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:
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:
The probability that the company will find 2 or fewer defective products in this batch is:
In which:
Then:
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:
In which:
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).
Then:
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
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:
In which:
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).
Then:
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
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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