Question 4 of 6: Stable/Capable Processes, Capability Indices, and Fraction Nonconforming from $C_p$/$C_{pk}$

Question 4: Stable/Capable Processes, Capability Indices, and Fraction Nonconforming from $C_p$/$C_{pk}$ (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) Stable vs. capable processes; capability analysis for a non-normal characteristic

A stable process is one that is in statistical control — only common-cause variation is present, so its output distribution (mean, spread, shape) is not changing over time and future behaviour can be predicted from past data. A capable process is a separate property: it is one whose stable, natural spread comfortably fits within the specification limits, so that (nearly) all output meets requirements. A process can be stable but not capable (perfectly consistent, but consistently producing too much scrap because its natural spread is wider than the tolerance), and, less usefully, can appear momentarily "capable" while still being unstable — which is not a trustworthy state, because an unstable process's spread is not predictable and today's apparent capability offers no guarantee about tomorrow's.

When the quality characteristic is not normally distributed, the standard $C_p/C_{pk}$ formulas (which assume the process spread is well described by $\pm3\sigma$ around the mean) are not valid, because the true percentage beyond a $\pm3\sigma$-based limit for a skewed distribution can be very different from the normal-theory $0.27\%$. Capability analysis instead uses a percentile-based approach: fit the actual (non-normal) distribution to the data — either a specific known family (e.g., Weibull for life data, lognormal for a positively-skewed dimension) or a flexible family such as the Johnson or Pearson system that can match the sample's skewness and kurtosis — and replace the assumed-symmetric $6\sigma$ spread with the empirical or fitted spread between the $0.135^{\text{th}}$ and $99.865^{\text{th}}$ percentiles (the same tail area a normal $\pm3\sigma$ would enclose): $C_p=(USL-LSL)/(X_{0.99865}-X_{0.00135})$, with a matching one-sided version for $C_{pk}$ measured from the median or mean to each limit. A simple non-parametric alternative (Clements' method) uses the empirical percentiles of the raw data directly, without assuming any distributional family at all.

(b) Control as a prerequisite for capability analysis; $C_p$, $C_{pk}$, $C_{pm}$; the $C_p$–$C_{pk}$ relation

Yes, the process should be in statistical control before a capability analysis is trusted. Capability indices are meant to describe the process's inherent, repeatable spread and centring so that the resulting fraction-nonconforming estimate predicts future performance. If the process is not in control, the sample mean and standard deviation used to compute $C_p/C_{pk}$ are contaminated by special-cause variation (a transient shift, an outlier, a trend) that will not necessarily recur in the same way, so the resulting index describes only what happened during that particular unstable window and has no predictive validity going forward — the very idea of "the process is capable of X% nonconforming" presumes a stable, repeatable distribution to make that statement about.

$C_p=(USL-LSL)/6\sigma$ measures only the process's potential capability — how well the natural spread fits within the tolerance band — and takes no account of where the process mean actually sits. $C_{pk}=\min\!\big[(USL-\mu)/3\sigma,\ (\mu-LSL)/3\sigma\big]$ measures actual capability: it also penalizes the process for being off-centre, using only the distance to the nearer limit. $C_{pm}=(USL-LSL)/\big(6\sqrt{\sigma^2+(\mu-T)^2}\big)$ goes a step further, replacing $\sigma$ with the total mean-squared deviation from the target $T$ (not just the spec limits), so it penalizes both excess spread and off-target centring in a single index, directly reflecting the Taguchi loss-function view that being off-target costs something even before a limit is crossed.

Because $C_{pk}$ always measures the same spread as $C_p$ but from the nearer (and therefore more restrictive) limit, $C_{pk}\le C_p$ always, with equality if and only if the process is perfectly centred on the midpoint of the specification range ($\mu=T$). The two are formally related by $C_{pk}=C_p(1-k)$, where $k=|T-\mu|/\big[(USL-LSL)/2\big]$ is the fractional off-centring of the mean relative to the tolerance half-width.

(c) Fraction nonconforming, $C_{pm}$, and the effect of centring, from $C_p=1.33$, $C_{pk}=1.05$

Given. $C_p=1.33$, $C_{pk}=1.05$; normal process; two-sided spec limits; target $T=(USL+LSL)/2$.

Find. The process fraction nonconforming; $C_{pm}$; and how the fraction nonconforming would change if the mean were centred.

Approach. Use $k=1-C_{pk}/C_p$ to get the mean's fractional off-centring, convert that into a $\sigma$-distance to each specification limit ($3C_{pk}$ to the near limit, $3(2C_p-C_{pk})$ to the far limit), read the tail probabilities off the standard normal table, then repeat with the mean exactly on target for the comparison.

1. Off-centring fraction and distance to each limit (in $\sigma$). $$k=1-\frac{C_{pk}}{C_p}=1-\frac{1.05}{1.33}=\boxed{0.2105},$$ $$\text{near limit: }3C_{pk}=3(1.05)=\boxed{3.15\sigma},\qquad \text{far limit: }3(2C_p-C_{pk})=3(2.66-1.05)=\boxed{4.83\sigma}.$$
2. Fraction nonconforming. $$p=[1-\Phi(3.15)]+\Phi(-4.83)\approx0.000816+0.0000007=\boxed{0.000817\ (\approx817\ \text{ppm})}.$$ The far side ($4.83\sigma$) contributes almost nothing; essentially all of the nonconformance comes from the side the mean has drifted toward.
3. $C_{pm}$. With $T=(USL+LSL)/2$, the mean's offset from target in $\sigma$ units is $|\mu-T|/\sigma=3(C_p-C_{pk})=3(1.33-1.05)=0.84$, so $$C_{pm}=\frac{C_p}{\sqrt{1+\left(\dfrac{\mu-T}{\sigma}\right)^2}}=\frac{1.33}{\sqrt{1+0.84^2}}=\frac{1.33}{1.306}=\boxed{1.018}.$$
4. Fraction nonconforming if the process were re-centred. Centring the mean makes $C_{pk}=C_p=1.33$, so both limits sit $3(1.33)=3.99\sigma$ from the mean: $$p_{\text{centred}}=2[1-\Phi(3.99)]\approx2(0.0000330)=\boxed{0.0000661\ (\approx66\ \text{ppm})}.$$ Centring the mean — with the spread ($\sigma$, hence $C_p$) completely unchanged — cuts the fraction nonconforming from about 817 ppm to about 66 ppm, better than a 12-fold reduction, because the single dominant near-limit tail (3.15$\sigma$) is replaced by two much smaller, balanced tails (3.99$\sigma$ each).