251solngr4-05a 4/14/05
Graded Assignment 4
(Open this document in 'Page Layout' view!)
Name:
Class days and time:
Assume that x ~ N (3,11) . Do the following:
Assignment will not be accepted without diagrams!
a. P0.50  x  4.00 
b. P0.21  x  0.21
c. F 0.04   Px  0.04 
d. F 4.00   Px  4.00 
e. P3.00  x  36.00 
f. P0  x  2.00 
g. P 4.00  x  4.00 
h. Find a symmetrical interval about the mean with 66% probability.
i. Find x.34 (the 66th percentile).
j. Find x.66 (What percentile is this?).
Before we start: You are supposed to know the following things about the Standardized Normal
Distribution. In the following rules z represents a variable with the Standardized Normal Distribution. Any
other random variable is x .
1) The Distribution is symmetrical about zero.
This means that you never have to look up a probability between, say -2 and zero on a
conventional Normal table. You instead look up the probability between zero and 2.
P2  z  0  P0  z  2
2) The area under the Normal curve, like the area under the curve for all continuous
distributions, is 1.
Since the zero is the halfway point under the curve, this implies the following:
Pz  0  .5 and Pz  0  .5
3) The standard notation to say that a given distribution is Normal with a certain mean and
standard deviation is ~ N  ,   .
This means that, if z represents a variable with the Standardized Normal Distribution,
we write z ~ N 0, 1 and, if x is Normally distributed with a mean of 3 and a standard
deviation of 11, we write x ~ N 3, 11 .
4) Any probability for a Normally distributed variable can be found using the Standardized
x
Normal Distribution by using the transformation z 
, even if x is zero.

5) There is no such thing as a probability that is above 1 or below zero.
Solution: Material in italics below is a description of the diagrams you were asked to make or a general
explanation and will not be part of your written solution. The x and z diagrams should look similar. If
you know what you are doing, you only need one diagram for each problem. General comment - I can't
give you much credit for an answer with a negative probability because there is no such thing!!!
251solngr4-05a 4/14/05
4  3
 0.50  3
z
 P 0.23  z  0.09 
a. P0.50  x  4.00   P 
11 
 11
 P0.23  z  0  P0  z  0.09   .0910  .0359  .1269
Make a diagram! For x ~ N (3,11) make a Normal curve centered at 3 and shade the area from 0.5 to 4;
for z make a Normal curve centered at zero and shade the area from -0.23 to 0.09. Because the area is on
both sides of the mean, you are adding. Add a vertical line at 3 to the x graphs and a vertical line at zero
to the z graphs!
Normal Curv e with Mean 3 and Standard Dev iation 11
Normal Curv e with Mean 0 and Standard Dev iation 1
The Area Between 0.5 and 4 is 0.1261
The Area Between -0.23 and 0.09 is 0.1268
0.04
0.4
0.3
Density
Density
0.03
0.02
0.01
0.00
0.2
0.1
-50
-25
0
Da ta A x is
25
0.0
50
-5.0
-2.5
0.0
Da ta A x is
2.5
5.0
0.21  3 
  0.21  3
z
 P 0.29  z  0.25 
b. P 0.21  x  0.21  P 
11 
 11
 P0.29  z  0  P0.25  z  0  .1141  .0987  .0154
For x ~ N (3,11) make a Normal curve centered at 3 and shade the area from -0.21 to 0.21; for z make a
Normal curve centered at zero and shade the area from -0.29 to -0.25. Because the area is on one side of
the mean you are subtracting. If you really try to draw this to scale, you will find out that the area is tiny.
If you can read the numbers over the graphs on the next page, you will find that the probabilities are
different by about 7% due to the fact that when z is close to zero, rounding errors have relatively large
effects.
N
ormal Curv e with Mean 3 and Standard Dev iation 11
Norm al Curv e with Mean 0 and Standard Dev iation 1
The Area Between -0.21 and 0.21 is 0.0147
The Area Between -0.29 and -0.25 is 0.0154
0.04
0.4
0.3
Density
Density
0.03
0.02
0.01
0.00
0.2
0.1
-50
-25
0
Da ta A x is
25
50
0.0
-5.0
-2.5
0.0
Da ta A x is
2.5
5.0
251solngr4-05a 4/14/05
0.04  3 

