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 distribution type

This is a binomial - distribution problem. Let \(n = 36\) (number of tablets sampled), \(p=0.02\) (probability of a defective tablet), and \(q = 1 - p=0.98\). The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times q^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\).

Step2: Calculate probability of \(X = 0\)

\[

$$\begin{align*} P(X = 0)&=C(36,0)\times(0.02)^{0}\times(0.98)^{36}\\ &=1\times1\times(0.98)^{36}\\ &\approx0.4832 \end{align*}$$

\]

Step3: Calculate probability of \(X = 1\)

\[

$$\begin{align*} C(36,1)&=\frac{36!}{1!(36 - 1)!}=\frac{36!}{1!35!}=36\\ P(X = 1)&=C(36,1)\times(0.02)^{1}\times(0.98)^{35}\\ &=36\times0.02\times(0.98)^{35}\\ &\approx0.3697 \end{align*}$$

\]

Step4: Calculate probability of acceptance

The shipment is accepted if \(X = 0\) or \(X = 1\). So \(P(X\leq1)=P(X = 0)+P(X = 1)\).
\[

$$\begin{align*} P(X\leq1)&=0.4832 + 0.3697\\ &=0.8529 \end{align*}$$

\]

Answer:

0.8529