252solnA1d 9/7/07 (Open this document in 'Page Layout' view!)
A. Parameter Estimation
1. Review of the Normal Distribution
A1
2. Point and Interval Estimation
3. A Confidence Interval for the Mean when the Population Variance is Known.
4. A Confidence Interval for the Mean when the Population Variance is not Known.
A2, text 8.21, 8.50 (8.21, 8.50) – Answers from both editions will be provided for 8.21. 8.95 on CD (8.93)
Graded Assignment 1 (Will be posted)
5. Deciding on Sample Size when working with a Mean
A3, 8.36 (8.36)
6. A Confidence Interval for a Proportion.
Text 8.22, 8.23, 8.24, 8.58, 8.93a,c on CD (8.22, 8.23,8.25, 8.58, 8.91a,c)
7. A Confidence Interval for a Variance.
Text 9.72 (9.67), A4
8. (A Confidence Interval for a Median.)
Optional - A5 -- solution is posted.
------------------------------------------------------------------------------------------------------------------------------------------------------------------
Solution to Problem A1 with diagrams is in this document.
PROBLEM A1: Let x ~ N 5, 6
Find:
a. P
x  3
b. Px  5
c. Px  6
d. Px  8
e. Px  2
f. P3  x  6
g. P6  x  8
h. P1  x  3
i. P3  x  27 
j. A symmetrical region about the mean with 60% of the probability
k. The 30th percentile of the distribution
l. The 70th percentile
m. x .05
n.
x.07
Solution: x ~ N 5, 6 . Make diagrams! Do not make diagrams of x with zero in the middle. Make up your
mind! If you are diagramming x , put the mean in the middle; if you are diagramming z put zero in the
x
middle. Remember z 
and that this equation implies that if we have z and need a value of x , as in

parts j-n, we use x    z . The diagrams below were generated by Minitab using a new version of a
program written by Jon Cryer at the University of Iowa. Because exact values of z were used to compute
the probabilities, the probabilities for x and for values of z are not always the same.
252solnA1d 9/7/07 (Open this document in 'Page Layout' view!)
A vertical line at the mean should be added to all x diagrams. A vertical line at zero should be added
to all z diagrams.
35

a) Px  3  P z 
  Pz  0.33   Pz  0  P0.33  z  0  .5  .1293  .3707
6 

For z make a diagram. Draw a Normal curve with a mean at 0. Indicate zero by a vertical line! Shade
the entire area below -0.33. Because this is on one side of zero, we must subtract the area between -0.33
and zero from the area below zero. If you wish, make a completely separate diagram for x . Draw a Normal
curve with a mean at 5. Indicate the mean by a vertical line! Shade the area below 3. This are is on one
side of the mean (5), so we subtract the area between 3 and the mean from the area below the mean. The
area below the mean in a symmetrical distribution is always .5000.
55

b) Px  5  P z 
  Pz  0  .5000
6 

For z make a diagram. Draw a Normal curve with a mean at 0. Indicate zero by a vertical line! Shade
the entire area below zero. We already know that this is .5000. If you wish, make a completely separate
diagram for x . Draw a Normal curve with a mean at 5. Indicate the mean by a vertical line! Shade the
area below 5. Since 5 is the mean, the area below the mean in a symmetrical distribution is always .5000.
252solnA1d 9/7/07 (Open this document in 'Page Layout' view!)
65

c) Px  6  P z 
  Pz  0.17   Pz  0  P0  z  0.17   .5  .0675  .5675
6 

For z make a diagram. Draw a Normal curve with a mean at 0. Indicate zero by a vertical line! Shade
the entire area below 0.17. Because this is on both sides of zero, we must add the area between 0.17 and
zero to the area below zero. If you wish, make a completely separate diagram for x . Draw a Normal curve
with a mean at 5. Indicate the mean by a vertical line! Shade the area below 6. This area is on both sides
of the mean (5), so we add the area between 6 and the mean to the area below the mean. The area below the
mean in a symmetrical distribution is always .5000.
85

d) Px  8  P z 
  Pz  0.50   Pz  0  P0  z  0.50   .5  .1915  .3085
6 

For z make a diagram. Draw a Normal curve with a mean at 0. Indicate zero by a vertical line! Shade
the entire area above 0.50. Because this is on one side of zero, we must subtract the area between 0.50 and
zero from the area above zero. If you wish, make a completely separate diagram for x . Draw a Normal
curve with a mean at 5. Indicate the mean by a vertical line! Shade the area above 8. This are is on one
side of the mean (5), so we subtract the area between 8 and the mean from the area above the mean. The
area above the mean in a symmetrical distribution is always .5000.
252solnA1d 9/7/07 (Open this document in 'Page Layout' view!)
25

