The quality manager for Frozen Pizzas, Inc., is concerned about the quality of the lot of several thousand pizzas which his company produced this week, and is now ready to ship. If he uses an acceptance sampling plan of n = 8 and c = 1, what is the probability that, if this lot is 10 percent defective, it will be accepted for shipment without inspection?
A) .9619
B) .9428
C) .8131
D) .4305
E) .1869

Question

Quizplus · Accepted Answer

We can use the binomial distribution to calculate the probability that a sample of 8 pizzas will have no more than one defective pizza if the true proportion of defective pizzas is 10%. 
Probability of accepting a lot without inspection = P(no more than 1 defective pizza in the sample of 8)
= P(0 defective) + P(1 defective)
= C(8,0)*(0.1)^0*(0.9)^8 + C(8,1)*(0.1)^1*(0.9)^7
= 0.9^8 + 8*0.1*0.9^7
= 0.8131
 
Therefore, the probability of accepting the lot without inspection is 0.8131 or 81.31%. The best choice is C.

The Quality Manager for Frozen Pizzas, Inc