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determining compliance with bathing water quality standards? ... Commission to keep percent compliance rather than calculation of 95th percentiles in any new.
Letters in Applied Microbiology 2002, 34, 283–286

Does calculation of the 95th percentile of microbiological results offer any advantage over percentage exceedence in determining compliance with bathing water quality standards? P.R. Hunter School of Medicine, Health Policy and Practice, University of East Anglia, Norwich NR4 7TJ, UK 2001/222: received 24 July 2001, revised 21 December 2001 and accepted 7 January 2002

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Aims: Draft WHO guidance is likely to suggest a calculated 95th percentile method, rather than percentage exceedence, in assessing compliance of bathing waters with microbiological standards. This study set out to determine whether this was an appropriate development. Methods and Results: A series of Monte Carlo studies compared five non-parametric methods for calculating the 95th percentile with the parametric method and compares results with percentage exceedence. It is shown that the Hazen method gives the closest proximity to the parametric method for calculating 95th percentile values. However, the difference between 95th percentile results and percentage exceedence, as currently used, is trivial compared to uncertainty due to sample variation. Conclusions: It is concluded that a calculated 95th percentile for beach classification offers little advantage compared to percentage exceedence, other than offering a false sense of certainty. Furthermore, the additional calculation needed in determining 95th percentile values will demand electronic calculation, increase the chance of calculation errors and make the results less understandable to beach managers and the general public. Significance and Impact of the Study: This study should encourage the European Commission to keep percent compliance rather than calculation of 95th percentiles in any new bathing water directive.

INTRODUCTION Current microbiological standards for the classification of recreation water in the UK are based on testing for Total Coliform and Faecal Coliform counts in 100 ml of water (HMSO 1991). Under these regulations, for a bathing water to be classified as BW1 (the highest standard) 95% of water samples must have counts of less than 10 000 Total Coliforms/100 ml and less than 2000 Faecal Coliforms. The value of relying solely on coliform was brought into question by work showing that self-reported diarrhoea in volunteer bathers was seen in people swimming in water where the microbiological counts were within the standard (Kay et al. 1994). This study also showed that gastrointestinal health Correspondence to: P.R. Hunter, Professor of Health Protection, School of Medicine, Health Policy and Practice, University of East Anglia, Norwich NR4 7TJ, UK (e-mail: [email protected]). ª 2002 The Society for Applied Microbiology

effects were most closely correlated with faecal streptococci counts. In large part as a result of the Kay study, there have been moves to change national and international standards for recreational water quality. The World Health Organization is currently revising its guidance on recreational water quality (WHO 1998). It is likely that new standards will focus on faecal streptococcal counts. It is also likely that the WHO will adopt a 95th percentile level of a set of measurements, rather than a proportion of samples exceeding a standard value, in order to determine classification of the bathing water (WHO 1998; Bartram and Rees 2000). There is, as yet, no clear guidance on how the 95th percentile should be calculated. This paper considers the possible options for calculating this 95th percentile and also considers whether calculating the 95th percentile offers any real advantage over a percentage compliance approach.

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Approaches to calculating the 95th percentile of a set of data For normally distributed data, the 95th percentile can easily be calculated from a knowledge of the mean and standard deviation by the following formula: Parametric: P ¼ m + sz where P is the 95th percentile value, m is the mean, s is the standard deviation and z the standard normal variable for the 95th percentile (1Æ6449) (Bartram and Rees 2000). However, bacteriological counts in water are not normally distributed and are highly skewed. Only after logarithmic transformation are the results normally distributed (Bartram and Rees 2000). In this case, m becomes the log geometric mean and s the standard deviation of the log values of the actual counts. The anti-log of P then gives the value of the 95th percentile. A further problem, and one less easily dealt with, is the large proportion of enterococci counts that are less than the limit of detection. These are typically recorded as