e) Px  2  P z 
  Pz  0.50   P0.50  z  0  Pz  0  .1915  .5  .6915
6 

For z make a diagram. Draw a Normal curve with a mean at 0. Indicate zero by a vertical line! Shade
the entire area above -0.50. Because this is on both sides of zero, we must add the area between -0.50 and
zero to the area above zero. If you wish, make a completely separate diagram for x . Draw a Normal curve
with a mean at 5. Indicate the mean by a vertical line! Shade the area above 2. This area is on both sides
of the mean (5), so we add the area between 2 and the mean to the area above the mean. The area above the
mean in a symmetrical distribution is always .5000.
65
 35
z
f) P3  x  6  P
  P0.33  z  0.17   P0.33  z  0  P0  z  0.17 
6 
 6
 .1293  .0675  .1968
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the area between -0.33 and 0.17. Because this is on both sides of zero, we must add the area between
-0.33 and zero to the area between zero and 0.17. If you wish, make a completely separate diagram for x .
Draw a Normal curve with a mean at 5. Indicate the mean by a vertical line! Shade the area between 3
and 6. These numbers are on either side of the mean (5), so we add the area between 3 and the mean to the
area between the mean and 6.
252solnA1d 9/7/07 (Open this document in 'Page Layout' view!)
85
 65
z
g) P6  x  8  P
  P0.17  z  0.50   P0  z  0.50   P0  z  0.17 
6 
 6
 .1915  .0675  .1240
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the area between 0.17 and 0.50. Because this is on one side of zero, we must subtract the area
between zero and 0.17 from the larger area between zero and 0.50. If you wish, make a completely separate
diagram for x . Draw a Normal curve with a mean at 5. Indicate the mean by a vertical line! Shade the
area between 6 and 8. These numbers are on the same side of the mean (5), so we subtract the area between
the mean and 6 from the larger area between the mean and 8.
35
1 5
z
h) P1  x  3  P
  P0.67  z  0.33  P0.67  z  0  P0.33  z  0
6
6 

 .2486  .1293  .1193
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the area between -0.67 and -0.33. Because this is on one side of zero, we must subtract the area
between -0.33 and zero from the larger area between -0.33 and zero. If you wish, make a completely
separate diagram for x . Draw a Normal curve with a mean at 5. Indicate the mean by a vertical line!
Shade the area between 1 and 3. These numbers are on the same side of the mean (5), so we subtract the
area between 3 and the mean from the larger area between 1 and the mean.
252solnA1d 9/7/07 (Open this document in 'Page Layout' view!)
27  5 
 35
z
i) P3  x  27   P
  P0.33  z  3.67   P0.33  z  0  P0  z  3.67 
6 
 6
 .1293  .4999  .6292
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate zero by a vertical line! Shade
the area between -0.33 and 3.67. Because this is on both sides of zero, we must add the area between -0.33
and zero to the area between zero and 3.67. If you wish, make a completely separate diagram for x . Draw
a Normal curve with a mean at 5. Indicate the mean by a vertical line! Shade the area between 3 and 27.
This area is on both sides of the mean (5), so we add the area between 3 and the mean to the area between
the mean and 27.
Remember that the reverse of the formula z 
x

is x    z .
j) A symmetrical region about the mean with 60% of the probability.
The diagram: If you do a diagram for z , it will show two points, z .20 and z .80 . z .80 (which has 80%
above it!) is below zero and z .20 is above. Since zero is halfway between these two points, the diagram
will show 60% split between the two sides of zero, so that 30% is between z .80 and zero, and 30% is
between zero and z .20 . The probability below z .80 and the probability above z .20 are both 20%.
From the diagram, we want two points z .20 and z .80 so that Pz .80  z  z.20   .6000 . The
upper point, z .20 will have P0  z  z.20   .3000 , and by symmetry z.80   z.20 . From the interior of
the Normal table the closest we can come is P0  z  0.84   .2995 which means z .20  0.84 , and our
interval for z is -0.84 to 0.84. The interval for x can then be written
x    z .20  5  0.846  5  5.04 or -0.04 to 10.04.
10 .04  5 
  0.04  5
P 0.04  x  10 .04   P 
z
  P0.84  z  0.84 
6
6


