Median Rank Failure Probability

Before a set of broken bars can be plotted, each one needs a probability attached to it, and this is the awkward little problem that has to be solved first. You have thirty strengths. Sort them. The weakest is rank 1 and the strongest is rank 30. What failure probability does each one represent?

The naive answer, $i/N$, fails immediately: it puts the strongest specimen at exactly 100% failed, and $\ln[\ln(1/(1-P_f))]$ is infinite there. The strongest bar in every batch would fall clean off the plot, and it is not a spare data point — it is one of thirty hard-won results. The estimator has to pull both ends in.

Bernard and Bos-Levenbach published the usual answer in 1953: $P_{f,i} = (i - 0.3)/(N + 0.4)$. It approximates the median of the $i$-th order statistic of a uniform sample — the exact quantity requires the incomplete beta function, and this closed form is accurate to well under 1% for any sample size worth plotting. The median rather than the mean is chosen because the distribution of each order statistic is skewed, and the median is the more robust centre for a quantity you are about to take two logarithms of.

Feel what the correction does. For ten specimens, the weakest plots at $0.7/10.4 = 6.73\%$ instead of 10%, and the strongest at $9.7/10.4 = 93.27\%$ instead of 100%. Both ends are pulled toward the middle, and the strongest bar stays on the paper.

Other plotting positions exist and give visibly different slopes on small samples. $(i-0.5)/N$ is the Hazen position; $i/(N+1)$ is the mean rank. On a sample of ten the choice can move a fitted $m$ by several percent, which is small against the ±30% confidence interval but large enough to make two moduli non-comparable if they were fitted differently. Say which you used. It costs one sentence in a report and saves an argument later.

The most useful thing this equation does is tell you how big a test programme has to be, and the answer is uncomfortable. To get a measured data point at 1% failure probability you need the weakest specimen to land there, which takes about 70 bars. For 0.1% it takes about 700. Nobody breaks 700 ceramic bars. That is why every quoted one-in-a-thousand or one-in-a-million survival stress in this field is an extrapolation of a fitted line into a region where no data exists, and why the absence of a threshold in the two-parameter model matters so much: the extrapolation has nothing to stop it and nothing to check it against.

There is one honest escape, and it is not statistical. Proof testing: load every single part to a chosen stress before it goes into service. Everything with a flaw bad enough to fail at that stress breaks on the bench instead of in the machine, and the survivors have a genuine, demonstrated truncation of their distribution — a real lower bound rather than a fitted one. It is expensive, it consumes some good parts, and for materials susceptible to slow crack growth it has to be done carefully because the proof load itself can extend a subcritical crack. It is also the only way to get a guaranteed minimum strength out of a material whose model does not have one.