Math Problem Statement

A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 45 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If one shipment of 7000 aspirin tablets actually has a 4% rate of defects, what is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected? The probability that this whole shipment will be accepted is (Round to four decimal places as needed.)

Solution

To determine the probability of accepting the shipment of aspirin tablets under the given conditions, we can model the situation using the binomial distribution. In this case, we have:

  • Total tablets tested, n=45n = 45
  • Probability of a defect, p=0.04p = 0.04
  • Number of defects allowed for acceptance, k1k \leq 1 (either 0 or 1 defect).

Step 1: Calculate the Probability of 0 Defects

Using the binomial formula for 0 defects:

P(X=0)=(n0)p0(1p)nP(X = 0) = \binom{n}{0} p^0 (1 - p)^{n}

Substituting in the values:

P(X=0)=(450)(0.04)0(0.96)45P(X = 0) = \binom{45}{0} (0.04)^0 (0.96)^{45} P(X=0)=11(0.96)45P(X = 0) = 1 \cdot 1 \cdot (0.96)^{45} P(X=0)=(0.96)450.2288P(X = 0) = (0.96)^{45} \approx 0.2288

Step 2: Calculate the Probability of 1 Defect

Using the binomial formula for 1 defect:

P(X=1)=(n1)p1(1p)n1P(X = 1) = \binom{n}{1} p^1 (1 - p)^{n-1}

Substituting in the values:

P(X=1)=(451)(0.04)1(0.96)44P(X = 1) = \binom{45}{1} (0.04)^{1} (0.96)^{44} P(X=1)=450.04(0.96)44P(X = 1) = 45 \cdot 0.04 \cdot (0.96)^{44} P(X=1)450.040.23880.4294P(X = 1) \approx 45 \cdot 0.04 \cdot 0.2388 \approx 0.4294

Step 3: Total Probability of Acceptance

Now, we sum the probabilities of accepting the shipment (0 or 1 defect):

P(Accepted)=P(X=0)+P(X=1)0.2288+0.42940.6582P(\text{Accepted}) = P(X = 0) + P(X = 1) \approx 0.2288 + 0.4294 \approx 0.6582

Final Answer

The probability that this whole shipment will be accepted is approximately 0.6582 (rounded to four decimal places).

Conclusion

Thus, almost all such shipments will not be accepted, as the acceptance probability is below 0.7, indicating that many shipments could still be rejected.

If you have any questions or need further details, let me know! Here are some related questions you might find interesting:

  1. How does changing the defect rate affect the acceptance probability?
  2. What would be the probability of acceptance if the sample size were increased to 60 tablets?
  3. How can this sampling method be improved to increase acceptance rates?
  4. What are the implications of a high defect rate on quality control?
  5. How is the binomial distribution applied in other quality control scenarios?

Tip: When working with probabilities in quality control, it's essential to understand the trade-offs between sample size and acceptance criteria.

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution

Formulas

P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)

Theorems

Binomial Distribution Theorem

Suitable Grade Level

Grades 11-12