# Here we use the Stat 2016 data to 
# illustrate how normal modelling works
# Curator/Coder: Kno Tsao
# 參考資料：2016 統計學課網
# http://faculty.ndhu.edu.tw/%7Echtsao/edu/16/stat/stat.html

#  Given the class 
# average=46.52,  s.d.= 27.3
# Q1: 我考得如何？ How do I do in this exam? Good or bad?
# say, my score=60, 46, 20, 10

# Assume X ~ N(46.52, 27.3^2) is reasonable/acceptable
# F(x)=P(X <= x), X ~ N(46.52, 27.3^2)
x<-c(60,46,20,10)
pnorm(x,46.52,27.3)

# 幾分會過？ What would be the passing threshold, if 
# the instructor decides to fail, say 1/4, 1/3 of 
# of the class?
# Solve x s.t. F(x)=P(X<x)=q, 1/4, 1/3, ...
#  x=F^{-1}(q)

q<-c(1/4,1/3,1/2,3/4)
qn<-qnorm(q,46.52,27.3)

# Note that F(F^{-1}(x))=x
pnorm(qn,46.52,27.3)

# Generate a random sample of size n=63
# from a N(46.52,(27.3)^2)
x<-rnorm(63,46.52,27.3)
hist(x)
summary(x); sd(x)

# Transformations on X
# X2 is transformed to make the distribution <skewed to the right>.
# This sqrt-x10 transformation make the lower scores higher while 
# maintains all scores lie in [0,100]
x2<-sqrt(x)*10
hist(x2)

# X3 is transformed to make the <skewed to the left> 
# while maintainting all scores in [0,100]
x3<-(x^2)/100
hist(x3)

# It is possible that the generated data having 
# negative values. 
# It happens despite the probability is quite small
pnorm(0,46.52,27.3)
# In such scenario, the common practice is to add a 
# constant such that all (new) scores are positive
x<-rnorm(63,46.52,27.3)
# You can find out the smallest value of x by min 
min(x)
# if negative, substract min(x) from all x
x<-x-min(x)
# the new x are all greater or equal to zero
x2<-sqrt(x)*10
hist(x2)