Question
A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 46 tablets, then accepf 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 5% 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
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Answer to a math question A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 46 tablets, then accepf 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 5% 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
Eliseo
4.6107 Answers
Let's denote:
-p
-q = 1 - p
The probability of having one defective tablet in a sample of 46 is given by the binomial distribution formula:
P(X = 1) = \binom{46}{1} p^1 q^{45} = 46 \cdot 0.05 \cdot (1 - 0.05)^{45}
The probability of having zero defective tablets in a sample of 46 is:
P(X = 0) = \binom{46}{0} p^0 q^{46} = 1 \cdot (1 - 0.05)^{46}
Thus, the probability of accepting the whole batch is the sum of these two probabilities:
P(\text{Accept}) = P(X = 0) + P(X = 1)
Now, we can calculate this probability:
P(\text{Accept}) = (1 - 0.05)^{46} + 46 \cdot 0.05 \cdot (1 - 0.05)^{45}
P(\text{Accept}) = 0.95^{46} + 46 \cdot 0.05 \cdot 0.95^{45}
\boxed{P(\text{Accept})\approx0.3232}
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The probability of having one defective tablet in a sample of 46 is given by the binomial distribution formula:
The probability of having zero defective tablets in a sample of 46 is:
Thus, the probability of accepting the whole batch is the sum of these two probabilities:
Now, we can calculate this probability:
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