APPENDIX C 

 SAMPLE CALCULATION 



In fitting a log-normal distribution to the data on wave heights and wave periods given 

 in Tables 1 and 2, a difficulty arises due to the fact that the lower limit of the lowest class 

 is zero and inasmuch as the logarithm of zero is minus infinity, it is not possible to assign a 

 mean value of the logarithm of the height or period for the lowest class. One way of circum- 

 venting this difficulty would be to use the logarithm of the algebraic mean of the class limits. 

 \ less arbitrary solution, used here, is to omit the relatively few values falling into the low- 

 est class and treat the remaining truncated data by the standard statistical method 11 for fit- 

 ting a truncated normal distribution. In the statistical sense used here, a truncation means 

 that only values larger than a specified lower limit are used. To fit a log-normal distribution 

 to the truncated data, requires the calculation of the mean value and the standard deviation 

 from the truncated data. 



The method and tables of Reference 11 are applied as indicated below. In the calcu- 

 lations, the symbols used are a for standard deviation and y and z for parameters needed to 

 enter Table IX of Reference 11, z being an estimate of the point of truncation. 



Following the procedure outlined on page 29 of Reference 10, we have from Table 6: 



(2Neu 2 )(2/V) (459.864) (12,362) 



y 



2{2Nco) 2 2(2071.605) 2 



y = 0.6623 



and from Table IK, s = -1.293 at y = 0.6623 and g(z) = 0.6736. 



2/V<u 2071.6(0.6736) 



o&s = g{z) = = 0.1129 



IN 12,362 



From Table II of Reference 11, at z = -1.293, we obtain 



Theoretical percent of truncation = 9.80 



Mean value ofw = w = -ss = 1.293 (0.1129) = 0.1460 



Mean value of h = h SQ = h T + u = 0.6990 + 0.1460 = 0.8450 



The value of h corresponding to P = 0.975 = A 975 = A 50 + 1.96 (0.1129) = 1.0663 

 The value of h corresponding to P = 0.025 = ^ 2>5 = ^ 50 - 1-96 (0.1129) = 0.6237 



49 



