Free sample questions

6 questions from this set, shown in full with the correct answer and the reasoning behind it. Read them as a study resource.

1. Q1.A team has delivered its results, handed the control plan to the process owner, and is capturing lessons learned before disbanding. Which stage is this?

   • A.Adjourning✓ Correct answer
   • B.Performing
   • C.Norming
   • D.Storming

   Explanation

   Adjourning covers closure: transferring ownership, capturing learning and recognising the members' contribution before the team dissolves. Performing is the primary foil: the team was performing until the deliverable was complete, and the closure activities are what mark the transition out of it.

   Reference: CSSBB BoK III

2. Q2.A survey of 200 customers on a 0 to 10 recommendation scale returns 120 scores of 9 or 10, 50 scores of 7 or 8, and 30 scores of 6 or below.
   
   What is the net promoter score?

   • A.60
   • B.75
   • C.90
   • D.45✓ Correct answer

   Explanation

   Net promoter score subtracts the detractor percentage from the promoter percentage and ignores passives entirely:
   
   $$NPS = \frac{120}{200} \times 100 - \frac{30}{200} \times 100 = 60 - 15 = 45$$
   
   60 is the primary foil: it is the promoter percentage on its own, which is the most common error because the passive group is discarded and invites the reader to stop there.

   Reference: CSSBB BoK II

3. Q3.A plant runs three shifts with known performance differences and wants an overall defect rate with each shift properly represented. Which method fits?

   • A.Cluster sampling
   • B.Convenience sampling
   • C.Stratified sampling✓ Correct answer
   • D.Simple random sampling

   Explanation

   Stratification guarantees representation of each known subgroup and reduces the variance of the overall estimate. Cluster sampling is the primary foil: it is a legitimate probability method and is cheaper, but selecting a single shift makes the estimate hostage to which shift was drawn, which is exactly the risk the known differences create.

   Reference: CSSBB BoK V

4. Q4.A gage study estimates repeatability at $\sigma_e = 0.30$ and reproducibility at $\sigma_o = 0.40$.
   
   What is the combined gage R and R standard deviation?

   • A.0.50✓ Correct answer
   • B.0.70
   • C.0.35
   • D.0.12

   Explanation

   The two components combine as variances, not as standard deviations:
   
   $$\sigma_{GRR} = \sqrt{0.30^2 + 0.40^2} = \sqrt{0.25} = 0.50$$
   
   0.70 is the primary foil: it adds the standard deviations directly, which always overstates the combined variation.

   Reference: CSSBB BoK V

5. Q5.A Weibull analysis of field failures returns a shape parameter of $\beta = 0.6$. What does this indicate?

   • A.An increasing failure rate, typical of wear-out
   • B.A poor fit, so a normal distribution should be used
   • C.A decreasing failure rate, typical of infant mortality✓ Correct answer
   • D.A constant failure rate, typical of random failures

   Explanation

   The Weibull shape parameter reads directly as failure rate behaviour: $\beta < 1$ decreasing, $\beta = 1$ constant and equivalent to the exponential, $\beta > 1$ increasing. A value of 0.6 therefore points at burn-in, manufacturing escapes or installation damage, not at wear-out, so extending preventive replacement would make matters worse.

   Reference: CSSBB BoK V

6. Q6.A process has $USL = 110$, $LSL = 90$, a mean of 104 and $\sigma = 2$.
   
   What are $C_p$ and $C_{pk}$?

   • A.$C_p = 1.67$ and $C_{pk} = 2.33$
   • B.$C_p = 1.67$ and $C_{pk} = 1.67$
   • C.$C_p = 1.67$ and $C_{pk} = 1.00$✓ Correct answer
   • D.$C_p = 1.00$ and $C_{pk} = 1.67$

   Explanation

   Compute the potential index first, then penalise for the off-centre mean:
   
   $$C_p = \frac{110-90}{6(2)} = 1.67 \qquad C_{pk} = \min\left(\frac{110-104}{6}, \frac{104-90}{6}\right) = 1.00$$
   
   The gap between the two is the cost of being off centre: simply recentring this process would lift $C_{pk}$ to 1.67 without reducing variation at all.

   Reference: CSSBB BoK V

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