251y0611 2/17/06
ECO251 QBA1
FIRST EXAM
February 16 and 17, 2006
ECO251 QBA1
Name: ____Key_____________
Student Number : _____________________
Class Hour: _____________________
Remember – Neatness, or at least legibility, counts. In most non-multiple-choice questions an answer
needs a calculation or short explanation to count.
Part I. (7 points)
(Source: Harvey J. Brightman) The following numbers are a sample and represent the pulse rates of 10
well-conditioned athletes.
31, 33, 36, 37, 37, 47, 44, 41, 38, 39.
Compute the following: Show your work!
a) The Median (1)
b) The Standard Deviation (3)
c) The 31st percentile (2)
d) The Coefficient of variation (1)
Solution:
Index
x
1
2
3
4
5
6
7
8
9
10
x2
31
961
33 1089
36 1296
37 1369
37 1369
47 2209
44 1936
41 1681
38 1444
39 1521
383 14875
x in order
31
33
36
37
37
38
39
41
44
47
a) Median: pn  1  .511  5.5  a.b The middle numbers are the 5th and 6th number, which are 37 and
x5  x 6
 37 .5.
2
 x5  .5( x 6  x5 )  37  .538  37   37  0.5  387 .5
38. The median is the average of two middle numbers. x.50 
Or x1 p  xa  .b( xa1  xa ) . So x.50
b) Standard Deviation: x 
 x  383  38.3 , s   x
2
n
10
2
 nx 2
n 1

14875  10 38 .32
206 .1

 22 .90 .
9
10  1
So s  22.9  4.7854 . Note that saying a standard deviation is zero is equivalent to saying that all values
of x are the same.
c) 31st percentile: pn  1  .3111  3.41 . So a  3 and .b  0.41
x1 p  xa  .b( xa1  xa ) so x1.31  x.69  x3  0.41( x 4  x3 )  36  0.41(37  36)  36.41
d) C 
s 4.7854

