If I have an standard deviation of 32bb/100 and a win rate of 9.8bb/100 in a 11K hands sample, what are my chances of having a 100 hands session with a winning rate of 20bb/100.
I know the calculation is quite simple but I just cant be sure about it. I will make an example and plase correct me if I am wrong.
Baseball. A pitcher had ERA of 3.25 with SD of 1.89: Based on his info only, if I want to know the probability of having 9innings with 6 or less Earned runs is (using Normal Distribution):
6 - 3.25 = 2.75/1.89 = 1.455 SDs. Looking in a graph or using the excell function .....uhm, =DISTR.NORM.ESTAND(1.455) = 92.7% Sorry that s the name of the excel function in spanish.
The point is that chances 92.7% that this pitcher will allow 6ER or less in his next 9 innings.
Am I correct?? Please help me!!!Statistics: Posted by AlexMR — Sun Sep 11, 2005 11:49 am
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