3) + P(X \geq 4) = 1](https://tex.z-dn.net/?f=P%28X%20%5Cleq%203%29%20%2B%20P%28X%20%5Cgeq%204%29%20%3D%201)
We want ![P(X \geq 4)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%204%29)
So
![P(X \geq 4) = 1 - P(X \leq 3)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%204%29%20%3D%201%20-%20P%28X%20%5Cleq%203%29)
In which
![P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)](https://tex.z-dn.net/?f=P%28X%20%5Cleq%203%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29)
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-5.17}*(5.17)^{0}}{(0)!} = 0.0057](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-5.17%7D%2A%285.17%29%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.0057)
![P(X = 1) = \frac{e^{-5.17}*(5.17)^{1}}{(1)!} = 0.0294](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7Be%5E%7B-5.17%7D%2A%285.17%29%5E%7B1%7D%7D%7B%281%29%21%7D%20%3D%200.0294)
![P(X = 2) = \frac{e^{-5.17}*(5.17)^{2}}{(2)!} = 0.0760](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20%5Cfrac%7Be%5E%7B-5.17%7D%2A%285.17%29%5E%7B2%7D%7D%7B%282%29%21%7D%20%3D%200.0760)
![P(X = 3) = \frac{e^{-5.17}*(5.17)^{3}}{(3)!} = 0.1309](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20%5Cfrac%7Be%5E%7B-5.17%7D%2A%285.17%29%5E%7B3%7D%7D%7B%283%29%21%7D%20%3D%200.1309)
So
![P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0057 + 0.0294 + 0.0760 + 0.1309 = 0.242](https://tex.z-dn.net/?f=P%28X%20%5Cleq%203%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%3D%200.0057%20%2B%200.0294%20%2B%200.0760%20%2B%200.1309%20%3D%200.242)
Finally
![P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.242 = 0.758](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%204%29%20%3D%201%20-%20P%28X%20%5Cleq%203%29%20%3D%201%20-%200.242%20%3D%200.758)
75.8% probability that a 31 square foot metal sheet has at least 4 defects.