 0.1249 or 12.49%
x
38 .3
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251y0611 2/17/06
How this was done on Minitab – Version 1
Data x  was placed in c2 (column 2).
MTB > describe c2
Descriptive Statistics: C2
Variable
C2
N
10
N*
0
Mean
38.30
SE Mean
1.51
StDev
4.79
Minimum
31.00
MTB > let c4 = c2*c2
MTB > sort c2 c6;
Compute x 2 in column 4.
SUBC> by c2.
x in order is put in column 6.
Q1
35.25
Median
37.50
Q3
41.75
Maximum
47.00
MTB > sum c2
Sum of C2
Sum of C2 = 383
MTB > sum c4
Sum of C4
Sum of C4 = 14875
MTB > print c2 c4 c6
Data Display
Row
1
2
3
4
5
6
7
8
9
10
C2
31
33
36
37
37
47
44
41
38
39
C4
961
1089
1296
1369
1369
2209
1936
1681
1444
1521
C6
31
33
36
37
37
38
39
41
44
47
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251y0611 2/17/06
Part II. (At least 35 points – 2 points each unless marked - Parentheses give points on individual
questions. Brackets give cumulative point total.) Exam is normed on 50 points.
1.
(Brightman) At an urban university, there are 7000 undergraduates whose ages are between 18
and 23, 2000 undergraduates between 24 and 29 years old, 1000 undergraduates between 30
1nd 35 years old and 1000 who are older than 35.
a) Without doing any math, explain in plain English whether the mean will be below, the
same as or above the median and why. (2)
Answer: The 1000 people above 35 will pull the mean way up above the median. This
distribution is skewed to the right.
b) Where will the mode be relative to the mean and median? (1)
[3]
Answer: Since the median is usually between the mean and the mode, the mode must lie to
the left of the mean and mode. If you made a diagram, it would show, left to right, mean
median mode.
2.
I have the average time of 10 randomly picked runners in the Boston Marathon.
a) Is this a parameter or a statistic?
Answer: This describes a sample and must be a statistic. A parameter describes a population.
b) What symbol should you use to indicate this mean?
Answer: The symbol for a sample mean is x .
[5]
3.
For a rather shapeless distribution with one mode, a mean of 100 and a standard deviation of
2, we can say that the percent of data falling between 80 and 120 is
a) At least 90%
b) At most 90%
c) 100%
d) *At least 99%
e) At most 99%
f) None of the above.
[7]
Explanation: 120 is 100 + 10(2). 80 is 100 - 10(2). The Tchebyschev inequality says that the
fraction of measurements more than k  10 standard deviations from the mean is, at most,
1  1
 1100 . So the fraction between 80 and 120 is at least 1  1100  99100  99% .
k2
102
4.
For a mound-shaped (symmetrical) distribution with one mode, a mean of 100 and a standard
deviation of 2, we can say that the percent of data falling above 96 is
a) *About 97.5%
b) About 95%
c) Almost 100%
d) About 68%
e) None of the above
Explanation: 96 is 100 - 2(2), i.e. 2 standard deviations below 100. The Empirical Rule says
that 95% of data points will fall within 2 standard deviations of the mean. 5% will be in the
tails of the distribution, because of symmetry, half (2.5%) will be above 100 + 2(2) = 104. The
total is thus 95% + 2.5% = 97.5%.
5.
The drawing of inferences about an unknown whole from a known part is
a) Deductive reasoning
b) *Inductive reasoning
c) Census taking
d) Sampling
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251y0611 2/17/06
e)
None of the above.
[11]
6.
Observations about a discrete quantitative variable
a) Can be made in only two categories
b) *Can assume values only at specific points of a scale of values with inevitable gaps
between these points.
c) Can assume values at all points of a scale of values with no breaks in between possible
values.
d) Cannot be meaningfully multiplied or divided.
e) Both b) and d) are true.
[13]
7.
Mark the variables below as qualitative or categorical (A), quantitative and continuous (B1) or
quantitative and discrete (B2) (1 each)
a) GPA
B1
b) Number of credits earned
B2
c) Major area of study
A
d) Grade obtained in a statistics course.
A
[17]
Explanation: The grade obtained in a statistics course is ordinal data, which is a form of
qualitative data.
8.
If I double all of the incomes in a sample of 1000 people, mark below which of the statistics
will change.
a) Pearson’s measure of skewness
b) The coefficient of variation.
c) *The mean
d) All of the above will change
e) None of the above will change
.
[19]
Explanation: Only the mean will change. Pearson’s measure and the coefficient of variation
are both dimension-free quantities, so both the numerator and denominator will double.
4
251y0611 2/17/06
TABLE 2-13
Given below is the stem-and-leaf display representing the amount of detergent used in gallons (with
leaves in 10ths of gallons) in a month by 25 drive-through car wash operations in Phoenix. (Ng p57)
9 | 147
10 | 02238
11 | 135566777
12 | 223489
13 | 02
9.
In table 12.3, if a percentage histogram is constructed using 9.0 to 9.9 as the first class, what
percent will be in the 12-12.9 class?
[21]
Solution: 24%. The frequency table would read
Class
f rel
f
9- 9.9
10-10.9
3
3
5
5
11-11.9
9
12-12.9
6
13-13.9
2
Sum n  25
25
 12 %
25 
20 %
25  36 %
6  24 %
25
2  8%
25
9
10. In table 12.3 find the median amount of detergent used.
position  pn  1  .526   13.0  a.b . The 13th number is 11.6.
[23]
11. Using the data in table 12.3. Assume that the data is to be presented in 6 classes, show how
you would decide what class interval to use and list the classes below with their frequencies.
(5)
[28]
13 .2  9.1
 0.6833 . If you use
Solution: The numbers go from 9.1 to 13.2. The calculation is thus
6
1, your last class will be empty, so try .7 or .75. or .8. But note that your classes will not cover all
the data if you use .7 and start at 9.0!
9 | 147
10 | 02238
11 | 135566777
12 | 223489
13 | 02
A
B
C
D
E
F
9.0
9.8
10.6
11.4
12.2
13.0
Class
to under
to under
to under
to under
to under
to under
9.8
10.6
11.4
12.2
13.0
13.8
Frequency
3_
4_
3_
7_
6_
2_
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251y0611 2/17/06
