To solve this problem, we first need to understand that the Poisson distribution is a good approximation of the binomial distribution when the number of trials is large and the probability of success is small.

In this case, the number of trials is 25 (the number of screws in a package) and the probability of a screw being defective is 0.01.

The parameter λ (lambda) for the Poisson distribution is the expected number of successes, which is the product of the number of trials and the probability of success. So, λ = 25 * 0.01 = 0.25.

The company will have to replace a package if there are 2 or more defective screws. So, we need to find the probability of having 0, 1 or 2 defective screws and subtract this from 1 to find the probability of having 2 or more defective screws.

The probability mass function of the Poisson distribution is P(X=k) = λ^k * e^-λ / k!, where k is the number of successes, e is the base of the natural logarithm (approximately 2.71828), and k! is the factorial of k.

So, P(X=0) = 0.25^0 * e^-0.25 / 0! = e^-0.25 ≈ 0.7788.
P(X=1) = 0.25^1 * e^-0.25 / 1! = 0.25 * e^-0.25 ≈ 0.1947.
P(X=2) = 0.25^2 * e^-0.25 / 2! = 0.0625 * e^-0.25 / 2 ≈ 0.0264.

The probability that the company must replace a package is 1 - P(X=0) - P(X=1) - P(X=2) ≈ 1 - 0.7788 - 0.1947 - 0.0264 ≈ 0.

So, the closest answer is (a) 0.01.

Question

To solve this problem, we first need to understand that the Poisson distribution is a good approximation of the binomial distribution when the number of trials is large and the probability of success is small.

In this case, the number of trials is 25 (the number of screws in a package) and the probability of a screw being defective is 0.01.

The parameter λ (lambda) for the Poisson distribution is the expected number of successes, which is the product of the number of trials and the probability of success. So, λ = 25 * 0.01 = 0.25.

The company will have to replace a package if there are 2 or more defective screws. So, we need to find the probability of having 0, 1 or 2 defective screws and subtract this from 1 to find the probability of having 2 or more defective screws.

The probability mass function of the Poisson distribution is P(X=k) = λ^k * e^-λ / k!, where k is the number of successes, e is the base of the natural logarithm (approximately 2.71828), and k! is the factorial of k.

So, P(X=0) = 0.25^0 * e^-0.25 / 0! = e^-0.25 ≈ 0.7788.
P(X=1) = 0.25^1 * e^-0.25 / 1! = 0.25 * e^-0.25 ≈ 0.1947.
P(X=2) = 0.25^2 * e^-0.25 / 2! = 0.0625 * e^-0.25 / 2 ≈ 0.0264.

The probability that the company must replace a package is 1 - P(X=0) - P(X=1) - P(X=2) ≈ 1 - 0.7788 - 0.1947 - 0.0264 ≈ 0.

So, the closest answer is (a) 0.01.

Knowee AI · Accepted Answer

To solve this problem, we first need to understand that the Poisson distribution is a good approximation of the binomial distribution when the number of trials is large and the probability of success is small.

In this case, the number of trials is 25 (the number of screws in a package) and the probability of a screw being defective is 0.01.

The parameter λ (lambda) for the Poisson distribution is the expected number of successes, which is the product of the number of trials and the probability of success. So, λ = 25 * 0.01 = 0.25.

The company will have to replace a package if there are 2 or more defective screws. So, we need to find the probability of having 0, 1 or 2 defective screws and subtract this from 1 to find the probability of having 2 or more defective screws.

The probability mass function of the Poisson distribution is P(X=k) = λ^k * e^-λ / k!, where k is the number of successes, e is the base of the natural logarithm (approximately 2.71828), and k! is the factorial of k.

So, P(X=0) = 0.25^0 * e^-0.25 / 0! = e^-0.25 ≈ 0.7788.
P(X=1) = 0.25^1 * e^-0.25 / 1! = 0.25 * e^-0.25 ≈ 0.1947.
P(X=2) = 0.25^2 * e^-0.25 / 2! = 0.0625 * e^-0.25 / 2 ≈ 0.0264.

The probability that the company must replace a package is 1 - P(X=0) - P(X=1) - P(X=2) ≈ 1 - 0.7788 - 0.1947 - 0.0264 ≈ 0.

So, the closest answer is (a) 0.01.

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