SD=0.025
N (people) = 20
X (average) = 7.5
My average is 8.
I want to know how many percent I am likely to be better than compare to N .
N (people) = 20
X (average) = 7.5
My average is 8.
I want to know how many percent I am likely to be better than compare to N .
In[]:=
mean=7.5;standardDeviation=0.025;observation=8;Zscore=
observation-mean
standardDeviation
Out[]=
20.
A z-score of 20 is extremely unlikely, and the question is a measurement of ‘how unlikely’
In[]:=
p1=Plot[PDF[NormalDistribution[mean,standardDeviation],x],{x,7,8},PlotRange->{{6.9,8.1},{0,20}}]
Out[]=