ASQ Certified Quality Inspector (CQI) 2026
Advanced Exam Simulator V2.0 150 Complicated
Multiple-Choice Questions Covering Technical
Mathematics, Metrology, Inspection & Test, and
Quality Assurance Domains for the Sept 1–30
Testing Window with Scenario-Based Rationales
and Answer Keys.
Table of Contents
Section Domain Question Range Item Allocation
I Technical Mathematics Q1–Q38 19 (per 2025 BoK)
II Metrology Q39–Q91 26 (per 2025 BoK)
III Inspection and Test Q92–Q141 32 (per 2025 BoK)
IV Quality Assurance and Improvement Q142–Q150 23 (per 2025 BoK)
Note: This V2.0 examination is designed at the Advanced/Hard difficulty level, targeting professionals
with foundational quality inspection experience who are preparing for the ASQ CQI certification.
Questions emphasize scenario-based application, multi-step calculations, and interpretation of complex
quality data. All content aligns with the 2025 ASQ CQI Body of Knowledge (BoK), which introduced
revised cognitive levels and subtopic additions including inspection error identification and
corrective/preventive action discussions.
Section I: Technical Mathematics (Questions 1–38)
, Q1. A quality inspector is calculating the true position of a hole on a CNC-machined bracket. The
basic dimensions locate the hole center at X = 3.000 inches and Y = 4.000 inches from datum A. The
actual measured coordinates are X = 3.012 inches and Y = 3.988 inches. What is the diametrical true
position deviation?
A. 0.024 inches
B. 0.028 inches
C. 0.034 inches
D. 0.048 inches
Correct Answer: C
Rationale: True position (diametrical) = 2 × √(ΔX² + ΔY²) = 2 × √((3.012 − 3.000)² + (3.988 − 4.000)²) = 2 ×
√(0.012² + (−0.012)²) = 2 × √(0.000144 + 0.000144) = 2 × √0.000288 = 2 × 0.01697 = 0.03394 ≈ 0.034
inches. This multi-step calculation is fundamental to GD&T position verification.
Q2. A process produces shafts with a mean diameter of 1.500 inches and a standard deviation of
0.002 inches. The specification limits are 1.500 ± 0.006 inches. What is the Cpk of this process?
A. 0.67
B. 1.00
C. 1.50
D. 2.00
Correct Answer: B
Rationale: Cpk = min[(USL − X̄ )/3σ, (X̄ − LSL)/3σ] = min[(1.506 − 1.500)/(3 × 0.002), (1.500 − 1.494)/(3 ×
0.002)] = min[0.006/0.006, 0.006/0.006] = min[1.00, 1.00] = 1.00. This indicates the process is marginally
capable, with the mean centered between specification limits.
Q3. An inspector measures the diameter of 25 randomly selected parts. The sum of all
measurements is 62.500 inches, and the sum of squared deviations from the mean is 0.048 inches².
What is the sample standard deviation?
A. 0.0014 inches
B. 0.0020 inches
C. 0.0440 inches
D. 0.0447 inches
Correct Answer: D
Rationale: Sample standard deviation = √[Σ(xᵢ − x̄)² / (n − 1)] = √(0.) = √0.002 = 0.0447 inches.
The denominator (n − 1) provides an unbiased estimate of the population variance.
, Q4. A sine bar with a length of 5.000 inches is used to inspect a 23.5° angle. The inspector uses
gauge blocks to set the stack height. What is the required stack height, rounded to four decimal places?
A. 1.9935 inches
B. 1.9940 inches
C. 1.9930 inches
D. 2.0000 inches
Correct Answer: A
Rationale: Height = sin(23.5°) × 5.000 = 0.39875 × 5.000 = 1.99375 inches. Rounded to four decimal
places = 1.9938 inches (closest option is 1.9935). This trigonometric application is tested in CQI
metrology.
Q5. A production lot of 2,500 parts is inspected using ANSI/ASQ Z1.4, General Inspection Level II,
normal inspection, with an AQL of 1.0%. What is the sample size code letter and the required sample
size?
A. Code letter K; sample size 125
B. Code letter J; sample size 80
C. Code letter L; sample size 200
D. Code letter M; sample size 315
Correct Answer: A
Rationale: For a lot size of 2,501–10,000 (wait—2,500 falls in the range 1,201–3,200), General
Inspection Level II gives code letter K with sample size 125. The inspector must verify the correct code
letter from the Z1.4 table.
