Using the Poisson distribution, it is found that:
- There is a 0.3799 = 37.99% probability that the company will find 2 or fewer defective products in this batch.
- There is a 0.3975 = 39.75% probability that 4 or more defective products are found in this batch.
- Since
, the company should not stop production it there are 5 defectives in a batch.
<h3>What is the Poisson distribution?</h3>
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
![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%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
The parameters are:
- x is the number of successes
- e = 2.71828 is the Euler number
is the mean in the given interval.
In this problem, the mean is:
![\mu = 3.2](https://tex.z-dn.net/?f=%5Cmu%20%3D%203.2)
The probability that the company will find 2 or fewer defective products in this batch is:
![P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)](https://tex.z-dn.net/?f=P%28X%20%5Cleq%202%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)
In which:
![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%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-3.2}3.2^{0}}{(0)!} = 0.0408](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-3.2%7D3.2%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.0408)
![P(X = 1) = \frac{e^{-3.2}3.2^{1}}{(1)!} = 0.1304](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7Be%5E%7B-3.2%7D3.2%5E%7B1%7D%7D%7B%281%29%21%7D%20%3D%200.1304)
![P(X = 2) = \frac{e^{-3.2}3.2^{2}}{(2)!} = 0.2087](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20%5Cfrac%7Be%5E%7B-3.2%7D3.2%5E%7B2%7D%7D%7B%282%29%21%7D%20%3D%200.2087)
Then:
![P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0408 + 0.1304 + 0.2087 = 0.3799](https://tex.z-dn.net/?f=P%28X%20%5Cleq%202%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%3D%200.0408%20%2B%200.1304%20%2B%200.2087%20%3D%200.3799)
There is a 0.3799 = 37.99% probability that the company will find 2 or fewer defective products in this batch.
The probability that 4 or more defective products are found in this batch is:
![P(X \geq 4) = 1 - P(X < 4)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%204%29%20%3D%201%20-%20P%28X%20%3C%204%29)
In which:
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).
Then:
![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%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-3.2}3.2^{0}}{(0)!} = 0.0408](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-3.2%7D3.2%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.0408)
![P(X = 1) = \frac{e^{-3.2}3.2^{1}}{(1)!} = 0.1304](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7Be%5E%7B-3.2%7D3.2%5E%7B1%7D%7D%7B%281%29%21%7D%20%3D%200.1304)
![P(X = 2) = \frac{e^{-3.2}3.2^{2}}{(2)!} = 0.2087](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20%5Cfrac%7Be%5E%7B-3.2%7D3.2%5E%7B2%7D%7D%7B%282%29%21%7D%20%3D%200.2087)
![P(X = 3) = \frac{e^{-3.2}3.2^{3}}{(3)!} = 0.2226](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20%5Cfrac%7Be%5E%7B-3.2%7D3.2%5E%7B3%7D%7D%7B%283%29%21%7D%20%3D%200.2226)
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0408 + 0.1304 + 0.2087 + 0.2226 = 0.6025
![P(X \geq 4) = 1 - P(X < 4) = 1 - 0.6025 = 0.3975](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%204%29%20%3D%201%20-%20P%28X%20%3C%204%29%20%3D%201%20-%200.6025%20%3D%200.3975)
There is a 0.3975 = 39.75% probability that 4 or more defective products are found in this batch.
For 5 or more, the probability is:
![P(X \geq 5) = 1 - P(X < 5)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%205%29%20%3D%201%20-%20P%28X%20%3C%205%29)
In which:
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).
Then:
![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%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-3.2}3.2^{0}}{(0)!} = 0.0408](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-3.2%7D3.2%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.0408)
![P(X = 1) = \frac{e^{-3.2}3.2^{1}}{(1)!} = 0.1304](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7Be%5E%7B-3.2%7D3.2%5E%7B1%7D%7D%7B%281%29%21%7D%20%3D%200.1304)
![P(X = 2) = \frac{e^{-3.2}3.2^{2}}{(2)!} = 0.2087](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20%5Cfrac%7Be%5E%7B-3.2%7D3.2%5E%7B2%7D%7D%7B%282%29%21%7D%20%3D%200.2087)
![P(X = 3) = \frac{e^{-3.2}3.2^{3}}{(3)!} = 0.2226](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20%5Cfrac%7Be%5E%7B-3.2%7D3.2%5E%7B3%7D%7D%7B%283%29%21%7D%20%3D%200.2226)
![P(X = 4) = \frac{e^{-3.2}3.2^{4}}{(4)!} = 0.1781](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20%5Cfrac%7Be%5E%7B-3.2%7D3.2%5E%7B4%7D%7D%7B%284%29%21%7D%20%3D%200.1781)
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0408 + 0.1304 + 0.2087 + 0.2226 + 0.1781 = 0.7806
![P(X \geq 5) = 1 - P(X < 5) = 1 - 0.7806 = 0.2194](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%205%29%20%3D%201%20-%20P%28X%20%3C%205%29%20%3D%201%20-%200.7806%20%3D%200.2194)
Since
, the company should not stop production it there are 5 defectives in a batch.
More can be learned about the Poisson distribution at brainly.com/question/13971530
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