When purchasing bulk orders of batteries, a toy manufacturer uses this acceptance sampling plan: Randomly select and test 56 batteries and determine whether each is within specifications. The entire shipment is accepted if at most 2 batteries do not meet specifications. A shipment contains 5000 batteries, and 11% of them do not meet specifications. 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.)
The company will accept what % of the shipments and will reject what % of the shipments, so:

We can model this situation using the **binomial distribution**. Here’s how we can approach it step by step:

### Step 1: Define the binomial distribution parameters
- The population size (shipment size) is 5000 batteries.
- The sample size is \( n = 56 \) (the number of randomly selected batteries).
- The probability of a battery failing (i.e., not meeting specifications) is \( p = 0.11 \) (since 11% of the batteries fail).
- The number of defective batteries in the sample is denoted by \( X \), which follows a binomial distribution:  
  \[
  X \sim \text{Binomial}(n = 56, p = 0.11)
  \]
  
We are asked to find the probability that the shipment will be accepted, which happens when at most 2 batteries are defective in the sample (i.e., \( X \leq 2 \)).

### Step 2: Calculate the binomial probability
The probability mass function of a binomial distribution is given by:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
where \( k \) is the number of defective batteries.

We need to calculate \( P(X \leq 2) \), which is the sum of the probabilities for 0, 1, and 2 defective batteries:
\[
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)
\]

Let's compute this probability.

### Step 3: Interpretation of Results
Once we compute \( P(X \leq 2) \), we can interpret how frequently shipments are accepted or rejected. The percentage of shipments accepted will be approximately \( P(X \leq 2) \times 100 \)%, and the percentage of shipments rejected will be the complement, i.e., \( 100\% - P(X \leq 2) \times 100 \).

Let me calculate that probability for you.The probability that the whole shipment will be accepted is approximately **0.0461** (or 4.61%).

This means that about **4.61%** of the shipments will be accepted, and about **95.39%** of the shipments will be rejected. Therefore, most shipments will be rejected under this acceptance sampling plan.

Would you like further details or have any questions?

Here are five related questions to expand on this concept:
1. How would the acceptance rate change if the sample size were increased?
2. What happens to the probability of acceptance if the defect rate decreases to 5%?
3. How does the binomial distribution relate to quality control in other industries?
4. Can you calculate the expected number of defective batteries in a sample of 56?
5. What is the probability of rejecting a shipment if 3 defective batteries are allowed?

**Tip:** When dealing with small acceptance thresholds, even slight changes in sample size or defect rate can have a significant impact on the acceptance probability.

When purchasing bulk orders of batteries, a toy manufacturer uses this acceptance sampling plan: Randomly select and test 56 batteries and determine whether each is within specifications. The entire shipment is accepted if at most 2 batteries do not meet specifications. A shipment contains 5000 batteries, and 11% of them do not meet specifications. What is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected?

Determine the probability of accepting a bulk battery shipment using a binomial distribution-based acceptance sampling plan. Learn how to calculate the probability that no more than 2 defective batteries appear in a sample of 56 when 11% of the 5000 batteries are defective. Suitable for Grades 11-12 students.

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