 Pz  0.27 
c. F 0.04   Px  0.04   P  z 
11 

 Pz  0  P0.27  z  0  .5  .1064  .3936
For x ~ N (3,11) make a Normal curve centered at 3 and shade the area below (left of) 0.04; for z make a
Normal curve centered at zero and shade the area below -0.27. Because the area is on one side of the mean
you are subtracting.
N
ormal Curv e with Mean 3 and Standard Dev iation 11
Norm al Curv e with Mean 0 and Standard Dev iation 1
The Area to the Left of 0.04 is 0.3939
The Area to the Left of -0.27 is 0.3936
0.04
0.4
0.3
Density
Density
0.03
0.02
0.01
0.00
0.2
0.1
-50
-25
0
Da ta A x is
25
0.0
50
-5.0
-2.5
0.0
Da ta A x is
2.5
5.0
4.00  3 

 Pz  0.09 
d. F 4.00   Px  4.00   P  z 
11 

 Pz  0  P0  z  0.09   .5  .0359  .5359
For x ~ N (3,11) make a Normal curve centered at 3 and shade the (entire!) area below (left of) 4; for z
make a Normal curve centered at zero and shade the area below 0.09. Because the area is on both sides of
the mean, you are adding.
N
ormal Curv e with Mean 3 and Standard Dev iation 11
Norm al Curv e with Mean 0 and Standard Dev iation 1
The Area to the Left of 4 is 0.5362
The Area to the Left of 0.09 is 0.5359
0.04
0.4
0.3
Density
Density
0.03
0.02
0.01
0.00
0.2
0.1
-50
-25
0
Da ta A x is
25
0.0
50
-5.0
-2.5
0.0
Da ta A x is
2.5
5.0
40 .00  3 
 3.00  3
z
 P0  z  3.36   .4996 This number comes from
e. P3.00  x  40.00   P 
11 
 11
the lower part of the table. Because this area begins at the mean, we do not add or subtract. For
x ~ N (3,11) make a Normal curve centered at 3 and shade the area between 3 and 40 - almost the entire
area to the right of 3; for z make a Normal curve centered at zero and shade the area between zero and
3.36.
N
ormal Curv e with Mean 3 and Standard Dev iation 11
Normal Curv e with Mean 0 and Standard Dev iation 1
The Area Between 3 and 36 is 0.4987
The Area Between 0 and 3 is 0.4987
0.04
0.4
0.3
Density
Density
0.03
0.02
0.01
0.00
0.2
0.1
-50
-25
0
Da ta A x is
25
50
0.0
-5.0
-2.5
0.0
Da ta A x is
2.5
5.0
251solngr4-05a 4/14/05
2.00  3 
0  3
z
 P 0.27  z  0.09 
f. P0  x  2.00   P 
11 
 11
 P0.27  z  0  P0.09  z  0  .1064  .0359  .0705
For x ~ N (3,11) make a Normal curve centered at 3 and shade the area from 0 to 2; for z make a
Normal curve centered at zero and shade the area from -0.27 to -0.09. Because the area is on one side of
the mean you are subtracting. If you can read the notation on the graphs below, you will see that both
answers disagree slightly with the solution above. The first discrepancy is due to a rounding error and the
second is due to the fact that the Normal table is limited to 4 decimal places. Both are larger because the
values of z are close to zero.
N
ormal Curv e with Mean 3 and Standard Dev iation 11
Norm al Curv e with Mean 0 and Standard Dev iation 1
The Are a Between 0 and 2 is 0.0713
The Area Between -0.27 and -0.09 is 0.0706
0.04
0.4
0.3
Density
Density
0.03
0.02
0.01
0.00
0.2
0.1
-50
-25
0
Da ta A x is
25
0.0
50
-5.0
-2.5
0.0
Da ta A x is
2.5
5.0
4.00  3 
  4.00  3
z
 P 0.64  z  0.09 
g. P 4.00  x  4.00   P 
11
11 