 2P0  z  0.84   2(.2995 )  .5990  .60
Check:
252solnA1d 9/7/07 (Open this document in 'Page Layout' view!)
k) The 30th percentile of the distribution.
The diagram: The diagram for z will show one point, z.70   z.30 which has 30% below it (and 70%
above it!) and is below zero, since zero is the 50 th percentile. Since zero has 50% below it, the diagram will
show only 50% – 30% = 20% between zero and  z .30 .
From the diagram, we want one point z.70   z.30 so that Pz  z.70   .3000 . We can also see
that Pz.70  z  0  .2000 . Because z.70   z.30 , P0  z  z.30   .2000 . From the interior of the
Normal table the closest we can come to .2000 is P0  z  0.52   .1985 . So z .30  0.52 or z.70  0.52 ,
and the value of x can then be written x.70    z.70  5  0.526  5  3.12  1.88 .
1.88  5 

Check: To check this Px  1.88   P  z 
 Pz  0.52 
6 

 Pz  0  P0.52  z  0  .5  .1985  .3015  .30
l) The 70th percentile.
The diagram: The diagram for z will show one point, z .30 which has 70% below it (and 30% above it!)
and is above zero because zero is the 50th percentile. Since zero has 50% below it, the diagram will show
only 50% – 30% = 20% between zero and z .40 .
From the diagram, we want one point z .30 so that Pz  z.30   .7000 or P0  z  z.30   .2000 .
Remember, the 70th percentile has 70% below it! From the interior of the Normal table the closest we can
come to .2000 is P0  z  0.52   .1985 . This means that z .30  0.52 , so the value of x can then be
written x    z .30  5  0.526  5  3.12  8.12 .
8.12  5 

Check: Px  8.12   P  z 
 Pz  0.52   Pz  0  P0  z  0.52   .5  .1985  .6985  .70
6 

252solnA1d 9/7/07 (Open this document in 'Page Layout' view!)
m) x .05
x .05 is the point on the Normal distribution with 5% above it. It is thus the 95 th percentile. If we look at the
t table, we can find z .05  1.645 . Then x.05    z.05  5  1.645 6  14.87 . Note that we could have
found the values of z in j-n in the same way.
n) x.07
x.07 is the point on the Normal distribution with 7% above it. It is thus the 93 rd percentile.
The diagram: The diagram for z will show one point, z .07 , which has 7% above it (and 93% below it!)
and is above zero because zero has 50% above it. Since zero has 50% below it, the diagram will show 50%
– 7% = 43% between zero and z .07 .
From the diagram, we want one point z .07 so that Pz  z.07   .0700 or P0  z  z.07   .4300 .
Remember, x.07 has 7% above it! From the interior of the Normal table the closest we can come to .4300
is P0  z  1.48   .4306 . This means that z.07  1.48 , so the value of x can then be written
x.07    z.07  5  1.486  5  8.88  13 .88 .
Check:
or
13 .88  5 

