To solve this problem, we will use the Poisson approximation for the binomial distribution. The Poisson distribution is often used as an approximation of the binomial distribution when the number of trials is large and the probability of success is small.

Step 1: Identify the parameters for the Poisson distribution. The mean (λ) of a Poisson distribution is np, where n is the number of trials and p is the probability of success. In this case, n = 25 (the number of screws in a package) and p = 0.01 (the probability that a screw is defective). So, λ = np = 25 * 0.01 = 0.25.

Step 2: Calculate the probability that more than 1 screw is defective. The company must replace a package if more than 1 screw is defective. So, we need to find the probability that X 1, where X is a random variable with a Poisson distribution with mean 0.25.

The probability that X 1 is 1 minus the probability that X ≤ 1. The probability that X ≤ 1 is the sum of the probabilities that X = 0 and X = 1.

Using the formula for the Poisson distribution, we have:

P(X = 0) = e^-λ * λ^0 / 0! = e^-0.25 * 0.25^0 / 1 = e^-0.25 = 0.7788.

P(X = 1) = e^-λ * λ^1 / 1! = e^-0.25 * 0.25^1 / 1 = e^-0.25 * 0.25 = 0.1947.

So, P(X ≤ 1) = P(X = 0) + P(X = 1) = 0.7788 + 0.1947 = 0.9735.

Therefore, P(X 1) = 1 - P(X ≤ 1) = 1 - 0.9735 = 0.0264.

So, the probability that the company must replace a package is approximately 0.0264. The correct answer is (d) 0.0264.

Question

To solve this problem, we will use the Poisson approximation for the binomial distribution. The Poisson distribution is often used as an approximation of the binomial distribution when the number of trials is large and the probability of success is small.

Step 1: Identify the parameters for the Poisson distribution. The mean (λ) of a Poisson distribution is np, where n is the number of trials and p is the probability of success. In this case, n = 25 (the number of screws in a package) and p = 0.01 (the probability that a screw is defective). So, λ = np = 25 * 0.01 = 0.25.

Step 2: Calculate the probability that more than 1 screw is defective. The company must replace a package if more than 1 screw is defective. So, we need to find the probability that X > 1, where X is a random variable with a Poisson distribution with mean 0.25.

The probability that X > 1 is 1 minus the probability that X ≤ 1. The probability that X ≤ 1 is the sum of the probabilities that X = 0 and X = 1.

Using the formula for the Poisson distribution, we have:

P(X = 0) = e^-λ * λ^0 / 0! = e^-0.25 * 0.25^0 / 1 = e^-0.25 = 0.7788.

P(X = 1) = e^-λ * λ^1 / 1! = e^-0.25 * 0.25^1 / 1 = e^-0.25 * 0.25 = 0.1947.

So, P(X ≤ 1) = P(X = 0) + P(X = 1) = 0.7788 + 0.1947 = 0.9735.

Therefore, P(X > 1) = 1 - P(X ≤ 1) = 1 - 0.9735 = 0.0264.

So, the probability that the company must replace a package is approximately 0.0264. The correct answer is (d) 0.0264.

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