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Question
a pharmaceutical company receives large shipments of aspirin tablets. the acceptance sampling plan is to randomly select and test 48 tablets, then accept the whole batch if there is only one or none that doesnt meet the required specifications. if one shipment of 7000 aspirin tablets actually has a 3% rate of defects, what is the probability that the 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 % of the shipments and will reject % of the shipments, so (round to two decimal places as needed.)
Step1: Identify the binomial probability formula
The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 48$ (number of tablets tested), $p=0.03$ (probability of a tablet being defective), and $1 - p = 0.97$ (probability of a tablet being non - defective). The batch is accepted if $k = 0$ or $k = 1$.
Step2: Calculate $P(X = 0)$
$C(48,0)=\frac{48!}{0!(48 - 0)!}=1$. Then $P(X = 0)=C(48,0)\times(0.03)^{0}\times(0.97)^{48}=1\times1\times(0.97)^{48}\approx0.2391$.
Step3: Calculate $P(X = 1)$
$C(48,1)=\frac{48!}{1!(48 - 1)!}=\frac{48!}{1!47!}=48$. Then $P(X = 1)=C(48,1)\times(0.03)^{1}\times(0.97)^{47}=48\times0.03\times(0.97)^{47}\approx0.3589$.
Step4: Calculate the probability of acceptance
$P(X\leq1)=P(X = 0)+P(X = 1)\approx0.2391 + 0.3589=0.5980$.
Step5: Calculate the acceptance and rejection percentages
The acceptance percentage is $0.5980\times100 = 59.80\%$. The rejection percentage is $(1 - 0.5980)\times100=40.20\%$.
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The probability that this whole shipment will be accepted is $0.5980$.
The company will accept $59.80\%$ of the shipments and will reject $40.20\%$ of the shipments, so many will be rejected.