The cumulative percentage for a value of x is given by (1 – e-λx) and so the value of x corresponding to a required cumulative percentage (P) is x = –ln(1 – P) / λ. Thus, a 99.9% upper limit
could be calculated by 6.91 / λ. If λ is estimated by the mean of the measurements, this is 6.91 times the mean.
More information about the Poisson and Exponential distributions can be found in the NIST/SEMATECH e-Handbook of Statistical
Methods.3 Information about Solver, POISSON, and EXPONDIST is provided in Excel's Help option. Minitab provides the option to add an
exponential distribution plot to a histogram.
1. US Food and Drug Administration. International Conference on Harmonization; Guidance on specifications: Test procedures
and acceptance criteria for biotechnological/biological products. Federal Register; 1999 Aug 19 [cited 2006 October 13];6(159):44934.
2. Deming WE. Some theory of sampling. NY: Dover Publications; 1966.
3. NIST/SEMATECH e-handbook of statistical methods. 2003 June 1 (updated 2006 July 18) [cited 2006 October 13]. Available
4. Howe WG. Two-sided tolerance limits for normal populations-some improvements. J of Am Statistical Assoc 1969;64:610-20.
5. Natrella MG. National Bureau of Standards handbook 91: Experimental statistics. Washington, DC: US Department of Commerce;
Excel procedures to apply all these methods can be obtained from firstname.lastname@example.org
Terry Orchard is a statistician at Bespak Europe, Blackhill Dr., Featherstone Road, Wolverton Mill South, Milton Keynes, Buckinghamshire,
MK12 5TS, England, + 44 (0)1908.552600, fax +44(0)1908.552613,