 P0.64  z  0  P0  z  0.09   .2389  .0359  .2748
For x ~ N (3,11) make a Normal curve centered at 3 and shade the area from -4 to 4; for z make a
Normal curve centered at zero and shade the area from -0.64 to 0.09. Because the area is on both sides of
the mean, you are adding.
N
ormal Curv e with Mean 2 and Standard Dev iation 11
Norm al Curv e with Mean 0 and Standard Dev iation 1
The Area Between -4 and 4 is 0.2794
The Area Between -0.64 and 0.09 is 0.2748
0.04
0.4
0.3
Density
Density
0.03
0.02
0.01
0.00
0.2
0.1
-50
-25
0
Da ta A x is
25
50
0.0
-5.0
-2.5
0.0
Da ta A x is
2.5
5.0
251solngr4-05a 4/14/05
In general, for the following problems remember that z .02 is a point with 2% above it and: 1) The values of
z you need here must came from the z-table (Table of the Standardized Normal Distribution), not the ttable because they aren't on the t-table. You can find values like z .25 or z .40 on the t-table, but I didn't ask
for them. 2) Numbers like z .24 are values of z not probabilities, you can't find them by taking 0.24 or .76
on the z part of the table and then reading a probability and claiming that it is a value of z . 3) If the
request is for an interval, two solutions are needed, but if the request is for a percentile, only one number is
acceptable. An answer for a percentile including  will not get full credit. Note also that since, in the
Normal distribution, the mean is the 50th percentile, a percentile below 50 is below the mean and a
percentile above 50 is above the mean.
h. Find a symmetrical interval about the mean with 66% probability.
Solution: Make a diagram. The diagram for z will show a central area with a probability of 66%. It is
split in two by a vertical line at zero into two areas with probabilities of 33%. The tails of the distribution
each have a probability of 50% - 33% = 17%. From the diagram, we want two points z .17 and z .83 so
that Pz .83  z  z .17   .6600 . The upper point z .17 will have P0  z  z .17  
66 %
 .3300 ,
2
and by symmetry z.83   z.17 . From the interior of the Normal table the closest we can come to .3300 is
P0  z  0.95   .3289 or P0  z  0.96   .3315 . Though either z .17  0.95 or z .17  0.96 is an
acceptable answer, in fact 0.95 is closer, so I will use z .17  0.95 , and our interval for z is -0.95 to 0.95.
Since x ~ N (3,11) , the diagram for x (if we bother) will show 66% probability split in two 33%
regions on either side of 3, with 17% above x.17 and 17% below x.83 . The interval for x can then be
written x    z .17  3  0.9511  3  10 .45 or -7.45 to 13.45.
To check this:
13 .45  3 
  7.45  3
z
 P 0.95  z  0.95 
P7.45  x  13 .45   P 
11
11 

 P0.95  z  0  P.0  z  0.95   2P0  z  0.95   2(.3289 )  .6578  66%.
Normal Curv e with Mean 0 and Standard Dev iation N
1ormal Curv e with Mean 3 and Standard Dev iation 11
The Area Between -0.95 and 0.95 is 0.6579
The Area Between -7.45 and 13.45 is 0.6579
0.04
0.4
0.03
Density
Density
0.3
0.2
0.01
0.1
0.0
0.02
-5.0
-2.5
0.0
Da ta A x is
2.5
5.0
0.00
-50
-25
0
Da ta A x is
25
50
i. Find x.34 (the 66th percentile).
Solution: Make a diagram. The diagram for z will show an area with a probability of 66% below z .34 .
It is split by a vertical line at zero into two areas. The lower one has a probability of 50% and the upper
one a probability of 66% - 50% =16%. The upper tail of the distribution has a probability of 34%, so that
the entire area above 0 adds to 50%. From the diagram, we want one point z .34 so that Pz  z 34   .66
or P0  z  z 34   .1600 . The closest we can come is P0  z  0.41  .1591 . We can say z.34  0.41
though 0.42 would be acceptable.
Since x ~ N (3,11) , the diagram for x would show 66% probability split in two regions on
either side of 3 with probabilities of 50% below 3 and 16% above 3 and below x.34 , and with 34% above
x.34 . z.34  0.41 , so the value of x can then be written x    z .34  3  0.4111  3  4.51  7.51.
251solngr4-05a 4/14/05
To check this:
7.51  3 

