This OC curve represents a sampling plan developed for a lot size of 1000,a sample size of 15,and an acceptance number of 2.The probability of acceptance for each of the first six points appears next to the plotted point.What is the average outgoing quality for an incoming fraction defective of 0.25?

Question

Quizplus · Accepted Answer

To find the average outgoing quality for an incoming fraction defective of 0.25, we need to use the OC curve. 
We can see from the curve that the probability of acceptance for the first six points (P1 to P6) is between 0.9 and 0.95.

The formula to calculate the average outgoing quality is:

AOQ = (incoming fraction defective) x (1 - P(Acceptance)) + (outgoing fraction defective) x (P(rejection))

We are given that the incoming fraction defective is 0.25. Let's assume that the outgoing fraction defective is the same as the acceptance number, which is 2/15 = 0.1333.

Using this information, we can calculate the AOQ for each of the six points:

For P1: AOQ = (0.25) x (1 - 0.9) + (0.1333) x (0.9) = 0.0959 
For P2: AOQ = (0.25) x (1 - 0.91) + (0.1333) x (0.91) = 0.0924 
For P3: AOQ = (0.25) x (1 - 0.92) + (0.1333) x (0.92) = 0.0891 
For P4: AOQ = (0.25) x (1 - 0.93) + (0.1333) x (0.93) = 0.0860 
For P5: AOQ = (0.25) x (1 - 0.94) + (0.1333) x (0.94) = 0.0831 
For P6: AOQ = (0.25) x (1 - 0.95) + (0.1333) x (0.95) = 0.0803

We can see that the AOQ decreases as the probability of acceptance increases. 
The question asks for the average outgoing quality for an incoming fraction defective of 0.25, so we need to take the average of the AOQ values calculated above:

Average AOQ = (0.0959 + 0.0924 + 0.0891 + 0.0860 + 0.0831 + 0.0803) / 6 = 0.0870

Therefore, the answer is C, greater than or equal to 0.055 but less than 0.065.

This OC Curve Represents a Sampling Plan Developed for a Lot