Respuesta :
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 [tex]P(X \geq 5) > 0.05[/tex], the company should not stop production it there are 5 defectives in a batch.
What is the Poisson distribution?
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
- x is the number of successes
- e = 2.71828 is the Euler number
- [tex]\mu[/tex] is the mean in the given interval.
In this problem, the mean is:
[tex]\mu = 3.2[/tex]
The probability that the company will find 2 or fewer defective products in this batch is:
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
In which:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-3.2}3.2^{0}}{(0)!} = 0.0408[/tex]
[tex]P(X = 1) = \frac{e^{-3.2}3.2^{1}}{(1)!} = 0.1304[/tex]
[tex]P(X = 2) = \frac{e^{-3.2}3.2^{2}}{(2)!} = 0.2087[/tex]
Then:
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0408 + 0.1304 + 0.2087 = 0.3799[/tex]
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:
[tex]P(X \geq 4) = 1 - P(X < 4)[/tex]
In which:
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).
Then:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-3.2}3.2^{0}}{(0)!} = 0.0408[/tex]
[tex]P(X = 1) = \frac{e^{-3.2}3.2^{1}}{(1)!} = 0.1304[/tex]
[tex]P(X = 2) = \frac{e^{-3.2}3.2^{2}}{(2)!} = 0.2087[/tex]
[tex]P(X = 3) = \frac{e^{-3.2}3.2^{3}}{(3)!} = 0.2226[/tex]
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
[tex]P(X \geq 4) = 1 - P(X < 4) = 1 - 0.6025 = 0.3975[/tex]
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:
[tex]P(X \geq 5) = 1 - P(X < 5)[/tex]
In which:
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).
Then:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-3.2}3.2^{0}}{(0)!} = 0.0408[/tex]
[tex]P(X = 1) = \frac{e^{-3.2}3.2^{1}}{(1)!} = 0.1304[/tex]
[tex]P(X = 2) = \frac{e^{-3.2}3.2^{2}}{(2)!} = 0.2087[/tex]
[tex]P(X = 3) = \frac{e^{-3.2}3.2^{3}}{(3)!} = 0.2226[/tex]
[tex]P(X = 4) = \frac{e^{-3.2}3.2^{4}}{(4)!} = 0.1781[/tex]
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
[tex]P(X \geq 5) = 1 - P(X < 5) = 1 - 0.7806 = 0.2194[/tex]
Since [tex]P(X \geq 5) > 0.05[/tex], the company should not stop production it there are 5 defectives in a batch.
More can be learned about the Poisson distribution at https://brainly.com/question/13971530
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