I asked 3 days ago a question on the same topic, Calculate talent in Bayesian Resume Rating
Bayesian Resume Rating for sports math is explained in PDF https://www.jellyjuke.com/uploads/5/8/0/2/58022979/mathematical_explanation_of_the_bayesian_resume_rating_10-23-18.pdf
The talent is calculated by formula:
In the example it gives 1.5 and I managed to calculate it computationally.
The standard deviation is given by formula:
If I made the same substitution for Pi(x) than suggested by the author I get 0.36 instead the result that appears in pdf: 0.6
I tried calculating programically with python and Wolfran Mathematica and get the same result. I think mine is wrong since it is observable in pdf plot
I don't know if I have an error with parameters or another substition is needed for Pi(x). I will share my Wolfran Mathematica code if somebody wants to check:
CCfunc[x_?NumericQ] := E^(-x^2/2)/Sqrt[2*Pi]
Nfunc[x_?NumericQ] := CCfunc[x]NIntegrate[CCfunc[y],{y,-Infinity,(x+1.62)/Sqrt[2*1.6^2+0.65^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(x-0.62)/Sqrt[2*1.6^2+0.6^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(x+0.38)/Sqrt[2*1.6^2+0.6^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(x-0.52)/Sqrt[2*1.6^2+0.56^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(x-0.1)/Sqrt[2*1.6^2+0.61^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(x+0.05)/Sqrt[2*1.6^2+0.6^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(x-0.32)/Sqrt[2*1.6^2+0.61^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(x-0.16)/Sqrt[2*1.6^2+0.6^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(x+1.93)/Sqrt[2*1.6^2+0.67^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(x+1.1)/Sqrt[2*1.6^2+0.63^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(x-0.49)/Sqrt[2*1.6^2+0.59^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(x+1.07)/Sqrt[2*1.6^2+0.63^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(x-0.32)/Sqrt[2*1.6^2+0.61^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(0.75-x)/Sqrt[2*1.6^2+0.59^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(-1.1-x)/Sqrt[2*1.6^2+0.63^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(0.16-x)/Sqrt[2*1.6^2+0.6^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(x-0.4)/Sqrt[2*1.6^2+0.59^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(x-0.87)/Sqrt[2*1.6^2+0.6^2]}]NIntegrate[CCfunc[y],{y,-Infinity,(x-1.57)/Sqrt[2*1.6^2+0.61^2]}]
nume = N[Normal[Integrate[(1.5-x)^2*Nfunc[x], {x, -Infinity, Infinity},GenerateConditions -> False]], 4]
den = N[Normal[Integrate[Nfunc[x], {x, -Infinity, Infinity},GenerateConditions -> False]], 4]
nume/den
I apologize! You have caught a mistake in my standard deviation equation. Like any standard deviation calculation, the whole thing needs to be in a square root. You got 0.36. Once you take the square root, you get 0.60, just like I did. I will fix it on my website. Thank you for catching that mistake!