# A quick review of using pnorm and qnorm
#
#   For a  Standard Normal distribution, N(0,1)
#
# find P(X<1.34)
pnorm( 1.34 )
# find P( X > 1.34)
1 - pnorm( 1.34 )  # first way to do this
pnorm( 1.34, lower.tail=FALSE)
#
# find P( -0.43 < X < 1.73 )
pnorm( 1.73 ) - pnorm( -0.43 )
#
# find P( X < -0.75 or X > 1.14 )
#            first way
pnorm( -0.75 ) + ( 1 - pnorm(1.14))
#            second way
pnorm( -0.75 ) + pnorm(1.14, lower.tail=FALSE)
#            third way
1 - ( pnorm(1.14)-pnorm( -0.75) )
#
# find the value of x such that 
#    P( X < x ) = 0.23
qnorm( 0.23 )
# find the value of x such that 
#    P( X > x ) = 0.23
qnorm( 1 - 0.23 )  # first way
qnorm( 0.23, lower.tail=FALSE)  # second way

# there is another type of problem that we can
# do because the normal distribution is symmetric.
# find x such that 
#  P( -x < X < x ) = 0.58
#  we can find -x by
qnorm( (1 - 0.58)/2)
# we can find x by 
qnorm( 0.21, lower.tail= FALSE)
# or 
qnorm( 0.79 )

#
#  or find x such that 
#  P( X <-x or X>x ) = 0.075
# Because we have a symmetric distribution we 
# also know that this means 
# that P( X < -x ) = 0.075/2 = 0.0375
qnorm( 0.0375 )  # which gives the value for -x
# or we could do
qnorm(0.0375, lower.tail=FALSE) # to get x

######################################################
#  Now, let us try some for non-standard             #
#  normal distributions.  We will do each            #
#  problem twice, once converting to the             #
#  problem to a standard normal, and then the        #
#  second time using the extra parameters of the     #
#  pnorm and qnorm functions                         #  
######################################################
#
#  For a normal population with mean=54.3 and
#  standard deviation 12.7, i.e., N(54.3,12.7)
#  find P( X < 63 )?
#  Convert to a z-score
z <- (63 - 54.3)/ 12.7
z
#  then use pnorm
pnorm( z )
#
# or, just use the expanded pnorm command
pnorm( 63, mean=54.3, sd=12.7)

#######################
# for a N(74.2, 6.54) population find
# P( X > 68)
#
# convert to z-score
z <- (68 - 74.2)/12.7
z
#  use pnorm
1 - pnorm( z )  # looking to the left,
# or
pnorm( z, lower.tail = FALSE)  # looking to the right

# or use the expanded pnorm function
1 - pnorm( 68, mean=74.2, sd=12.7 ) # looking left
# or
pnorm( 68, mean=74.2, sd=12.7, lower.tail=FALSE ) # looking right
#
########################
# for a N(3.2, 0.73) distribution
#  find P( 2.8 < X < 3.6 )
#
# convert both values to z-scores
z_left <- (2.8 - 3.2)/ 0.73
z_left
z_right <- ( 3.6 - 3.2 )/ 0.73
z_right
# use the pnorm functions
pnorm( z_right ) - pnorm( z_left )
#
#  or use the expanded pnorm function
pnorm( 3.6, mean=3.2, sd=0.73) - 
    pnorm( 2.8, mean=3.2, sd=0.73)


#########################
# For a N( 153, 14 ) distribution find
# P( X < 140 or X > 170)
#
# convert both values to z-scores
z_left <- ( 140 - 153 ) / 14
z_left
z_right <- ( 170 - 153 ) / 14
z_right
# use pnorm
pnorm( z_left) + (1-pnorm( z_right) )
#    or we coud do that as
pnorm( z_left ) + pnorm( z_right, lower.tail=FALSE)
#
# or we could use the expanded pnorm funtion
pnorm( 140, mean=153, sd=14) +
     ( 1 - pnorm( 170, mean=153, sd=14))
#  which we could have done as
pnorm( 140, mean=153, sd=14) +
     pnorm( 170, mean=153, sd=14, lower.tail=FALSE)


#
################### qnorm
#
# for a N(13.2,8.4) distribution 
# find the value of x such that
# P( X < x ) is 0.315
# for a N(0,1) the z value would be
z <- qnorm( 0.315 )
z
# so we convert that to our N(13.2,8.4)
# via
x <- z*8.4 + 13.2
x
#
# or we could use the expanded qnorm function
qnorm( 0.315, mean=13.2, sd=8.4 )

################################
#  for a N(4.3, 8.2) distribution
# find the value of x such  that 
#  P( X > x ) = 0.29
# for a N(0,1) the z value would be
z1 <- qnorm( 1 - 0.29 )
z1
#  which we could have found by 
z2 <- qnorm( 0.29, lower.tail=FALSE)
z2
# and then we need to move that value back 
# to our N( 4.3, 8.2)
x <- z1*8.2 + 4.3
x
# 
# or we could use the expanded qnorm as
qnorm( 1-0.29, mean=4.3, sd=8.2)
# or as
qnorm( 0.29, mean=4.3, sd=8.2, lower.tail=FALSE)