Fabric width is stable but off-centre. Specifications are 148–152 cm, process mean is 151 cm and σ is 0.5 cm. What is Cpk?
A 0.33
B 2.00
C 0.67
D 1.33
Answer: C. 0.67. Cpk takes the smaller of the distances from the mean to the two limits, divided by 3σ. The upper distance is 1 cm.
Of 20 rolls in a lot, four are nonconforming. Two are sampled without replacement and c = 0. What is the acceptance probability?
A 3/5
B 4/19
C 16/25
D 12/19
Answer: D. 12/19. The two conforming draws have probabilities 16/20 and 15/19. Their product is 12/19.
In a historical baseline, 4% of units are nonconforming. For sample size 100, estimate the three-sigma upper p-chart limit.
A 4.0%
B 12.0%
C 9.9%
D 6.0%
Answer: C. 9.9%. The limit is 0.04 + 3√(0.04 × 0.96/100), or about 0.099.
Independent measurement variance is 0.09 cm² and observed total variance is 0.25 cm². What process variance is estimated?
A 0.34 cm²
B 0.09 cm²
C 0.40 cm²
D 0.16 cm²
Answer: D. 0.16 cm². Under independence, total variance is the sum of measurement and process components. Subtract 0.09 from 0.25.
Uniform inspection units have an average of 4 defects. Find the upper three-sigma c-chart limit using the Poisson model.
A 10 defects
B 8 defects
C 4 defects
D 16 defects
Answer: A. 10 defects. The centre is 4 and the standard deviation is √4 = 2, giving a limit of 4 + 3 × 2.
One 400 m² lot scores 96 points and another 150 m² lot scores 51 points. The maximum is 30 points per 100 m². Which lot exceeds it?
A Only the 400 m² lot
B Both lots
C Only the 150 m² lot
D Neither lot
Answer: C. Only the 150 m² lot. The rates are 24 and 34 points per 100 m², so the smaller-area lot exceeds the stated limit.
When a stable process mean sits exactly midway between both specification limits, how are Cp and Cpk related?
A Cp is smaller than Cpk
B Cpk equals Cp
C Cpk is always zero
D Cpk is twice Cp
Answer: B. Cpk equals Cp. The mean is centred between limits, so the nearest distance is exactly half the specification width.
Acceptance sampling passes a lot at the designated poor quality level. What is the probability of this event called?
A Producer’s risk
B Repeatability
C Consumer’s risk
D Appraisal cost
Answer: C. Consumer’s risk. Consumer’s risk is the probability of accepting a lot at the designated poor quality level.
To monitor fabric faults, a u chart uses a baseline of 0.10 defects/m². For an inspected area of 90 m², what is the three-sigma upper limit per m²?
A 0.40 defects/m²
B 0.13 defects/m²
C 0.10 defects/m²
D 0.20 defects/m²
Answer: D. 0.20 defects/m². The bound is u-bar + 3√(u-bar/area) = 0.10 + 3√(0.10/90) = 0.20.
Twelve out of 150 pieces failed in one shift and eight out of 50 failed in another. What pooled failure fraction applies across all pieces?
A 8%
B 16%
C 10%
D 12%
Answer: C. 10%. Combine 20 failures and 200 inspected pieces. An unweighted average of the two shift rates would be inappropriate.
Assume independent draws from a large lot where p = 0.10. With sample size five and c = 0, what is the acceptance probability?
A 0.59049
B 0.50000
C 0.40951
D 0.10000
Answer: A. 0.59049. All five items must conform. Multiply 0.90 by itself five times.
Hold specifications and mean fixed while reducing process σ by half. How does Cp change?
A It doubles
B It halves
C It stays constant
D It becomes negative
Answer: A. It doubles. Capability width Cp is inversely proportional to σ for fixed specification width.
Suppose fabric weight is normally distributed around 150 g/m² with σ = 3 g/m². Approximately what fraction lies within 144–156 g/m²?
