FE Industrial & Systems Practice Question Solution included after investigation
CASE E-090 Β· FE

The 5% Defect Lot That Was Accepted 73.6% of the Time, Not 95%

A 5% defective process did not imply a 95% lot-acceptance probability. Under an n=20, c=1 plan, acceptance required observing zero or one defective item.

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Case E-090 engineering failure visual
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2

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3

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4

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5

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Case Brief

SAME PRINCIPLES. HIGHER STANDARDS.
PROBLEM STATEMENT

A single-sampling inspection plan selects n=20 items from a large lot and accepts the lot if at most c=1 defective item is found. Assume independent items and a true defective fraction p=0.05. A note states that the lot will be accepted 95% of the time because 95% of items are good. Determine the actual lot-acceptance probability using the binomial model.

Quality Industrial Systems Design Verification Root Cause Analysis Safe Correction
E-090
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CASE CLOSED Β· ENGINEERING VERDICT

What actually failed β€” and what should change.

What failed

Single-item conformance probability was reported as lot acceptance probability.

What was missed

The sampling plan accepts only particular binomial sample-count outcomes.

Corrected model

Sum binomial probabilities from zero defectives through the acceptance number c.

QUICK ANSWER

For n=20, c=1, and p=0.05, P(accept)=P(X≀1)=0.95^20+20(0.05)(0.95^19)=0.73584, or about 73.58%.

ENGINEERING TAKEAWAY

For a binomial single-sampling plan with sample size n and acceptance number c, calculate P(accept)=Ξ£ from x=0 to c of C(n,x)p^x(1-p)^(n-x).

CASE QUESTIONS

Questions this case should settle.

How is acceptance probability calculated for a single-sampling plan?

Sum the binomial probabilities from zero defectives through the acceptance number c.

What happens if the acceptance number is reduced from 1 to 0?

Acceptance probability decreases because only samples with zero defectives pass.

NEXT FE PRACTICE QUESTION

The Part That Lost $45, Not $60

Quality Β· 5–10 min

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