#  topic 11 c
#
     # first load the papfelton() function
     #
     #  this will actually loaded both papfelton() and 
     #  qapfelton()
source("../apfelton.R")
     #  Now solve the problems
papfelton( 3 )          # solves P(X<3)
1 - papfelton( 3.4 )    # solves P(X>3.4)
     # or, using the alternative approach
papfelton( 3.4, lower.tail=FALSE)  # also solves P(X>3.4)
papfelton( 1.63 ) - papfelton( -1.45 ) #solves P(-1.45<X<1.63)
papfelton(0.54) + (1-papfelton(2.04))  #solves P(X<0.54 or X>2.04)
     # or, using the alternative approach
papfelton(0.54)+papfelton(2.04,lower.tail=FALSE)
     #
     # now solve the "backwards" problem if finding a 
     # value, y, such that P(X<=y)=0.35.  We use
     # qapfleton() to do this.
qapfelton( 0.35 )
     # find y such that P(X>=y) = 0.17.
     # We could do this as
qapfelton( 1-0.17 )
     # or we could use the lower.tail=FALSE parameter
qapfelton( 0.17, lower.tail=FALSE)
     #
     #  start using pblumenkopf()
source("../blumenkopf.R")
     # find P( X < -2.4 )
pblumenkopf( -2.4 )
     # find P( X > 2.4 )
1 - pblumenkopf( 2.4 )   # the old, short way
pblumenkopf( 2.4, lower.tail=FALSE) # the way that is more clear
     # find P(-1.335 < X < 0.827 )
pblumenkopf( 0.827 ) - pblumenkopf( -1.335 )
     # find P( X <1.2 or X > 2.13 )
pblumenkopf( 1.2 ) + (1 - pblumenkopf(2.13) )    #old
pblumenkopf( 1.2 ) + pblumenkopf(2.13, lower.tail=FALSE)  #new
     # Find y such that P(X < y ) = 0.513
qblumenkopf( 0.513 )
     # Find y such that P( X > y ) = 0.823
qblumenkopf( 1 - 0.823 )  # old way
qblumenkopf( 0.823, lower.tail=FALSE )   # better way
     # Find y such that P( X < -y  or X > y ) = 0.123
qblumenkopf( 0.123/2, lower.tail=FALSE)
     # Find y such that P( -y < X < y ) = 0.950
qblumenkopf( (1-0.950)/2, lower.tail=FALSE)