To solve this problem, we need to compute the probability of accepting a shipment based on the acceptance sampling plan. The steps involve using the binomial distribution to model the number of defective batteries in a random sample.

Step 1: Parameters of the Binomial Distribution

• The total number of batteries sampled, $n = 37$.
• The probability that a randomly selected battery is defective is $p = 0.02$ (2% of the batteries do not meet specifications).
• The shipment will be accepted if at most 3 defective batteries are found in the sample of 37.

So, we are looking for the probability that the number of defective batteries, $X$, in a sample of 37 follows a binomial distribution: $X \sim \text{Binomial}(n = 37, p = 0.02)$ We want to find $P(X \leq 3)$, which is the probability that there are at most 3 defective batteries in the sample.

Step 2: Binomial Distribution Formula

The probability mass function (PMF) for the 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, $n$ is the sample size, and $p$ is the probability of a defective battery.

We need to compute $P(X \leq 3)$, which is the sum of the probabilities for $X = 0$, $X = 1$, $X = 2$, and $X = 3$: $P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

Step 3: Compute the Individual Probabilities

1. For $P(X = 0)$: $P(X = 0) = \binom{37}{0} (0.02)^0 (0.98)^{37} = (1) \times (1) \times (0.98)^{37} \approx 0.4877$

2. For $P(X = 1)$: $P(X = 1) = \binom{37}{1} (0.02)^1 (0.98)^{36} = 37 \times 0.02 \times (0.98)^{36} \approx 0.3741$

3. For $P(X = 2)$: $P(X = 2) = \binom{37}{2} (0.02)^2 (0.98)^{35} = \frac{37 \times 36}{2} \times 0.0004 \times (0.98)^{35} \approx 0.1120$

4. For $P(X = 3)$: $P(X = 3) = \binom{37}{3} (0.02)^3 (0.98)^{34} = \frac{37 \times 36 \times 35}{6} \times 0.000008 \times (0.98)^{34} \approx 0.0203$

Step 4: Sum the Probabilities

Now, we sum the individual probabilities: $P(X \leq 3) = 0.4877 + 0.3741 + 0.1120 + 0.0203 = 0.9941$

Step 5: Conclusion

Part 1: Probability of Acceptance

The probability that the whole shipment will be accepted is approximately: $P(X \leq 3) \approx 0.9941$ So, the probability is 0.9941, or 99.41%.

Part 2: Acceptance and Rejection Rates

• Acceptance rate: 99.41% of shipments will be accepted.
• Rejection rate: $100\% - 99.41\% = 0.59\%$.

Thus, the company will accept 99.41% of the shipments and reject 0.59% of the shipments.

Since the rejection rate is very small, almost all of the shipments will be accepted.

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Summary

• Probability of acceptance: 0.9941 or 99.41%.
• Percentage of accepted shipments: 99.41%.
• Percentage of rejected shipments: 0.59%.
• Conclusion: Almost all of the shipments will be accepted.

Would you like further clarification on any of these steps?