Math Problem Statement

A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 55 ​tablets, then accept the whole batch if there is only one or none that​ doesn't meet the required specifications. If one shipment of 5000 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? Question content area bottom Part 1 The probability that this whole shipment will be accepted is    enter your response here

Solution

To solve this problem, we need to calculate the probability that, in a random sample of 55 tablets from a shipment with a 4% defect rate, we find at most one defective tablet. If that condition is met, the shipment will be accepted.

  1. Define Variables:

    • n=55n = 55: sample size (number of tablets tested)
    • p=0.04p = 0.04: probability of a tablet being defective
    • XX: number of defective tablets in the sample
  2. Distribution: Since each tablet's defect status is independent of others, we can model XX, the number of defective tablets in the sample, as a binomial random variable: XBinomial(n=55,p=0.04)X \sim \text{Binomial}(n = 55, p = 0.04)

  3. Probability of Acceptance: The shipment is accepted if there are 0 or 1 defective tablets in the sample. Thus, we need to calculate: P(X1)=P(X=0)+P(X=1)P(X \leq 1) = P(X = 0) + P(X = 1)

  4. Binomial Probability Formula: The probability of kk defects in a sample of nn tablets, with defect rate pp, is: P(X=k)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} where (nk)\binom{n}{k} is the binomial coefficient.

  5. Calculate Probabilities:

    • For P(X=0)P(X = 0): P(X=0)=(550)(0.04)0(0.96)55=(0.96)55P(X = 0) = \binom{55}{0} (0.04)^0 (0.96)^{55} = (0.96)^{55}
    • For P(X=1)P(X = 1): P(X=1)=(551)(0.04)1(0.96)54=55×0.04×(0.96)54P(X = 1) = \binom{55}{1} (0.04)^1 (0.96)^{54} = 55 \times 0.04 \times (0.96)^{54}
  6. Compute the Values: Now, using the values for P(X=0)P(X = 0) and P(X=1)P(X = 1), we sum them up to find P(X1)P(X \leq 1).

I'll calculate this for you now.It seems like I can’t do more advanced data analysis right now. Please try again later.

To help with this problem, I can walk you through the remaining calculation if you’d like, or you can use the formula and plug in values manually with a calculator. Let me know if you’d like further help on the exact steps.

Would you like details on this, or do you have any other questions? Here are some related questions to consider:

  1. How does the binomial distribution apply to quality control sampling?
  2. What are alternative ways to calculate probabilities without using binomial formulas?
  3. How would changing the sample size from 55 to 100 impact the acceptance probability?
  4. What defect rates would lead to near-certain acceptance or rejection?
  5. How could you determine a minimum sample size needed to detect defects effectively?

Tip: In quality control, increasing the sample size generally improves the sensitivity of defect detection but may also increase the rejection rate for minor defect levels.

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution
Acceptance Sampling

Formulas

P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
P(X <= 1) = P(X = 0) + P(X = 1)

Theorems

Binomial Theorem

Suitable Grade Level

Grades 11-12