and 1078 = 107!2 ci/pci x 10" m*/ha. Both the input data and the results of the regression analysis based on Equation 4 are shown in Figure 9. The log-log regression equation, $ = ax fits the cumulative inventory, IPu, estimates quite well, as shown by the graph and a correlation coefficient of r = 0.9999. A similar relationship would be expected for the regression of soil concentrations on distance from ground zero, but this regression line would, of course, have a negative slope. In other words, the results shown in Figure 9 provide a basis for expecting that the frequency distributions of plutonium concentrations in randomly selected soil samples will be more or less lognormally distributed. As soil is the source of plutonium in vegetation samples, the same reasoning should apply to plutonium concentrations in vegetation samples. Composite Means of Cy and Cc. As mentioned earlier, it is assumed that cattle grazing the inner: compound at random would (in time) consume a composite sample of vegetation Cc. = where rman and soil, taking from each stratum an amount of vegetation and soil proportional to the area of the stratum enclosed by the fence. This notion of composite sampling, or composite grazing if you prefer, is formulated as follows: exp (in C, + in fi), j=1,2, .. . 6. (5) C, is the average concentration of plutonium (pCi/g) C, is a normally distributed random number specified by the l day's intake of vegetation or soil, In in estimated population parameters (y = xg and o = sg) listed for stratum j in Table 4, and £3 is the area of stratum j divided by the total area of the inner compound, given as percentages, in Table 3. Applying the procedure indicated by Equation 5 to the arithmetic means listed in Table 4 yields Cc, = 71 pCi/g and Co = 579 pCi/g. are used below to estimate Tou’ These values Estimates of Ibu Based on Parameter Means The empirical and theoretical estimates made thus far can be used to estimate the value of I, . For example, Pu Ip, 7 (6,115 x 71) + (236 x 579) = 570,809 pCi/day where 6,115 = 1, = 163.5(410)°-73/(4.5 x 0.48), Equation 1, 495 (6)