Question 3 of 6: Process Capability — Non-normal Data, $C_p/C_{pk}/C_{pm}$, and Fraction Nonconforming

Question 3: Process Capability — Non-normal Data, $C_p/C_{pk}/C_{pm}$, 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) Capability assessment for a non-normal characteristic

The ordinary $C_p/C_{pk}$ indexes assume normality and are unreliable (often badly optimistic) when the data are skewed or heavy-tailed. Three standard approaches: (1) fit a transformation (e.g. Box-Cox or Johnson family) that makes the transformed data approximately normal, then compute $C_p/C_{pk}$ on the transformed scale and transform the spec limits accordingly; (2) fit the actual (non-normal) distribution to the data (e.g. Weibull, lognormal) and define "percentile-based" capability indexes that replace $\mu\pm3\sigma$ with the distribution's own $0.135$ and $99.865$ percentiles, e.g. $C_p=(USL-LSL)/(x_{0.99865}-x_{0.00135})$; (3) use a nonparametric/distribution-free tolerance interval based on order statistics when no parametric family fits well. All three replace the normal-theory $\pm3\sigma$ spread with a spread estimate appropriate to the data's actual shape.

(b) Statistical control as a precondition; $C_p$ vs. $C_{pk}$ vs. $C_{pm}$; their relation

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 the sample statistics ($\bar x$, $s$) do not estimate fixed, meaningful parameters and any index computed from them is not predictive of future output — it only describes the specific (unstable) sample taken.

$C_p=(USL-LSL)/6\sigma$ measures potential capability — how the process spread compares with the tolerance band, ignoring where the process is centred. $C_{pk}=\min\!\big[(USL-\mu)/3\sigma,\ (\mu-LSL)/3\sigma\big]$ measures actual capability by using the distance from the mean to the nearer spec limit, so it penalizes an off-centre process. $C_{pm}=(USL-LSL)/\big(6\sqrt{\sigma^2+(\mu-T)^2}\big)$ (the Taguchi capability index) further penalizes deviation of the mean from the target $T$ (which need not be the midpoint of the spec limits), directly reflecting the quadratic loss-function philosophy from Q1(a) rather than just the spec-limit boundary.

Relation: $C_{pk}\le C_p$ always, with equality if and only if the process is perfectly centred ($\mu=(USL+LSL)/2$); the gap $C_p-C_{pk}$ quantifies how far off-centre the process is, in units of $C_p$.

(c) Fraction nonconforming from $C_p=1.20$, $C_{pk}=1.15$

Given. $C_p=1.20$, $C_{pk}=1.15$; normal distribution; two-sided spec (LSL, USL unspecified numerically — solved in units of $\sigma$).

Find. $P(X\lt LSL)$ and $P(X\gt USL)$ separately.

Approach. Let $M=(USL+LSL)/2$ and $d=(USL-LSL)/2$, so $C_p=d/3\sigma$. $C_{pk}$ uses the distance to the nearer limit, $C_{pk}=(d-|\mu-M|)/3\sigma=C_p-|\mu-M|/3\sigma$. This gives the two spec limits' distances from $\mu$ directly in units of $\sigma$, without needing the actual numeric LSL/USL.

1. Offset from centre. $$\frac{|\mu-M|}{3\sigma}=C_p-C_{pk}=1.20-1.15=0.05$$
2. Distance (in $\sigma$) to the nearer limit (the one $C_{pk}$ is bound by). $$Z_{\text{near}}=3\,C_{pk}=3(1.15)=3.45$$
3. Distance (in $\sigma$) to the farther limit. The two limits are $d$ apart from $M$ on either side and $\mu$ is $0.05(3\sigma)=0.15\sigma$ off-centre, so the far side is $2C_p-C_{pk}$ in units of $C_p$-equivalent $Z$: $$Z_{\text{far}}=3(2C_p-C_{pk})=3(2(1.20)-1.15)=3(1.25)=3.75$$
4. Fraction nonconforming. $$P(\text{beyond nearer limit})=1-\Phi(3.45)=0.00028029$$ $$P(\text{beyond farther limit})=1-\Phi(3.75)=0.00008842$$ $$\boxed{P_{\text{near limit}}=0.0280\%,\quad P_{\text{far limit}}=0.0088\%}$$