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<br />'I I ,.-(. I, <br />v 0... v... <br /> <br />, <br /> <br />Finally, using the distribution above obtain the flow having a 1% exceedance probability and compare it to <br />that obtained using the frequency factor approach in Bulletin 17. <br /> <br />Using the zoom feature in Mathcad to get a good estimate of the value of In (flow) corresponding to an <br />exceedance probability of .01, I get a value of In (flow) = 9.34. The corresponding flow is: <br /> <br />ft3 ft3 <br />exp(9.34),- = 11384-. Bulletin 17 gets a value of 11,400 cfs using the frequency factor method. <br />sec sec <br /> <br />log pearson type 111 example problem <br />bulletin 17.mcd <br />last save 11/14/2003/12:01 PM <br /> <br />C:\MyFiles\Mathcad application areas <br />\Statistics\log Pearson Type III aka Gamma <br />distribution files\ <br />11/14/2003/12:07 PM <br /> <br />SofS <br />