12. In Problem 3.42 in the text, data on waiting times in a bank in a commercial district is given.
Assume that the 5-number summary is {0.38, 3.20, 4.50, 5.55, 10.03}. Use this to make a
horizontal box plot, but first find the interquartile range. Then do the following: An upper
fence is defined as Q3 + 1.5(IQR) and a lower fence is Q1 – 1.5(IQR). Indicate the fences by
vertical lines at the end of the whiskers in your box plot – do not let the whiskers extend
beyond the fences, but only show a fence if there is data beyond it. Any points beyond the
fence should be represented by dots. (4)
[32]
Drawing: IQR = 5.55 - 3.20 = 2.35. 1.5(IQR) = 3.525. Lower fence is 3.20 – 3.525 = -0.325 .
Upper fence is 5.55 + 3.525 = 9.075. Whiskers go from 0.38 to 3.20 and 5.55 to 9.075. Box goes
from 3.20 to 5.55. Median stripe is at 4.50. Using the word drawing tool, this is the closest I could
come.
-1
0
1
2
3
4
5
6
7
8
9
10
11
13. What characteristic do the mean, median and mode have that they do not share with the
variance and the interquartile range? (1)
[33]
Answer: They are all measures of central tendency.
14. The following data represents proven oil deposits in billions of barrels divided by region.
Show this as a Pareto chart. (5)
[37]
Billions of
Per Cent of
Barrels
Total
North America
54.8
5.3
Central and South America
Western Europe
Africa
Middle East
Far East and Oceania
Eastern Europe and CIS
95.2
17.2
74.9
683.6
44.0
59.0
9.3
1.6
7.3
66.5
4.3
5.7
Drawing: Your graph should be a bar graph with a line above the bars with the following points
marked:
Region
Location
of line
Height of
bar
Middle
East
66.5%
66.5%
C and S
America
75.8%
Africa
83.1%
E Europe
and CIS
88.8%
9.3%
7.3%
5.7%
N America
94.1%
5.3%
F East and
Oceania
98.4%
4.3%
W.Europe
100%
1.6%
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251y0611 2/17/06
ECO251 QBA1
FIRST EXAM
February 16 and 17, 2006
TAKE HOME SECTION
Name: _________________________
Student Number: _________________________
Throughout this exam show your work! Please indicate clearly what sections of the problem you are
answering and what formulas you are using. Turn this is with your in-class exam.
Part IV. Do all the Following (12+ Points). These are based on problems by Edward J. Kane. Show your
work!
1. The following numbers give us the frequency distributions of interest rate changes in the US between
1801 and 1961 (153 years). Treat these data as a sample. Personalize the data below by adding the last digit
of your student number to the first frequency and the second to last digit of your student number to the
second frequency. For example, Seymour Butz’s student number is 876509 so he adds 9 to first frequency
and 0 to the second frequency and uses {10, 2, 9, 25, 65, 35, 10, 5, and 1} (over 162 years). You may
check your work on the computer, but what is turned in should look as if it had all been done by hand.
Class
Frequency
a. Calculate the Cumulative Frequency (0.5)
b. Calculate the Mean (0.5)
-.95 to -.76
1
c. Calculate the Median (1)
-.75 to -.56
2
d. Calculate the Mode (0.5)
-.55 to -.36
9
e. Calculate the Variance (1.5)
-.35 to -.16
25
f. Calculate the Standard Deviation (1)
-.15 to .04
65
g. Calculate the Interquartile Range (1.5)
.05 to .24
35
h. Calculate a Statistic showing Skewness and
.25 to .44
10
.45 to .64
5
interpret it (1.5)
.65 to .84
1
i. Make an ogive of the data (Neatness
Counts!)(1)
j. Extra credit: Put a (horizontal) box plot below
the ogive chart using the same horizontal scale
(1)
Solution using the original numbers: If we use the original numbers and the computational method, we
get the following for the frequencies.
Frel
Row
Class
f f rel
F
1 -.95 to -.76
1
2 -.75 to -.56
2
3 -.55 to -.36
9
4 -.35 to -.16 25
5 -.15 to .04
65
6 .05 to .24
35
7 .25 to .44
10
8 .45 to .64
5
9 .65 to .84
1
Total
153
0.006536
0.013072
0.058824
0.163399
0.424837
0.228758
0.065359
0.032680
0.006536
1.00000
1
3
12
37
102
137
147
152
153
0.00654
0.01961
0.07843
0.24183
0.66667
0.89542
0.96078
0.99346
1.00000
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251y0611 2/15/06
If we use the computational method we get the following table.
x
Row
Class
f
fx
fx2
fx3
1 -.95 to -.76
1
2 -.75 to -.56
2
3 -.55 to -.36
9
4 -.35 to -.16 25
5 -.15 to .04
65
6 .05 to .24
35
7 .25 to .44
10
8 .45 to .64
5
9 .65 to .84
1
Total
153
-0.85
-0.65
-0.45
-0.25
-0.05
0.15
0.35
0.55
0.75
-0.85
-1.30
-4.05
-6.25
-3.25
5.25
3.50
2.75
0.75
-3.45
0.7225
0.8450
1.8225
1.5625
0.1625
0.7875
1.2250
1.5125
0.5625
9.2025
-0.614125
-0.549250
-0.820125
-0.390625
-0.008125
0.118125
0.428750
0.831875
0.421875
-0.581625
If we like to work, we may have used the definitional method instead.
Row
Class
f
1 -.95 to -.76 1
2 -.75 to -.56 2
3 -.55 to -.36 9
4 -.35 to -.16 25
5 -.15 to .04 65
6 .05 to .24
35
7 .25 to .44
10
8 .45 to .64
5
9 .65 to .84
1
Total
153
x
-0.85
-0.65
-0.45
-0.25
-0.05
0.15
0.35
0.55
0.75
xx
f x  x 
-0.85
-1.30
-4.05
-6.25
-3.25
5.25
3.50
2.75
0.75
-3.45
-0.827451
-0.627451
-0.427451
-0.227451
-0.027451
0.172549
0.372549
0.572549
0.772549
-0.82745
-1.25490
-3.84706
-5.68627
-1.78431
6.03922
3.72549
2.86275
0.77255
0.00000
From the f and fx columns of either display, we find n 
mean is x 
n
0.68468
0.78739
1.64443
1.29335
0.04898
1.04206
1.38793
1.63906
0.59683
9.12471
 f  153
 fx   3.45  0.022549 . We also find that
153
f x  x 3
f x  x 2
fx
 fx
2
-0.566535
-0.494048
-0.702913
-0.294173
-0.001345
0.179807
0.517071
0.938443
0.461082
0.037389
and
 fx
 9.2025 ,
 3.45 , so that the
 fx
3
 0.581625 ,
 f x  x   0 (except for a rounding error),  f x  x 2  9.12471, and  f x  x 3  0.0373887.
The width of all these intervals is 0.75  0.95   0.20 . (Note that, to be reasonable, the mean, median
and quartiles must fall between -0.95 and 0.85.)
a. Calculate the Cumulative Frequency (0.5): See the F column on the previous page.
fx  3.45
b. Calculate the Mean (0.5): We have already found x 