Px  7.51  P  z 
 Pz  0.41  Pz  0  P0  z  0.41  .5000  .1591  .6591  .66. So
11 

7.51 is about the 66th percentile. Remember that you should add a vertical line at 3(the mean) to the x
graphs and a vertical line at zero to the z graphs!
N
ormal Curv e with Mean 0 and Standard Dev iation N
1orm al Curv e with Mean 3 and Standard Dev iation 11
The Area to the Left of 0.41 is 0.6591
The Area to the Left of 7.51 is 0.6591
0.04
0.4
0.03
Density
Density
0.3
0.2
0.01
0.1
0.0
0.02
-5.0
-2.5
0.0
Da ta A x is
2.5
0.00
5.0
-50
-25
0
Da ta A x is
25
50
j. Find x.66 (What percentile is this?).
Solution: Since 66% is above this point, 100% - 66% = 34% is below this point and it is the 34th percentile.
Make a diagram. The diagram for z will show an area in the left tail with a probability of 34% to the
left of z .66 . Above zero there is an area of 50%. So between z .66 and zero there is a probability of 50% 34% = 16% (or 66% - 50% = 16%). A diagram for x ~ N (3,11) would be pretty much the same, except that
3 would replace zero, so that we could clearly see that we are looking for a number below 3. From the
diagram, we want one point z.66   z.34 so that Pz  z.66   .3400 . We can also see that
Pz.66  z  0  .1600 . Because z.66   z.34 , P0  z  z.34   .1600 . From the interior of the Normal
table the closest we can come is P0  z  0.41  .1591 or P0  z  0.42   .1628 . Though either of
these would be acceptable, 0.41 is closer, so let’s say z.34  0.41 or z.66  0.41 , so the value of
then be written x    z.66    z 34  3  0.4111  3  4.51  1.51.
x can
 1.51  3 

