Add training workflow, datasets, and runbook

training_data/curated/text/cd73bb080bb91d5a27efe183e93c714d8c1fee1aff85eedb07eb0c8a12c1ddcd.txt

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+458 Part IV: Additional Considerations 
+This approximation is quite accurate for option pricing purposes, since one is not 
+really interested in thousandths of a point where option prices are concerned. 
+Example: Suppose that XYZ is trading at 45 and we are interested in evaluating the 
+July 50 call, which has 60 days remaining until expiration. Furthermore, assume that 
+the volatility of XYZ is 30% and that the risk-free interest rate is currently 10%. The 
+theoretical value calculation is shown in detail, in order that those readers who wish 
+to program the model will have something to compare their calculations against. 
+page: 
+Initially, determine t, d1, and d2, by referring to the formulae on the previous 
+t = 60/365 = .16438 years 
+d _ In (45/50) + (.1 + .3 x .3/2) x .16438 
+1-
+.3 X ✓.16438 
+= -.10536 + (.145 X .16438) = __ 67025 .3 X .40544 
+d2 = -.67025 - .3 ✓.16438 = -.67025 - (.3 x .40544) = -.79189 
+Now calculate the cumulative normal distribution function for d1 and d2 by 
+referring to the above formulae: 
+dl = -.67025 
+l 1 
+y = l + (.2316419 I -.67025 I) = 1.15526 = ·86561 
+z = .3989423e--(-.67025 X -.67025)/2 
+= .3989423e-0·22462 = .31868 
+There are too many calculations involved in the computation of the fifth-order 
+polynomial to display them here. Only the result is given: 
+X = .74865 
+Since we are determining the cumulative normal distribution of a negative 
+number, the distribution is determined by subtracting x from l. 
+N(d1) = N(-.67O25) = l -x = l - .74865 = .25134 
+In a similar manner, which requires computing new values for x, y, and z, 
+N(d2) = N(-.79179) = 1- .78579 = .21421