 0.022549 .
n
153
c. Calculate the Median (1): position  pn  1  .5153   76.5 . This is above F  37 and below

 pN  F 
 w so
 f p 
F  102 , so the interval is the 5th, -0.15 to 0.04, which has a frequency of 65. x1 p  L p  
 .5153   37 
x1.5  x.5  0.15  
 0.20   0.15  0.6077 0.20   0.028462
65


d. Calculate the Mode (0.5): The mode is the midpoint of the largest group. Since 65 is the largest
frequency, the modal group is -0.15 to 0.04 and the mode is -0.05.
Note that to be reasonable, Q1  x50  Q3 and that Q1, x50 , Q3, x and the mode must be between
-0.95 and 0.85!
fx 2  nx 2 9.20250  153  0.022549 2 9.12471
2
e. Calculate the Variance (1.5): s 


 0.060031
n 1
152
152

or s
2
 f x  x 

n 1
2

9.12471
 0.060031 . The computer got 0.066310.
152
f. Calculate the Standard Deviation (1): s  0.060031  0.245012
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251y0611 2/15/06
g. Calculate the Interquartile Range (1.5): First Quartile: position  pn  1  .25154   38.5 . This is
above F  37 and below F  102 , so the interval is the 5th, -0.15 to 0.04, which has a frequency of 65.
 pN  F 
 .25153   37 
x1 p  L p  
 w gives us Q1  x1.25  x.75  0.15  
 0.20 
65


 f p 
 0.15  0.019231 0.20   0.146154 .
Third Quartile: position  pn  1  .75154   115 .50 . This is above F  102 and below F  137 , so the
interval is the 6th, 0.05 to 0.24 which has a frequency of 35.
 .75153   102 
x1.75  x.25  0.05  
 0.20   0.05  0.364286 0.20  0.122857 ..
35


