Question

a pharmaceutical company receives large shipments of aspirin tablets. the acceptance sampling plan is to randomly select and test 36 tablets, then accept the whole batch if there is only one or none that doesnt meet the required specifications. if one shipment of 4000 aspirin tablets actually has a 2% 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.)

Explanation:

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, $n = 36$, $p=0.02$, and $1 - p = 0.98$. The shipment is accepted if $k = 0$ or $k = 1$.

Step2: Calculate $P(X = 0)$

$C(36,0)=\frac{36!}{0!(36 - 0)!}=1$. Then $P(X = 0)=C(36,0)\times(0.02)^{0}\times(0.98)^{36}=1\times1\times(0.98)^{36}$.
Using a calculator, $(0.98)^{36}\approx0.4832$.

Step3: Calculate $P(X = 1)$

$C(36,1)=\frac{36!}{1!(36 - 1)!}=\frac{36!}{1!35!}=36$. Then $P(X = 1)=C(36,1)\times(0.02)^{1}\times(0.98)^{35}=36\times0.02\times(0.98)^{35}$.
$(0.98)^{35}\approx0.4931$, and $36\times0.02\times0.4931 = 36\times0.009862=0.3550$.

Step4: Calculate the probability of acceptance

The probability that the shipment is accepted is $P(X\leq1)=P(X = 0)+P(X = 1)$.
$P(X\leq1)=0.4832 + 0.3550=0.8382$.

Answer:

$0.8382$