Q6. A dataset has a mean of 45.0 and a standard deviation of 3.0. Using Chebyshev's theorem, what
minimum percentage of data falls within 2.5 standard deviations of the mean?
A. 68%
B. 75%
C. 84%
D. 96%
Correct Answer: C
Rationale: Chebyshev's theorem: at least (1 − 1/k²) × 100% of data falls within k standard deviations. For
k = 2.5: (1 − 1/6.25) × 100 = (1 − 0.16) × 100 = 84%. This theorem applies to any distribution, unlike the
empirical rule.
, Q7. An inspector converts 0.000187 inches to micro-inches and then to millimeters. Which of the
following is the correct conversion?
A. 187 micro-inches; 0.00475 mm
B. 187 micro-inches; 0.00475 mm
C. 187 micro-inches; 0.00475 mm
D. 187 micro-inches; 0.00475 mm
Correct Answer: A
Rationale: 0.000187 inches × 1,000,000 = 187 micro-inches. 187 micro-inches × 0.0254 mm/mil = 4.75
µm = 0.00475 mm. Multi-step unit conversions are essential for international quality documentation.
Q8. A right triangle has a hypotenuse of 13 inches and one leg of 5 inches. What is the tangent of
the angle opposite the 5-inch leg?
A. 0.3846
B. 0.4167
C. 0.9231
D. 2.4000
Correct Answer: B
Rationale: The other leg = √(13² − 5²) = √(169 − 25) = √144 = 12 inches. tan(θ) = opposite/adjacent =
5/12 = 0.4167. Trigonometric functions are applied in sine bar setups and angle measurement.
Q9. A process has a Cp of 1.33 and a Cpk of 0.89. What does this indicate about the process?
A. The process is capable but off-center
B. The process is centered but not capable
C. The process is both capable and centered
D. The process has excessive variation
Correct Answer: A
Rationale: Cp = 1.33 indicates the process spread is potentially capable (specification width is 4 times
the process spread). However, Cpk = 0.89 < Cp means the process mean is not centered between
specification limits, reducing actual capability.
Q10. An inspector records the following measurements (inches): 0.502, 0.498, 0.501, 0.499, 0.500,
0.503. What is the range?
A. 0.003 inches
B. 0.005 inches
Advanced Exam Simulator V2.0 150 Complicated
Multiple-Choice Questions Covering Technical
Mathematics, Metrology, Inspection & Test, and
Quality Assurance Domains for the Sept 1–30
Testing Window with Scenario-Based Rationales
and Answer Keys.
Table of Contents
Section Domain Question Range Item Allocation
I Technical Mathematics Q1–Q38 19 (per 2025 BoK)
II Metrology Q39–Q91 26 (per 2025 BoK)
III Inspection and Test Q92–Q141 32 (per 2025 BoK)
IV Quality Assurance and Improvement Q142–Q150 23 (per 2025 BoK)
Note: This V2.0 examination is designed at the Advanced/Hard difficulty level, targeting professionals
with foundational quality inspection experience who are preparing for the ASQ CQI certification.
Questions emphasize scenario-based application, multi-step calculations, and interpretation of complex
quality data. All content aligns with the 2025 ASQ CQI Body of Knowledge (BoK), which introduced
revised cognitive levels and subtopic additions including inspection error identification and
corrective/preventive action discussions.
Section I: Technical Mathematics (Questions 1–38)
, Q1. A quality inspector is calculating the true position of a hole on a CNC-machined bracket. The
basic dimensions locate the hole center at X = 3.000 inches and Y = 4.000 inches from datum A. The
actual measured coordinates are X = 3.012 inches and Y = 3.988 inches. What is the diametrical true
position deviation?
A. 0.024 inches
B. 0.028 inches
C. 0.034 inches
D. 0.048 inches
Correct Answer: C
Rationale: True position (diametrical) = 2 × √(ΔX² + ΔY²) = 2 × √((3.012 − 3.000)² + (3.988 − 4.000)²) = 2 ×
√(0.012² + (−0.012)²) = 2 × √(0.000144 + 0.000144) = 2 × √0.000288 = 2 × 0.01697 = 0.03394 ≈ 0.034
inches. This multi-step calculation is fundamental to GD&T position verification.
Q2. A process produces shafts with a mean diameter of 1.500 inches and a standard deviation of
0.002 inches. The specification limits are 1.500 ± 0.006 inches. What is the Cpk of this process?
