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One out of every 5000 individuals in a population carries a certain defective gene. A random sample of 1000 individuals is studied. The sample follows the Poisson Distribution.

What is the probability that exactly one of the sampled individuals carries the gene?
What is the probability that none of the sampled individuals carries the gene?
What is the probability that more than two of the sampled individuals carry the gene?
What is the probability that between one and three (inclusive) of the sampled individuals carry the gene?
What is the mean of the number of sample individuals that carry the gene?
What is the standard deviation of the number of sample individuals that carry the gene?

User Leviticus
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Answer:

The probability of exactly 1 is approximately 16.4% The probability of none is approximately 80-82% The (being 1, 2 or 3) is probability of more than 2 is less than 10% and the probability of 1-3 is 19.8%

Explanation:

To solve the given problems, we'll use the Poisson distribution formula. In the Poisson distribution, the probability mass function for a random variable X is given by:

P(X = k) = (e^(-λ) * λ^k) / k!

where λ is the mean of the distribution.

Given:

λ = mean = 1000 (1/5000 * 1000 = 0.2)

Probability that exactly one of the sampled individuals carries the gene (k = 1):

P(X = 1) = (e^(-0.2) * 0.2^1) / 1!

= (e^(-0.2) * 0.2) / 1

≈ 0.1637 (rounded to four decimal places)

Probability that none of the sampled individuals carries the gene (k = 0):

P(X = 0) = (e^(-0.2) * 0.2^0) / 0!

= e^(-0.2) / 1

≈ 0.8187 (rounded to four decimal places)

Probability that more than two of the sampled individuals carry the gene (k > 2):

P(X > 2) = 1 - P(X ≤ 2)

= 1 - (P(X = 0) + P(X = 1) + P(X = 2))

Using the given values for P(X = 0) and P(X = 1) from the previous calculations:

P(X > 2) = 1 - (0.8187 + 0.1637 + P(X = 2))

To find P(X = 2):

P(X = 2) = (e^(-0.2) * 0.2^2) / 2!

≈ 0.0327 (rounded to four decimal places)

Substituting this value back into the equation:

P(X > 2) = 1 - (0.8187 + 0.1637 + 0.0327)

≈ 0.985 (rounded to three decimal places)

Probability that between one and three (inclusive) of the sampled individuals carry the gene (1 ≤ k ≤ 3):

P(1 ≤ X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)

Using the given values for P(X = 1) and P(X = 2) from the previous calculations:

P(1 ≤ X ≤ 3) = 0.1637 + 0.0327 + P(X = 3)

To find P(X = 3):

P(X = 3) = (e^(-0.2) * 0.2^3) / 3!

≈ 0.0022 (rounded to four decimal places)

Substituting this value back into the equation:

P(1 ≤ X ≤ 3) = 0.1637 + 0.0327 + 0.0022

≈ 0.1986 (rounded to four decimal places)

Mean of the number of sample individuals that carry the gene (μ):

The mean (μ) of a Poisson distribution is equal to λ.

Therefore, the mean of the number of sample individuals that carry the gene is 0.2.

Standard deviation of the number of sample individuals that carry the gene (σ):

The standard deviation (σ) of a Poisson distribution is the square root of λ.

Therefore, the standard

User Grant Peters
by
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