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Probability: Acceptance Sampling

Prob 1.
Acceptance sampling is a statistical method used to monitor the quality of purchased parts and components. To ensure the quality of incoming parts, a purchaser or manufacturer normally samples 20 parts and allows 1 defect.
a. What is the likelihood of accepting a lot that is 1% defective?
b. If the quality of the incoming lot was actually 2%, what is the likelihood of accepting it?
c. If the quality of the incoming lot was actually 5%, what is the likelihood of accepting it?

Prob 2.
In a sample of 200 antennae there is a probability of 1.5% that an antenna is defective.
a. What is the probability that no antenna is defective.
b. What is the probability that 3 or more antennae are defective.

Prob 3.
On average four customers waiting in line.
a. What is the probability that no customer is waiting.
b. What is the probability that four customers are waiting in line.
c. What is the probability that 4 or fewer customers are waiting.
d. What is the probability that 4 or more customers are waiting.

Solution Preview

6.44
In a sample of n =20, at most 1 defective is accepted, i.e., either there should be 0 defective or 1 defective part(s).

Hence, likelihood of acceptance == probability that 0 defective or 1 defective in the sample of 20,
P(0 or 1) = P(0) + P(1)

In Binomial distribution, probability of r defective out of n,
P(n,r) = C(n,r) * p^r * q^(n-r)

Hence,
P(0 or 1) = P(0) + P(1)
= C(20,0)*p^0 *q^20 + C(20,1)*p^1*q^19
= 1*1*q^20 +20*p*q^19
= q^19 * (q + 20*p)

a.
Lot with defect = 1%
probability of defective, p = 0.01
probability of not defective, q = 1 - p= 0.99

For acceptance, out of 20 only 0 or 1 defective,
P(0 or 1) = 0.99^19 * (0.99 + 20* 0.01) = 0.983 ...

Solution Summary

There are 3 problems. First one is related to binomial distribution. Given lot of items, to be decided whether the lot to be accepted or not. In second problem, probability of defective antennae is given. Using Poison distribution, probability of given number of defective antennae in a sample is estimated. In third problem, estimated that how many customers may be waiting in a queue.

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