Statistics Question (Probability)?
A pharmaceutical company receives large shipments of ibuprofen tablets and uses this acceptance sampling plan: randomly select and test 23 tablets, then accept the whole batch if there is at most one that doesn’t meet the required specifications. If a particular shipment of thousands of ibuprofen tablets actually has a 12% rate of defects, what is the probability that this whole shipment will be accepted?
(Report answer as a decimal value accurate to four decimal places.)
P(accept shipment) =______?
Please help and explain! thanks
- 8 years agoFavorite Answer
First, let Y denote the number of defective tablets in the 23
So Y ~ Binomial(n=23, p=0.12)
P(accept shipment) = P(Y≤1)=P(Y=0)+P(Y=1)
P(Y=0)= (23 choose 0)*(0.12)^0*(1-0.12)^23=0.05286
P(Y=1)= (23 choose 1)*(0.12)^1*(1-0.12)^22=0.16578
This should be the answer.
What we have here is a binomial distribution with 23 Bernoulli trials, each having a probability of 0.12 of occuring.
You want at most one defective tablet, so both 0 and 1 defective tablets is o.k.
Then you apply the probability formula of a binomial distribution to both of these outcomes.
P(Y=y)=(n choose y)*p^y*(1-p)^(n-y)
In this case, n=23, p=0.12
Look at it this way, (n choose y) ensures that all the permutations are taken into count, of the 23, y are defective and 23-y are not defective.
- Anonymous8 years ago
Let x be the random variable denotes the no. of defective tablets.
Need to find P(x<=1)=?
It follows binomial distribution with n=12, p=0.12 and q=1-0.12=0.88
= 12c0*(0.12)^0*(0.88)^12 + 12c1*(0.12)^1*(0.88)^11
P(accept shipment) =P(x<=1)=0.5685(Ans.)Source(s): http://tutorteddy.com/