#  Topic 11d
#  Look at the Normal distribution
# 
#  First we will look at the standard normal, N(0,1)
#
#  We will just repeat the kind of questions that
#  we saw in earlier distributions.
#

############################################
#   Using the table, find
#For the standard normal distribution, N(0,1)
#  P(X < -0.47 ) = 
#  P(X <= 2.38 ) = 
#  P(X < 1.3765) =
#  P( X > 2.24 ) =
#  P( X > -1.39 ) =
#  P( -0.62 < X < 1.26 ) =
#  P( X < 0.34  or X>0.98 ) =
######################################

######################################
# now do the same using the pnorm() function

#  P(X < -0.47 ) = 
pnorm( -0.47 )

#  P(X <= 2.38 ) = 
pnorm( 2.38 )

#  P(X < 1.3765) =
pnorm( 1.3765 )

#  P( X > 2.24 ) =
1 - pnorm( 2.24 ) # Is one approach
pnorm( 2.24, lower.tail=FALSE )  # another way

#  P( X > -1.39 ) =
1 - pnorm( -1.39 ) # Is one approach
pnorm( -1.39, lower.tail=FALSE )  # another way

#  P( -0.62 < X < 1.26 ) =
pnorm(1.26) - pnorm(-0.62)

#  P( X < 0.34  or X>0.98 ) = 
pnorm(0.34) + (1-pnorm(0.98))  # is one way
pnorm(0.34) + 
    pnorm( 0.98, lower.tail=FALSE)# is another way

#######################################
##  go back to the table and use it to solve:
#
#  find the y value such that
#      P(X < y ) = 0.2358    y=
#      P(X < y ) = 0.7611    y=
#      P(X > y ) = 0.3520    y=
#      P(X > y ) = 0.9192    y=
#      P(X < y ) = 0.5981    y=
#        because a normal distribution is symmetric
#      P(X< -y or X>y ) = 0.0346  y=?
######################################

######################################
# now do the same using the qnorm() function
#      P(X < y ) = 0.2358    y=
qnorm( 0.2358 )

#      P(X < y ) = 0.7611    y=
qnorm( 0.7611 )

#      P(X > y ) = 0.3520    y=
qnorm( 1 - 0.3520 )  # is one way
qnorm( 0.3520, lower.tail=FALSE ) # is another way

#      P(X > y ) = 0.9192    y=
qnorm( 1 - 0.9192 )  # is one way
qnorm( 0.9192, lower.tail=FALSE ) # is another way

#      P(X < y ) = 0.5981    y=
qnorm( 0.5981 )

#        because a normal distribution is symmetric
#      P(X< -y or X>y ) = 0.0346  y=?
qnorm( 0.0346/2, lower.tail=FALSE)


################
## and before we leave the standard normal
## look at some special cases
## find P( X>0 and X<1 )
pnorm(1) - pnorm(0)
## find P( -1 < X < 1 )
pnorm(1)-pnorm(-1)
## find P( 1 < X < 2 )
pnorm(2) - pnorm(1)
## find P(0 < X < 2 )
pnorm(2)-pnorm(0)
## find P( -2 < X < 2 )
pnorm(2) - pnorm(-2)
## find P( -3 < X < 3 )
pnorm(3) - pnorm(-3)


#############################################
##  Now we will look at non-standard normal
##  distribution.  For each such non-standard
##  normal distribution we need to give the 
##  mean and the standard deviation of that
##  distribution.  We will do this in the 
##  N(mean, standard deviation) syntax.  Thus
##  N(25,7) is a normal distribution with 
##  mean=25 and standard deviation=7.
###############################################
##

# for a N(25,7) find P(X < 32)
# the long way is to compute the equivalent z value
# for a standard normal
z <- (32 - 25)/7
z
#  and then find P(X < z ) for a N(0,1)
pnorm( z )
#  neat way in R is to use the longer pnorm()
pnorm( 32, mean=25, sd=7)

# for a N(9,13) find P(X<32)
#  the long way
z <- (32-9)/13
z
pnorm( z )
#  the neat way
pnorm( 32, mean=9, sd=13)

# for N(13.7, 4.6) find P( X > 16.2 )
# the long way
z <- (16.2 - 13.7)/4.6
z
pnorm( z, lower.tail=FALSE)
#  or 
1 - pnorm( z )
# the neat way
pnorm( 16.2, mean=13.7, sd=4.6, lower.tail=FALSE)
# or
1 - pnorm(16.2, mean=13.7, sd=4.6)

# for N(-6.2, 1.34) find P( -7 < x < -5)
# the long way
z1 <- (-7 - -6.2)/1.34
z1
z2 <- (-5 - -6.2)/1.34
z2
pnorm( z2 ) - pnorm(z1)
#  the neat way
pnorm( -5, mean=-6.2, sd=1.34) -
  pnorm( -7, mean=-6.2, sd=1.34)

#  portions of the SAT are created so that the
#  distribution of scores is normal with mean=500
#  and standard deviation of 100.
#
# Is we select a random test taker, what is the
# probability that that person's score is
# more than 625?
# i.e., for a N(500,100) what is P( X > 625 )
#  the long way
z <- (625-500)/100
z
pnorm( z, lower.tail=FALSE)
# the neat way
pnorm( 625, 500, 100, lower.tail=FALSE)
#   Note that as long as you put the mean and
#   standard deviation in the right order you
#   do not need to name them.

##############################################
###   Now we will go the other way.  Find the 
###   y value such that p is the desired 
###   probability.

# for a N(45.7,8.6 ) distribution, find y such
#     that P(X < y ) = 0.333
# the long way...
#   what is the z value in a standard normal 
#   such that P(X < z) = 0.333
z <- qnorm( 0.333 )
z
# then translate that back to our N(45.7, 8.6)
z*8.6+45.7
#  the neat way
qnorm( 0.333, mean=45.7, sd=8.6)

# for a N(-9.2, 1.6 ) distribution, find y such
#     that P(X > y ) = 0.219
#   what is the z value in a standard normal 
#   such that P(X > z) = 0.219
z <- qnorm( 1 - 0.219 )
z
# then translate that back to our N(-9.2, 1.6 )
z*1.6 + -9.2
#  the neat way
qnorm( 0.219, mean=-9.2, sd=1.6, lower.tail=FALSE)

# for a N(134.2, 12.43 ) distribution, 
#     find y1 and y2 such
#     that P(X < y1 or X>y2 ) = 0.075
#     and where abs(mean - y1 ) = abs(mean-y2)
#   What is the z value in a standard normal 
#   such that P(-z < X < z) = 0.075
#
# this works because the normal distribution
#     is symmetric
z <- qnorm( 0.075/2, lower.tail=FALSE )
z
# then translate z back to our N(134.2, 12.43 )
y2 <- z*12.43 + 134.2  #  this is our y2
y2
# and translate -z back to our N(134.2, 12.43 )
y1 <- -z*12.43 + 134.2  # this is our y1
y1
#  note
abs( 134.2 - y1)
abs( 134.2 - y2)
# and

#  the neat way
qnorm( 0.075/2, mean=134.2, 
       sd=12.43, lower.tail=FALSE) # this is y2
qnorm( 0.075/2, mean=134.2, 
       sd=12.43 ) # this is y1

#  Going back to the SAT distribution, N(500,100),
# find the 73rd percentile.  That is just the 
# score that has 73% of the values be less than it.
qnorm( 0.73, 500,100)