An aircraft seam requires 25 rivets. The seam will have to be reworked if any of the rivets is defec...

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Part a: Finding the Probability of a Defective Rivet

1. Understand the given information: 15% of all seams need reworking, meaning there is at least one defective rivet in those seams. This implies that 85% of the seams do not need reworking because all their rivets are not defective.

2. Define the probabilities: Let \(p\) be the probability that a single rivet is defective, and \(q\) be the probability that a single rivet is not defective. Since these are complementary events, \(q = 1 - p\).

3. Calculate the probability for non-reworking seams: For a seam not to require reworking, all 25 rivets must not be defective. The probability of this happening is \(q^{25} = 0.85\), because 85% of the seams do not need reworking.

4. Solve for \(q\): To find \(q\), we take the 25th root of 0.85:

\[ q = \sqrt[25]{0.85} \]

5. Find \(p\): Since \(q = 1 - p\), we rearrange to solve for \(p\):

\[ p = 1 - q \]

6. Perform calculations: Calculating \(q\) as the 25th root of 0.85 and then \(p\) as \(1 - q\), we get the probability that a rivet is defective.

Part b: Probability of a Defective Rivet for 10% Reworking Requirement

1. Setup for 10% reworking requirement: Now we want only 10% of seams to require reworking, meaning 90% do not. This changes our target probability for seams not requiring reworking to \(q^{25} = 0.90\).

2. Calculate \(q\) for the new requirement: Similar to part a, we solve for \(q\) by taking the 25th root of 0.90:

\[ q = \sqrt[25]{0.90} \]

3. Solve for \(p\): With \(q\) known, \(p = 1 - q\) gives us the required probability of a rivet being defective to meet the new reworking rate.

Performing the Calculations

- For part a, finding \(q\) as the 25th root of 0.85:

\[ q = \sqrt[25]{0.85} \approx 0.99403 \]

Then, calculating \(p\):

\[ p = 1 - 0.99403 \approx 0.00597 \]

Thus, the probability that a rivet is defective is approximately 0.6%.

- For part b, calculating \(q\) as the 25th root of 0.90:

\[ q = \sqrt[25]{0.90} \approx 0.99616 \]

And then \(p\):

\[ p = 1 - 0.99616 \approx 0.00384 \]

To ensure only 10% of seams need reworking, the probability of a rivet being defective should be approximately 0.4%.

These steps outline a method to connect the probability of individual defects to the overall quality control standard in manufacturing processes, essential for maintaining high reliability in aircraft construction.