Question 3 of 6: Supplier's/Producer's Risk, $C_p/C_{pk}/C_{pm}$, and Fraction Nonconforming

Question 3: Supplier's/Producer's Risk, $C_p/C_{pk}/C_{pm}$, and Fraction Nonconforming (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Supplier's and producer's risk

In an acceptance-sampling/capability context, the producer's risk ($\alpha$) is the probability that a genuinely good lot or process (quality at or better than the Acceptable Quality Level) is nevertheless rejected by the sampling plan/decision rule — a Type I error from the producer's point of view, since a good lot is wrongly penalized. The supplier's (consumer's) risk ($\beta$) is the probability that a genuinely bad lot or process (quality at or worse than the Rejectable/Lot-Tolerance Quality Level) is nevertheless accepted — a Type II error, since the consumer wrongly receives poor-quality product. Both risks are read as two points on the same operating-characteristic (OC) curve of the sampling plan or decision rule: $\alpha=1-P_a(AQL)$ and $\beta=P_a(RQL)$, where $P_a(p)$ is the probability of acceptance as a function of true quality $p$.

(b) Statistical control precondition; $C_p$ vs. $C_{pk}$; relation to $C_{pm}$; non-normal capability

Yes, the process should be in statistical control before a capability analysis is performed. A capability index is a statement about what the process will continue to produce, projected from the sample data; if the process is not in control its mean and/or variance are not stable, so $\bar x$ and $s$ do not estimate fixed, meaningful parameters and any index computed from them only describes the specific unstable sample taken — it is not predictive of future output.

$C_p=(USL-LSL)/6\sigma$ measures potential capability — how the process spread compares with the tolerance band, ignoring where the process is centred; it can be high even for a badly off-target process. $C_{pk}=\min[(USL-\mu)/3\sigma,\ (\mu-LSL)/3\sigma]$ measures actual capability using the distance from the mean to the nearer spec limit, so it penalizes an off-centre process; $C_{pk}\le C_p$ always, with equality iff the process is perfectly centred at $(USL+LSL)/2$.

Relation between $C_p$ and $C_{pm}$. $C_{pm}=(USL-LSL)/\big(6\sqrt{\sigma^2+(\mu-T)^2}\big)$ (the Taguchi capability index) replaces $\sigma$ in $C_p$'s denominator with the root-mean-square deviation from the target $T$ (not necessarily the spec midpoint), so $C_{pm}=C_p/\sqrt{1+\big[(\mu-T)/\sigma\big]^2}$ — $C_{pm}\le C_p$ always, with equality iff $\mu=T$, and $C_{pm}$ falls off faster than $C_{pk}$ as the process drifts because it penalizes distance from the target directly (a quadratic, Taguchi-loss-consistent penalty) rather than only the distance to the nearer tolerance boundary.

Non-normal capability. As in Q1(a), the standard indices assume normality and are unreliable for skewed/heavy-tailed data; the usual fixes are (1) fit a normalizing transformation (Box-Cox, Johnson) and compute $C_p/C_{pk}$ on the transformed scale; (2) fit the characteristic's actual distribution and define percentile-based indices that replace $\mu\pm3\sigma$ with the fitted distribution's own 0.135 and 99.865 percentiles, e.g. $C_p=(USL-LSL)/(x_{0.99865}-x_{0.00135})$; or (3) use a nonparametric/distribution-free tolerance interval from order statistics when no parametric family fits well.

(c) Fraction nonconforming, $C_{pm}$, and the centered-process comparison

Given. $C_p=1.33,\ C_{pk}=1.05$; normal distribution; two-sided spec (LSL, USL unspecified numerically — solved in units of $\sigma$); target $T=(USL+LSL)/2$.

Find. $P(X\lt LSL)+P(X\gt USL)$; $C_{pm}$; how the total fraction nonconforming changes if the process is re-centered at $T$.

Approach. Let $d=(USL-LSL)/2$ so $C_p=d/3\sigma$, and $C_{pk}$ uses the distance to the nearer limit: $C_p-C_{pk}=|\mu-T|/3\sigma$. This fixes both spec limits' distances from $\mu$ in units of $\sigma$ without needing the numeric LSL/USL.

1. Distance (in $\sigma$) to the nearer limit. $$Z_{\text{near}}=3\,C_{pk}=3(1.05)=3.15$$
2. Distance (in $\sigma$) to the farther limit. The half-width is $3C_p=3.99\sigma$ on each side of $T$, and $\mu$ sits $3(C_p-C_{pk})=0.84\sigma$ off-centre toward the near limit, so $$Z_{\text{far}}=3(2C_p-C_{pk})=3(2(1.33)-1.05)=4.83$$
3. Fraction nonconforming. $$P_{\text{near}}=1-\Phi(3.15)=0.0816\%,\qquad P_{\text{far}}=1-\Phi(4.83)=0.0000683\%$$ $$\boxed{P(\text{nonconforming})=P_{\text{near}}+P_{\text{far}}=0.0817\%}$$
4. $C_{pm}$. The offset from target is $(\mu-T)/\sigma=3C_p-Z_{\text{near}}=3.99-3.15=0.84$, so $$C_{pm}=\frac{C_p}{\sqrt{1+\big[(\mu-T)/\sigma\big]^2}}=\frac{1.33}{\sqrt{1+0.84^2}}$$ $$\boxed{C_{pm}=1.018}$$
5. Centered-process comparison. If the mean were moved onto target ($\mu=T$), both limits would sit at the full half-width $Z=3C_p=3.99\sigma$ from the mean, giving $$P_{\text{centered}}=2\big[1-\Phi(3.99)\big]=0.00661\%$$ $$\boxed{\text{Centering the process would cut the nonconforming fraction from }0.0817\%\text{ to }0.00661\%\text{ (a}\approx12\times\text{ reduction)}}$$