Math Problem Statement

pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 52 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 3% 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 solve this problem, we can model it using the binomial distribution. The scenario involves testing 52 aspirin tablets from a shipment of 7,000, and the shipment is accepted if one or none of the selected tablets do not meet the required specifications. The defect rate in the shipment is 3%, or 0.03.

Step 1: Define the binomial distribution

Let XX represent the number of defective tablets in the sample of 52.

  • The probability of a defective tablet is p=0.03p = 0.03.
  • The number of trials (tablets tested) is n=52n = 52.
  • XX follows a binomial distribution: XBinomial(n=52,p=0.03)X \sim \text{Binomial}(n = 52, p = 0.03).

The probability mass function for the binomial distribution is given by:

P(X=k)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

where kk is the number of defective tablets found in the sample.

Step 2: Find the probability of accepting the shipment

The shipment will be accepted if there is at most one defective tablet in the sample. This corresponds to the event X1X \leq 1, i.e., 0 or 1 defective tablets in the sample.

We need to calculate:

P(X1)=P(X=0)+P(X=1)P(X \leq 1) = P(X = 0) + P(X = 1)

We can calculate P(X=0)P(X = 0) and P(X=1)P(X = 1) using the binomial formula.

Step 3: Calculate the probabilities

Let's compute P(X=0)P(X = 0) and P(X=1)P(X = 1):

P(X=0)=(520)(0.03)0(0.97)52=(0.97)52P(X = 0) = \binom{52}{0} (0.03)^0 (0.97)^{52} = (0.97)^{52}

P(X=1)=(521)(0.03)1(0.97)51=52×0.03×(0.97)51P(X = 1) = \binom{52}{1} (0.03)^1 (0.97)^{51} = 52 \times 0.03 \times (0.97)^{51}

Now, I will compute these values to find the final probability.The probability that the whole shipment will be accepted (i.e., there is at most one defective tablet in the sample) is approximately 0.5352.

Interpretation:

  • About 53.52% of shipments will be accepted.
  • Since the probability is not very high (it's less than 0.5), this suggests that many shipments with a 3% defect rate will be rejected.

Would you like more details or have further questions on this problem?

Here are some related questions you might find interesting:

  1. How would the probability change if the defect rate were lower or higher?
  2. What is the expected number of defective tablets in a sample of 52 tablets?
  3. How does the size of the sample (52 tablets) affect the acceptance probability?
  4. What would happen if the acceptance criteria were more stringent (e.g., requiring no defective tablets)?
  5. How can we calculate the probability of rejecting the shipment?

Tip: When working with binomial distributions, using the cumulative distribution function (CDF) can simplify calculations for problems like this one, where you're interested in the probability of an event happening a certain number of times or fewer.

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution
Acceptance Sampling

Formulas

Binomial probability formula: P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
Cumulative probability: P(X ≤ 1) = P(X = 0) + P(X = 1)

Theorems

Binomial Theorem

Suitable Grade Level

Undergraduate (Statistics or Probability course)