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Chapter 8*: SANITARY INSPECTION AND MICROBIOLOGICAL WATER QUALITY - page 34 / 52

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value for the 50-percentage point will be equal to the mean “m”. For completeness, it can be noted that the standard normal distribution is symmetrical, so that the values of z for the 20, 10 and 5 percentile points are respectively -0.8416, -1.2816 and -1.6449. This approach can only be applied when the data follow a normal distribution, whereas calculation of classical percentiles does not require normality. For bathing area A, in the example, the 50 percentile point corresponds to the value of “m” (597), while the 90 percentile point is estimated by the log-normal distribution method from the log mean 2.7758 and the log standard deviation 0.6934 (Table 8.9). Hence, the log 90 percentile point is 2.7758 + 1.2816 × 0.6934 = 3.6645 and the antilog is 4,618 cfu per 100 ml. This value is quite similar to that obtained by plotting the data on the log-normal-normal distribution probability paper (4,700 cfu per 100 ml). For bathing area B, the 50 percentile point corresponds to a value of m = 15 cfu per 100 ml while the 90 percentile point is 1.1933 + 1.2816 × 0.9883 = 1.1933 + 1.2666 = 2.4599 and its antilog is 288 cfu per 100 ml.

The classical statistical calculation of percentiles estimates the variability of the distribution of a set of results (after ordering them in ascending order) independently of whether they are normally distributed and indicates the concentration of micro- organisms that embraces a specific percentile. For example, 80 per cent and 95 per cent of the data will be below the value (cfu per 100 ml) of the 80th and 95th percentiles, respectively, and 20 per cent and 5 per cent of the data will be above those concentrations, respectively. Some computer programs will not calculate the 95th percentile unless 20 records are available, although they will do so if the individual records are specifically considered the midpoint of an interval. An example of the calculation by hand is given in Box 8.3.

Box 8.3 Calculation of percentiles

Pr = xi + (j - i) (xi+1

  • -

    xi)

where:

Pr is the percentile required (i.e. P50, P80, P90 or P95) xi is the concentration that corresponds with an i position in the ranking corresponding to that Pr j is the next position in the ranking (calculated as j = r (n + 1)/100) xi+1 is the next concentration in the ranking For bathing areas A and B of Table 8.8 and 8.9, the calculation of j is: 5 0 ( 1 2 + 1 ) / 1 0 0 = 6 5 0 / 1 0 0 = 6 . 5 = j f o r P 5 0 , a 90 (12 + 1) = 1,170/100 = 11.7 = j for P For bathing area A: xi = 590 is the concentration at position 6 while 600 is the concentration at position xi+1 n d 90 (= 7)

Then

P

50

P

90

= 590 + (6.5 - 6) (600 - 590) = 590 + 5 = 595 cfu per 100 ml, and = 3,390 + (11.7 - 11) × (6,700 - 3,390) = 3,390 + 2,317 = 5,707 cfu per 100 ml

P

50

P

90

  • For bathing area B, applying the same criteria: = 8 + (6.5 - 6) × (32 - 8) = 8 + 12 = 20 cfu per 100 ml, and = 140 + (11.7 - 11) × (1,600 - 140) = 140 + 1,022 = 1,162 cfu per 100 ml

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