To check this: Px  1.51  P  z 
 Pz  0.41  Pz  0  P0  z  0.41
11 

 1.51  3 

 .5000  .1591  .3409  .34. Or we could try Px  1.51  P  z 
 Pz  0.41
11 

 Pz  0  P0.41  z  0  .5  .1591  .6591  .66 . So -3.39 is about the 34th percentile or x.66 .
.
Normal Curv e with Mean 0 and Standard Dev iation N
1ormal Curv e with Mean 3 and Standard Dev iation 11
The Area to the Left of -0.41 is 0.3409
The Area to the Left of -1.51 is 0.3409
0.04
0.4
0.03
Density
Density
0.3
0.2
0.01
0.1
0.0
0.02
-5.0
-2.5
0.0
Da ta A x is
2.5
5.0
0.00
-50
-25
0
Da ta A x is
25
50
251solngr4-05a 4/14/05
Try it, You’ll like it!
In order to make diagrams like the one here, open Minitab. Make sure that, in the ‘editor’ pull down menu,
‘output editable’ and ‘enable commands’ are checked. If you don’t see these, put your cursor in the
‘session’ window and pull down ‘editor’ again. Put absolutely anything on the worksheet and use the ‘file’
pull-down menu and ‘save as’ (to any file you can put it in and with any name you want) to create a
connection between your commands and a file. Now copy the following instructions using Word and use
the ‘tools’ menu in Minitab to go to Notepad and open a new document and paste these instructions into it.
Save these as
normarea5A.txt in the same file that you saved the worksheet to.
Now call this macro from the session window by typing in %normarea5A.txt after the MTB> prompt.
------------------------------------------------------------------------------------------------------------------- ----------Here is the text of the macro.
GMacro
NormArea4 # For Release 14 - low density version of NormArea3
# Originally written by Jon Cryer: Jon-Cryer@uiowa.edu
Note Graphic display of normal curve areas
Note Finds and displays areas to the left or right of a given value
Note or between two values. (This macro uses C100-C116 and K100-K116)
Note
Note Enter the mean and standard deviation of the normal curve.
Set c100;
file 'terminal';
NObs 2.
# In this version, input data is in C6 - mean, std. dev., lower limit, upper limit,
# c6(5) is 0 if Area is between C6(3) and C6(4)
# c6(5) is 1 if Area is to the left of C6(4)
# c6(5) is 2 if Area is to the right of C6(3)
#Let C100(1)= c6(1)
#Let C100(2)= c6(2)
Let K100 = C100(1)
Let K101 = C100(2)
#Let K115 = C6(5)
Note
Note Do you want the area to the left of a value? (Y or N)
YesNo K115
If K115 = 1
Call Left
EndIf
If K115 = 1
Goto 1
EndIf
Note Do you want the area to the right of a value? (Y or N)
YesNo K115
If K115 = 1
Let K115 = 2
EndIf
If K115 = 2
Call Right
EndIf
If K115 = 0
Call Between
EndIf
Mlabel 1
Erase C100-C116 K100-K116
251solngr4-05a 4/14/05
EndMacro
GMacro # Want area to the left
Left
Note
Note Enter the value for which you want the area to the left.
Set c100;
file 'terminal';
NObs 1.
#Let c100(1) = c6(4)
Note
Note ...working...
Let K102 = C100(1)
Let K103 = K100-4*K101
let K104 = K100+4*K101
#In Cryer's original version k105 was .001*k101
Let K105 = .01*K101
Set C101
K103:K104/K105
End
PDF C101 C102;
Normal K100 K101.
Set C103
K103:K102/K105
End
Let K114 = N(C103)
Let C103(K114) = K102
PDF C103 C104;
Normal K100 K101.
CDF K102 C106; # Get area to report
Normal K100 K101.
Text C106 C106; # Convert to text for labeling
decimal 4.
Let K109 = C106
Let C107 = K102
Text C107 C107
Let K107 = C107
Let K110 = "The Area to the Left of "
KKCAT K110 K107 K112
Let K111 = " is "
KKCAT K112 K111 K112
KKCAT K112 K109 K112
# Get mean and stdev labeling
Let C108 = K100
Text C108 C108
Let K103 = C108 # Mean as text
Let C109 = K101
Text C109 C109
Let K104 = C109 # Stdev as text
Let K105 = "Normal Curve with Mean "
Let K106 = " and Standard Deviation "
KKCAT K105 K103 K111
KKCAT K111 K106 K111
KKCAT K111 K104 K111
Let K113 = -0.01/K101 # Set marker placement a little below zero
251solngr4-05a 4/14/05
Plot C102*C101 C104*C103;
NoLegend;
WTitle "Normal Curve Area";
Title K111;
Title K112;
Connect;
Size 2;
Color 1;
Area;
Type 0 1;
Color 1 4;
AXLabel 1 "Data Axis";
AXLabel 2 "Density";
Marker K102 K113;
Type 19;
Color 1;
Size 2.0;
Scale 2;
Min 0;
Overlay.
Return
EndMacro
GMacro # Want area to the right
Right
Note
Note Enter the value for which you want the area to the right.
Set c100;
file 'terminal';
NObs 1.
#Let c100(1) = c6(3)
Note
Note ...working...
Let K102 = C100(1)
Let K103 = K100-4*K101
let K104 = K100+4*K101
#In Cryer's original version k105 was .001*k101
Let K105 = .01*K101
Set C101
K103:K104/K105
End
PDF C101 C102;
Normal K100 K101.
Set C103
K102:K104/K105
End
Let K114 = N(C103)
Let C103(K114) = K104
PDF C103 C104;
Normal K100 K101.
CDF K102 C106; # Get area to report
