Question 1199632: The Ziteck Corporation buys parts from international suppliers. One part is currently being purchased from a Malaysian supplier under a contract that calls for at most 5% of the 10,000 parts to be defective. When a shipment arrives, Ziteck randomly samples 10 parts. If it finds 2 or fewer defectives in the sample, it keeps the shipment; otherwise, it returns the entire shipment to the supplier.
a)Assuming that the conditions for the binomial distribution are satisfied, what is the probability that the sample will lead Ziteck to keep the shipment if the defect rate is actually 0.05?
b)Suppose the supplier is actually sending Ziteck 10% defects. What is the probability that the sample will lead Ziteck to accept the shipment anyway?
c)Comment on this sampling plan (sample size and accept/reject point). Do you think it favors either Ziteck or the supplier? Discuss.
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! ### Part a) Probability that Ziteck keeps the shipment if the defect rate is 5%
This is a binomial distribution problem where:
- \( n = 10 \) (sample size),
- \( p = 0.05 \) (probability of a defective part),
- \( X \) is the number of defective parts in the sample.
Ziteck keeps the shipment if \( X \leq 2 \). The probability of this is:
\[
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)
\]
The probability mass function of a binomial distribution is:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
So:
1. \( P(X = 0) = \binom{10}{0} (0.05)^0 (0.95)^{10} \)
2. \( P(X = 1) = \binom{10}{1} (0.05)^1 (0.95)^9 \)
3. \( P(X = 2) = \binom{10}{2} (0.05)^2 (0.95)^8 \)
Summing these gives \( P(X \leq 2) \). Let's calculate.
The probability that Ziteck will keep the shipment if the defect rate is 5% is approximately **0.9885** (98.85%).
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### Part b) Probability that Ziteck keeps the shipment if the defect rate is 10%
If the defect rate is \( p = 0.10 \), we use the same approach, calculating \( P(X \leq 2) \) for \( n = 10 \) and \( p = 0.10 \).
The probability that Ziteck will keep the shipment if the defect rate is 10% is approximately **0.9298** (92.98%).
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### Part c) Commentary on the sampling plan
1. **Sample size**: A sample size of 10 is relatively small. While it simplifies the process of inspecting shipments, it may not provide a reliable estimate of the overall defect rate in large shipments, especially if the defect rate is close to the threshold.
2. **Acceptance/rejection point**: The decision rule (\( X \leq 2 \)) makes it highly likely that Ziteck will accept shipments, even if the defect rate is higher than 5%. For example, with a 10% defect rate, there is still a 92.98% chance of accepting the shipment.
3. **Favors supplier or Ziteck?**:
- The plan appears to favor the supplier, as it allows for a high likelihood of shipment acceptance even when the defect rate exceeds the contractual limit.
- From Ziteck's perspective, this increases the risk of receiving shipments with a defect rate that exceeds the acceptable threshold, potentially leading to quality issues in their operations.
**Recommendation**: Ziteck could consider increasing the sample size or lowering the acceptance threshold (\( X \leq 1 \)) to reduce the risk of accepting defective shipments. These changes would provide a better balance between supplier quality assurance and Ziteck's operational needs.
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