a pharmaceutical company receives large shipments of aspirin tablets. the acceptance sampling plan is to…

a pharmaceutical company receives large shipments of aspirin tablets. the acceptance sampling plan is to randomly select and test 58 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 4% 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.)

a pharmaceutical company receives large shipments of aspirin tablets. the acceptance sampling plan is to randomly select and test 58 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 4% 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.)

Answer

Explanation:

Step1: Identify the distribution

This is a binomial - distribution problem. Let (n = 58) (the number of tablets sampled), (p=0.04) (probability of a defective tablet), and (q = 1 - p=0.96) (probability of a non - defective tablet). 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)!}). The shipment is accepted if (X = 0) or (X = 1).

Step2: Calculate (P(X = 0))

[ \begin{align*} P(X = 0)&=C(58,0)\times(0.04)^{0}\times(0.96)^{58}\ &=1\times1\times(0.96)^{58}\ &\approx0.0993 \end{align*} ]

Step3: Calculate (P(X = 1))

[ \begin{align*} C(58,1)&=\frac{58!}{1!(58 - 1)!}=\frac{58!}{1!57!}=58\ P(X = 1)&=C(58,1)\times(0.04)^{1}\times(0.96)^{57}\ &=58\times0.04\times(0.96)^{57}\ &\approx0.2342 \end{align*} ]

Step4: Calculate the probability of acceptance

The probability of acceptance (P(X\leq1)=P(X = 0)+P(X = 1)). [ \begin{align*} P(X\leq1)&=0.0993 + 0.2342\ &=0.3335 \end{align*} ]

Answer:

0.3335