Math Problem Statement

Question content area top Part 1 A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 44 ​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 2​% 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. ​(Round to four decimal places as​ needed.)

Solution

This is a problem involving binomial probability since we are dealing with the probability of a certain number of defective tablets in a sample of 44, and we have a fixed probability of defects (2%) for each tablet.

Step-by-Step Solution:

  1. Define Variables:

    • n = 44 (the number of tablets sampled).
    • p = 0.02 (the probability that a single tablet is defective).
    • x = 0 or 1 (the acceptable number of defective tablets in the sample, since the batch is accepted if 1 or fewer tablets are defective).
  2. We will use the binomial distribution formula: P(X=x)=(nx)px(1p)nxP(X = x) = \binom{n}{x} p^x (1 - p)^{n - x} where:

    • P(X=x)P(X = x) is the probability of observing exactly xx defective tablets in the sample.
    • (nx)\binom{n}{x} is the binomial coefficient, given by n!x!(nx)!\frac{n!}{x!(n - x)!}.
  3. Calculate the probability for x=0x = 0: P(X=0)=(440)(0.02)0(0.98)44P(X = 0) = \binom{44}{0} (0.02)^0 (0.98)^{44} P(X=0)=1×1×(0.98)440.364169P(X = 0) = 1 \times 1 \times (0.98)^{44} \approx 0.364169

  4. Calculate the probability for x=1x = 1: P(X=1)=(441)(0.02)1(0.98)43P(X = 1) = \binom{44}{1} (0.02)^1 (0.98)^{43} P(X=1)=44×0.02×(0.98)430.374906P(X = 1) = 44 \times 0.02 \times (0.98)^{43} \approx 0.374906

  5. Add these probabilities together to find the total probability of accepting the batch: P(accepted)=P(X=0)+P(X=1)P(\text{accepted}) = P(X = 0) + P(X = 1) P(accepted)0.364169+0.374906=0.7391P(\text{accepted}) \approx 0.364169 + 0.374906 = 0.7391

Conclusion:

The probability that the whole shipment will be accepted is 0.7391 (rounded to four decimal places).

Will most shipments be accepted or rejected?

Given that the probability of acceptance is about 73.91%, most shipments with a 2% defect rate will be accepted.


Would you like a more detailed explanation or further questions on this topic?

Here are some related questions you might find interesting:

  1. How would the probability of acceptance change if the defect rate were higher (e.g., 5%)?
  2. How do you calculate the expected number of defective tablets in the sample?
  3. What changes would occur if the sample size was increased from 44 to 100?
  4. How can we use the normal approximation to the binomial distribution in this scenario?
  5. What is the likelihood of rejecting a batch if the defect rate is 2%?

Tip: When solving binomial probability problems, it's important to first identify the parameters (sample size, success probability, number of successes) and then apply the binomial distribution formula for exact values or the normal approximation for large sample sizes.

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution

Formulas

Binomial Probability Formula P(X = x) = C(n, x) * p^x * (1 - p)^(n - x)

Theorems

Binomial Theorem

Suitable Grade Level

Grades 11-12