Gordon O'Quinn
05/14/2024 · Primary School
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 3000 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.)
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
To determine the probability that the whole shipment will be accepted, we can model the number of defective tablets in the sample of 55 using the binomial distribution. The parameters for the binomial distribution are:
- \( n = 55 \) (the number of tablets sampled)
- \( p = 0.03 \) (the probability of a tablet being defective)
We want to find the probability of having at most 1 defective tablet in the sample, which means we need to calculate \( P(X \leq 1) \), where \( X \) is the number of defective tablets in the sample.
This can be calculated as:
\[
P(X = 0) + P(X = 1)
\]
Using the binomial probability formula:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
1. **Calculate \( P(X = 0) \)**:
\[
P(X = 0) = \binom{55}{0} (0.03)^0 (0.97)^{55} = 1 \cdot 1 \cdot (0.97)^{55}
\]
Calculating \( (0.97)^{55} \):
\[
(0.97)^{55} \approx 0.2240
\]
So,
\[
P(X = 0) \approx 0.2240
\]
2. **Calculate \( P(X = 1) \)**:
\[
P(X = 1) = \binom{55}{1} (0.03)^1 (0.97)^{54} = 55 \cdot 0.03 \cdot (0.97)^{54}
\]
Calculating \( (0.97)^{54} \):
\[
(0.97)^{54} \approx 0.2317
\]
So,
\[
P(X = 1) \approx 55 \cdot 0.03 \cdot 0.2317 \approx 55 \cdot 0.006951 \approx 0.3823
\]
3. **Combine the probabilities**:
Now, we can find \( P(X \leq 1) \):
\[
P(X \leq 1) = P(X = 0) + P(X = 1) \approx 0.2240 + 0.3823 \approx 0.6063
\]
Thus, the probability that the whole shipment will be accepted is approximately:
\[
\boxed{0.6063}
\]
### Conclusion
Since the probability of acceptance is about 0.6063, this means that slightly more than half of such shipments will be accepted. Therefore, while many shipments will be accepted, a significant number will also be rejected, as the acceptance rate is not overwhelmingly high.
Quick Answer
The probability that the whole shipment will be accepted is approximately 0.6063.
Answered by UpStudy AI and reviewed by a Professional Tutor
UpStudy ThothAI
Self-Developed and Ever-Improving
Thoth AI product is constantly being upgraded and optimized.
Covers All Major Subjects
Capable of handling homework in math, chemistry, biology, physics, and more.
Instant and Accurate
Provides immediate and precise solutions and guidance.
Try Now
Ask Tutors
Ask AI
10x
Fastest way to Get Answers & Solutions
By text
Enter your question here…
By image
Re-Upload
Submit