Question

A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 27 tablets. The entire shipment is accepted if at most 2 tablets do not meet the required specifications. If a particular shipment of thousands of aspirin tablets actually has a 6.0% rate of​ defects, what is the probability that this whole shipment will be​ accepted?

The probability that this whole shipment will be accepted is__

Answer 1

We know that number of samples,n is 27.

We also know that the probability of success,p (here, we are considering the rate of defect as the success %) is 6% or 0.06.

We know that the shipment will be accepted only if the defects are 0, 1 or 2 out of 27.

This is a binomial distribution with X~B(n,p) = B(27,0.06)

To find P(X=0),P(X=1) and P(X=2), we must use the formula of the distribution function.

P(X=r)= nCr * p^r * q^n-r

Substituting r= 0, 1 and 2, we get

P(X=0)= 0.18812739731

P(X=1)= 0.32421955706

P(X=2)= 0.26903324948

Thus, probability that the shipment is accepted= P(X=0)+P(X=1)+P(X=2)

= 0.78138020385 or 78.14%