Acceptance sampling is an important quality control technique, where a batch of
ID: 3156541 • Letter: A
Question
Acceptance sampling is an important quality control technique, where a batch of data is tested to determine if the proportion of units having a particular attribute exceeds a given percentage. Suppose that 12% of produced items are known to be nonconforming. Every week a batch of items is evaluated and the production machines are adjusted if the proportion of nonconforming items exceeds 16%. Use Table 1.
What is the probability that the production machines will be adjusted if the batch consists of 56 items?(Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)
What is the probability that the production machines will be adjusted if the batch consists of 112 items?(Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)
Acceptance sampling is an important quality control technique, where a batch of data is tested to determine if the proportion of units having a particular attribute exceeds a given percentage. Suppose that 12% of produced items are known to be nonconforming. Every week a batch of items is evaluated and the production machines are adjusted if the proportion of nonconforming items exceeds 16%. Use Table 1.
a.What is the probability that the production machines will be adjusted if the batch consists of 56 items?(Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)
Probability b.What is the probability that the production machines will be adjusted if the batch consists of 112 items?(Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)
ProbabilityExplanation / Answer
a) mean=np=56*0.12=6.72
sd=sqrt(npq)=2.4317
P(x>16.5) because we use continuity correction factor.
=1-NORMDIST(16.5,6.72,2.4317,TRUE)=1-0.9997=0
b) mean=112*0.12=13.44
sd=sqrt(npq)=3.4391
P(X>16.5)=1-NORMDIST(16.5,13.44,3.4391,TRUE)=1-0.8132=0.1868