Normal K100 K101.
Let C106 = 1 - C106
Text C106 C106; # Convert to text for labeling
decimal 4.
Let K109 = C106
251solngr4-05a 4/14/05
Let C107 = K102
Text C107 C107
Let K107 = C107
Let K110 = "The Area to the Right of "
KKCAT K110 K107 K112
Let K111 = " is "
KKCAT K112 K111 K112
KKCAT K112 K109 K112
# Get mean and stdev labeling
Let C108 = K100
Text C108 C108
Let K103 = C108 # Mean as text
Let C109 = K101
Text C109 C109
Let K104 = C109 # Stdev as text
Let K105 = "Normal Curve with Mean "
Let K106 = " and Standard Deviation "
KKCAT K105 K103 K111
KKCAT K111 K106 K111
KKCAT K111 K104 K111
Let K113 = -0.01/K101 # Set marker placement a little below zero
Plot C102*C101 C104*C103;
NoLegend;
WTitle "Normal Curve Area";
Title K111;
Title K112;
Connect;
Size 2;
Color 1;
Area;
Type 0 1;
Color 1 4;
AXLabel 1 "Data Axis";
AXLabel 2 "Density";
Marker K102 K113;
Type 19;
Color 1;
Size 2.0;
Scale 2;
Min 0;
Overlay.
Return
EndMacro
GMacro # Want area between two values
Between
Note
Note Enter the two values for which you want the area between.
Set c100;
file 'terminal';
NObs 2.
#let c100(1)= c6(3)
#let c100(2) = c6(4)
Note
Note ...working...
251solngr4-05a 4/14/05
Let K102 = C100(1)
Let K115 = C100(2)
Let K103 = K100-4*K101
let K104 = K100+4*K101
#In Cryer's original version k105 was .001*k101
Let K105 = .01*K101
Set C101
K103:K104/K105
End
PDF C101 C102; # Get values for full normal curve
Normal K100 K101.
Set C103
K102:K115/K105
End
Let K114 = N(C103)
Let C103(K114) = K115
PDF C103 C104;
Normal K100 K101.
CDF K102 C106; # Get area to report
Normal K100 K101.
CDF K115 C110;
Normal K100 K101.
Let C106 = C110-C106
Text C106 C106; # Convert area to text for labeling
decimal 4.
Let K109 = C106
Let C107 = K102 # Convert lower value to text
Text C107 C107
Let K107 = C107
Let C107 = K115
Text C107 C107 # Convert upper value to text
Let K116 = C107
Let K110 = "The Area Between "
Let K111 = " and "
Let K113 = " is "
KKCAT K110 K107 K112
KKCAT K112 K111 K112
KKCAT K112 K116 K112
KKCAT K112 K113 K112
KKCAT K112 K109 K112
# Get mean and stdev for labeling
Let C108 = K100
Text C108 C108
Let K103 = C108 # Mean as text
Let C109 = K101
Text C109 C109
Let K104 = C109 # Stdev as text
Let K105 = "Normal Curve with Mean "
Let K106 = " and Standard Deviation "
KKCAT K105 K103 K111
KKCAT K111 K106 K111
KKCAT K111 K104 K111
Let K113 = -0.01/K101 # Set marker placement a little below zero
Let K114 = -0.01/K101
Plot C102*C101 C104*C103;
251solngr4-05a 4/14/05
NoLegend;
WTitle "Normal Curve Area";
Title K111;
Title K112;
Connect;
Color 1;
Size 2;
Area;
Type 0 1;
Color 1 4;
AXLabel 1 "Data Axis";
AXLabel 2 "Density";
Marker K102 K113;
Type 19;
Color 1;
Size 2.0;
Marker K115 K114;
Type 19;
Color 1;
Size 2.0;
Scale 2;
Min 0;
Overlay.
Return
EndMacro
Here is what happened when I saved my nonsense file and called the normarea5A macro. The only things
that I wrote in the session window were the call to the macro and the responses to the prompts below.
————— 4/14/2005 9:02:13 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > Save "C:\Documents and Settings\rbove\My Documents\Minitab\notmuch.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\notmuch.MTW'
Existing file replaced.
Results for: notmuch1.MTW
MTB > WSave "C:\Documents and Settings\rbove\My
Documents\Minitab\notmuch1.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\notmuch1.MTW'
MTB > %normarea5A.txt
Executing from file: normarea5A.txt
Graphic display of normal curve areas
Finds and displays areas to the left or right of a given value
or between two values. (This macro uses C100-C116 and K100-K116)
Enter the mean and standard deviation of the normal curve.
DATA> 3
DATA> 11
Do you want the area to the left of a value? (Y or N)
251solngr4-05a 4/14/05
n
Do you want the area to the right of a value? (Y or N)
n
Enter the two values for which you want the area between.
DATA> .50
DATA> 4
...working...
Normal Curve Area
Now the computer produced the graph you see at the beginning of the paper. In
order to copy it into a word document, I used the ‘editor’ pulldown menu and
‘layout tool.’ The layout tool allowed me to use arrows to move my graphs and
use ‘rows’ to limit the number of rows to 1. This produced a rather steep
looking graph, but I hit the ‘finish’ button, which made it look a little
better. I used the ‘edit’ pulldown menu and ‘copy graph’ to copy the graph,
which I then pasted into a Word document. I clicked on the graph in the Word
document and used my cursor to pull up the bottom of the graph to improve its
shape.
© Roger Even Bove 2005