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# Expression for the OC curve

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A single test on a certain device costs more than the entire lot. It is therefore decreed that not more than three tests will be performed per lot. The following acceptance plan will therefore be employed: Test two units. If both function accept the lot. If both fail, reject the lot. Otherwise test another item. If it fails, reject the lot. If it functions accept.
i) Assuming the binomial distribution is applicable, write down the expression for the OC curve, i.e., write the expression for the probability of acceptance as a function of the lot fraction defective p. Evaluate it at lot fraction defective of p=2%.
ii) Write the expression for the OC curve of a single sampling plan having n=3 and c=1. Evaluate it when p=2%
iii) For the first plan the sample size is a random variable assuming the value n=2 or n=3. Compute the average sample size when the lot is 2% defective.

https://brainmass.com/statistics/data-collection/expression-oc-curve-16573

## SOLUTION This solution is FREE courtesy of BrainMass!

An OC curve is drawn as Prob of acceptance against the percent of defectives.

1) Prob of Acceptance if the lot has p fraction defectives :

= 2c2*(1-p)^2 + (1-p)^2*q ( note : q = 1-p )

2)C = 1 means that total defectives should not be greater than 1 out of 3.
Expression of OC Curve :
3c1* (1-.02)^2 * .02^1 + (1-.02)^3

3)Average Sample Size:

Prob of acceptance if N =2 : ( 1-.02)*(1-.02)= .9604
Prob of acceptance if N =3 : 3c2*(1-.02)^2*.02 = .0576

Average = 2*.9604 + 3*.0576 = 2.0936

Hope it is clear.

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