A 50%
B 95%
C 68%
D 99.7%
Answer: B. 95%. The limits lie two standard deviations on either side of the mean. Approximately 95% of a normal distribution lies there.
For 200 inspected units with p-bar = 0.03, find the expected np-chart centre line.
A 60 units
B 0.03 units
C 3 units
D 6 units
Answer: D. 6 units. The centre line for the number nonconforming is n × p-bar = 200 × 0.03.
After maintenance, twenty successive roll means trend upward without crossing the control limits. What response is justified?
A Investigate the sustained pattern
B Ignore every in-limit pattern
C Raise all product limits
D Discard the whole lot automatically
Answer: A. Investigate the sustained pattern. A sustained nonrandom trend can indicate a systematic change even when no individual point lies beyond its control limit.
Counting 8 defects in 80 m² and 12 in 200 m², what pooled defect rate belongs at the centre of a u chart based on these data?
A 0.1000 defects/m²
B 0.0800 defects/m²
C 0.0714 defects/m²
D 0.0600 defects/m²
Answer: C. 0.0714 defects/m². Divide 20 defects by the combined 280 m². A pooled rate weights each area correctly.
On an operating characteristic curve, a relatively good lot is rejected by the sampling plan. Whose risk does this represent?
A Producer’s risk
B Consumer’s risk
C Both at every quality
D Calibration bias
Answer: A. Producer’s risk. Producer’s risk is the probability of rejecting a lot at the designated good quality level.
Several observations cross a control limit, but all meet product specifications. What response is justified?
A Replace specifications with limits
B Investigate process instability
C No investigation is warranted
D Accept that process is stable
Answer: B. Investigate process instability. Control limits flag potential special causes even while individual units meet product specifications.
For equal repeated checks, one inspector has a very consistent 1 cm high reading while others match the standard. What should be corrected first?
A Every product tolerance
B The inspector’s measurement bias
C Total lot size only
D The dye fixation time
Answer: B. The inspector’s measurement bias. Consistent deviation from a reference indicates bias. Repeating a biased measurement does not remove the offset.
For a large lot with p = 0.10, a two-item sample is accepted with up to one nonconforming item. What is acceptance probability under independent draws?
A 0.90
B 0.81
C 0.99
D 0.19
Answer: C. 0.99. The lot is rejected only if both sampled items are nonconforming. That event has probability 0.10².
Subgroups use a known within-process σ of 3 g/m² and contain nine observations. What is the three-sigma upper limit for X-bar when process mean is 150 g/m²?
A 159 g/m²
B 153 g/m²
C 150.33 g/m²
D 151 g/m²
Answer: B. 153 g/m². The subgroup mean’s standard error is 3/√9 = 1 g/m². Add three times this to the mean.
Inspection flags 6 defects per 1,000 garments, and four separate defect opportunities are defined per garment. Find defects per million opportunities.
A 6,000
B 6
C 24,000
D 1,500
Answer: D. 1,500. There are 4,000 opportunities. Divide 6 by 4,000 and multiply by one million.
If baseline fraction is 0.05 and sample size is quadrupled, how does p-chart standard error change?
A It falls to one quarter
B It stays the same
C It doubles
D It halves
Answer: D. It halves. The standard error is proportional to 1/√n. Quadrupling n halves it.
Inspection effort rises from 200 to 300 hours. Internal failure cost falls by ₹60,000, and each additional inspection hour costs ₹400. What is the net reduction in combined costs?
A ₹20,000
B ₹40,000
C ₹100,000
D ₹60,000
Answer: A. ₹20,000. The 100 added hours cost ₹40,000. Subtract that increase from ₹60,000 of avoided failure cost.
The upper and lower limits for fabric weight are 170 and 150 g/m². Within-process σ is 2.5 g/m². Estimate Cp.
A 0.75
B 1.33
C 8.00
D 2.67
Answer: B. 1.33. Cp is the 20 g/m² specification width divided by 6 × 2.5 g/m².