Px  13 .88   P  z 
  Pz  1.48   Pz  0  P0  z  1.48   .5  .4306  .9306  .93
6


13 .88  5 

Px  13 .88   P  z 
  Pz  1.48   Pz  0  P0  z  1.48   .5  .4306  .0694  .07
6


252solnA1d 9/7/07 (Open this document in 'Page Layout' view!)
How the Diagrams Were Done
Two Minitab Macros, NormArea6c and NormArea6 are on the server under Minitab. Data and these
routines should be copied in the same file. The macros should appear with the suffix ‘.mac’ on them. If
they have ‘.txt’ on them, that suffix has to be used when they are called. For the creation of the 2005
diagrams, it seemed easier to use Normarea6a, which prompts the user. The macro Normarea6a and a
dummy worksheet were stored in a file in My Documents.
————— 9/7/2007 5:22:26 PM ————————————————————
Welcome to Minitab, press F1 for help.
#To get prompts the 'Editor' pulldown menu was used and 'enable commands' checked. The only thing that
had to be done was to load the dummy worksheet using the ‘File’ pull-down menu and to enter the Macro
command %normarea6. The macro then prompted for inout of the numbers. As an illustration parts of
problems a), d), and g) are shown below.
————— 9/7/2007 5:22:26 PM ————————————————————
Welcome to Minitab, press F1 for help.
#To get MTB prompts the 'Editor' pulldown menu was used and 'enable commands' checked.
MTB > WOpen "C:\Documents and Settings\RBOVE\My
Documents\Minitab\notmuch.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\notmuch.MTW'
Worksheet was saved on Thu Apr 14 2005
Results for: notmuch.MTW
MTB > %normarea6a
Executing from file: normarea6a.MAC
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> 0
DATA> 1
Do you want the area to the left of a value? (Y or N)
y
Enter the value for which you want the area to the left.
DATA> -0.33
...working...
Normal Curve Area
MTB > %normarea6a
Executing from file: normarea6a.MAC
252solnA1d 9/7/07 (Open this document in 'Page Layout' view!)
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> 5
DATA> 6
Do you want the area to the left of a value? (Y or N)
n
Do you want the area to the right of a value? (Y or N)
y
Enter the value for which you want the area to the right.
DATA> 8
...working...
Normal Curve Area
MTB > %normarea6a
Executing from file: normarea6a.MAC
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> 5
DATA> 6
Do you want the area to the left of a value? (Y or N)
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> 6
DATA> 8
...working...
Normal Curve Area
The graphs were arranged by using the ‘Editor’ pulldown menu and ‘layout tool.’ They were then saved
using the ‘File’ pull-down menu with ‘.jpg’ format.
252solnA1d 9/7/07 (Open this document in 'Page Layout' view!)
Normarea6c enables the user to run a number of problems at once. The data appears in the first
columns of the worksheet. A worksheet for Problems d-f appears below. NormArea 6c follows that to give
an explanation of the columns. It moves each row of the first 5 columns into column 6 and calls
NormArea6. Notice that when C5 is 1, the numbers in C3 are ignored. When C5 is 2, the numbers in C4
are ignored.
Results for: 252PrA1d-f.MTW
MTB > print c1-c5
#This is the data stored in the file 252PrA1d-f.MTW
Data Display
Row
1
2
3
4
5
6
C1
5
0
5
0
5
0
C2
6
1
6
1
6
1
C3
8.00
0.50
2.00
-0.50
3.00
-0.33
C4
9.00
9.00
9.00
9.00
6.00
0.17
C5
2
2
2
2
0
0
#Note that this is the original work and Normarea5c was used instead of Normarea6c.
#C1 contains means, C2 contains standard deviations
#C3 contains lower limits of intervals, C4 contains upper limits of intervals
#C5 contains 0 if Area is between points in C3 and C4
#C5 contains 1 if Area is to the left of point in c4
#C6 contains 2 if Area is to the right of point in C3
Assuming that Minitab knows where the Macro is, the whole thing is started off with ‘%NormArea5c.’ A
copy of the session follows.
————— 1/20/2005 11:38:08 PM ——————————————————
Welcome to Minitab, press F1 for help.
MTB > WOpen "C:\Documents and Settings\rbove\My
Documents\Minitab\252PrA1d-f.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252PrA1d-f.MTW'
Worksheet was saved on Thu Jan 20 2005
Results for: 252PrA1d-f.MTW
MTB > %NormArea5c
Executing from file: NormArea5c.MAC
Executing from file: NormArea5.MAC
Data Display
C6
5
6
8
...working...
9
2
Normal Curve Area
Data Display
C6
0.0
1.0
...working...
0.5
9.0
2.0
Normal Curve Area
Data Display
C6
5
6
2
...working...
9
2
Normal Curve Area
Data Display
C6
0.0
1.0
...working...
-0.5
9.0
2.0
252solnA1d 9/7/07 (Open this document in 'Page Layout' view!)
Normal Curve Area
Data Display
C6
5
6
3
...working...
6
0
Normal Curve Area
Data Display
C6
0.00
1.00
...working...
-0.33
0.17
0.00
Normal Curve Area
Every time the words ‘Normal Curve Area’ were printed out, a graph wass
produced. To arrange the graphs, use the ‘Editor’ menu and choose ‘Layout
Tool.’ The tool provides a dummy for the final arrangement of the graphs and
arrows can be used to move the graphs out of the layout. When an acceptable
layout is achieved, press ‘Finish.’ The result can now be copied or printed
out. Do not use the Minitab format for the graphs. I used jpg.