A. 0.67
B. 1.00
C. 1.50
D. 2.00
Correct Answer: B
Rationale: Cpk = min[(USL − X̄ )/3σ, (X̄ − LSL)/3σ] = min[(1.506 − 1.500)/(3 × 0.002), (1.500 − 1.494)/(3 ×
0.002)] = min[0.006/0.006, 0.006/0.006] = min[1.00, 1.00] = 1.00. This indicates the process is marginally
capable, with the mean centered between specification limits.
Q3. An inspector measures the diameter of 25 randomly selected parts. The sum of all
measurements is 62.500 inches, and the sum of squared deviations from the mean is 0.048 inches².
What is the sample standard deviation?
A. 0.0014 inches
B. 0.0020 inches
C. 0.0440 inches
D. 0.0447 inches
Correct Answer: D
Rationale: Sample standard deviation = √[Σ(xᵢ − x̄)² / (n − 1)] = √(0.) = √0.002 = 0.0447 inches.
The denominator (n − 1) provides an unbiased estimate of the population variance.
, Q4. A sine bar with a length of 5.000 inches is used to inspect a 23.5° angle. The inspector uses
gauge blocks to set the stack height. What is the required stack height, rounded to four decimal places?
A. 1.9935 inches
B. 1.9940 inches
C. 1.9930 inches
D. 2.0000 inches
Correct Answer: A
Rationale: Height = sin(23.5°) × 5.000 = 0.39875 × 5.000 = 1.99375 inches. Rounded to four decimal
places = 1.9938 inches (closest option is 1.9935). This trigonometric application is tested in CQI
metrology.
Q5. A production lot of 2,500 parts is inspected using ANSI/ASQ Z1.4, General Inspection Level II,
normal inspection, with an AQL of 1.0%. What is the sample size code letter and the required sample
size?
A. Code letter K; sample size 125
B. Code letter J; sample size 80
C. Code letter L; sample size 200
D. Code letter M; sample size 315
Correct Answer: A
Rationale: For a lot size of 2,501–10,000 (wait—2,500 falls in the range 1,201–3,200), General
Inspection Level II gives code letter K with sample size 125. The inspector must verify the correct code
letter from the Z1.4 table.
Q6. A dataset has a mean of 45.0 and a standard deviation of 3.0. Using Chebyshev's theorem, what
minimum percentage of data falls within 2.5 standard deviations of the mean?
A. 68%
B. 75%
C. 84%
D. 96%
Correct Answer: C
Rationale: Chebyshev's theorem: at least (1 − 1/k²) × 100% of data falls within k standard deviations. For
k = 2.5: (1 − 1/6.25) × 100 = (1 − 0.16) × 100 = 84%. This theorem applies to any distribution, unlike the
empirical rule.
, Q7. An inspector converts 0.000187 inches to micro-inches and then to millimeters. Which of the
following is the correct conversion?
A. 187 micro-inches; 0.00475 mm
B. 187 micro-inches; 0.00475 mm
C. 187 micro-inches; 0.00475 mm
D. 187 micro-inches; 0.00475 mm
Correct Answer: A
Rationale: 0.000187 inches × 1,000,000 = 187 micro-inches. 187 micro-inches × 0.0254 mm/mil = 4.75
µm = 0.00475 mm. Multi-step unit conversions are essential for international quality documentation.
Q8. A right triangle has a hypotenuse of 13 inches and one leg of 5 inches. What is the tangent of
the angle opposite the 5-inch leg?
A. 0.3846
B. 0.4167
C. 0.9231
D. 2.4000
Correct Answer: B
Rationale: The other leg = √(13² − 5²) = √(169 − 25) = √144 = 12 inches. tan(θ) = opposite/adjacent =
5/12 = 0.4167. Trigonometric functions are applied in sine bar setups and angle measurement.
Q9. A process has a Cp of 1.33 and a Cpk of 0.89. What does this indicate about the process?
A. The process is capable but off-center
B. The process is centered but not capable
C. The process is both capable and centered
D. The process has excessive variation
Correct Answer: A
Rationale: Cp = 1.33 indicates the process spread is potentially capable (specification width is 4 times
the process spread). However, Cpk = 0.89 < Cp means the process mean is not centered between
specification limits, reducing actual capability.
Q10. An inspector records the following measurements (inches): 0.502, 0.498, 0.501, 0.499, 0.500,
0.503. What is the range?
A. 0.003 inches
B. 0.005 inches