IQR  Q3  Q1  0.122857  0.146154   0.269011 .
h. Calculate a Statistic showing Skewness and interpret it (1.5): We had n  153 and
that the mean is x  0.022549 . We also found that
 f x  x 
3
k 3

 fx
2
 9.2025 ,
 fx
3
 fx
 3.45 , so
 0.581625 and
 0.037389. x.5  0.028462 and mo  0.05.
n
(n  1)( n  2)
 fx
3
 3x
 fx
2
 2nx 3


153
 0.581625  3 0.022549 9.2025  2153  0.022549 3
152 151 

 0.00666609 0.581625  0.622522  0.003508   0.00666609 .037389   0.000249 .
or k 3 
n
(n  1)( n  2)
or g1 
k3
or
s
3

 f x  x 
0.000249236
0.245012 3
3

153
0.037389   0.0002492
152 151 
The computer gets 0.000249236
 0.0169453
Pearson's Measure of Skewness SK 
Actually, this should be SK1 
3mean  mode 3 0.022549   0.05 

 .3361
std .deviation
0.245012
mean  mode   0.022549   0.05   .1120 or
std .deviation
0.245012
3mean  median 3 0.022549   0.028462 

 .0724
0.245012
std .deviation
Because of the positive sign, the measures all imply skewness to the right..
SK 2 
i. Make an ogive of the data (Neatness Counts!)(1): An ogive is a simple line graph with cumulative
frequency on the y-axis and the numbers -0.95 to 0.84 (actually 0 to 1 would be fine) on the x-axis. Each F
or Frel point is plotted against the upper limit of the class, x . In addition 1 empty class must be added at
the beginning. The first point on your graph should be (-0.95, 0); the next point is (-0.75, 3) or (-.75,
.00654). At x  0.85 the height of the line should be either 153 or 1. After this point the ogive is a
horizontal line.
j. Extra credit: Put a (horizontal) box plot below the ogive chart using the same horizontal scale (1)
The five-number summary is [-0.95 (or -0.85), -0.146154 , -0.28462, 0.122857, 0.85(or .75)]
IQR  Q3  Q1  0.122857  0.146154   0.269011 . 1.5IQR   0.403516 . If you use fences, they should
be at 0.146154  .403516  0.5577 and 0.122857  .403516  0.52537 . So the box extends from
(-.146154) to .122857, with a median marked by a horizontal line at -0.028462. The whiskers go from the
box to -.5577 and 0.52537 with dotted lines showing the full range.
9
251y0611 2/15/06
Solution using the original numbers with nines added to the first two frequencies: If we use these new
numbers and the computational method, we get the following for the frequencies.
Frel
Row
Class
f f rel
F
1 -.95 to -.76 10
2 -.75 to -.56 11
3 -.55 to -.36
9
4 -.35 to -.16 25
5 -.15 to .04
65
6 .05 to .24
35
7 .25 to .44
10
8 .45 to .64
5
9 .65 to .84
1
Total
171
0.058480
0.064327
0.052632
0.146199
0.380117
0.204678
0.058480
0.029240
0.005848
1.00000
10
21
30
55
120
155
165
170
171
0.05848
0.12281
0.17544
0.32164
0.70175
0.90643
0.96491
0.99415
1.00000
If we use the computational method we get the following table.
x
Row
Class
f
fx
fx2
fx3
1 -.95 to -.76 10
2 -.75 to -.56 11
3 -.55 to -.36
9
4 -.35 to -.16 25
5 -.15 to .04
65
6 .05 to .24
35
7 .25 to .44
10
8 .45 to .64
5
9 .65 to .84
1
Total
171
-0.85
-0.65
-0.45
-0.25
-0.05
0.15
0.35
0.55
0.75
-8.50 7.2250
-7.15 4.6475
-4.05 1.8225
-6.25 1.5625
-3.25 0.1625
5.25 0.7875
3.50 1.2250
2.75 1.5125
0.75 0.5625
-16.95 19.5075
-6.14125
-3.02088
-0.82013
-0.39063
-0.00813
0.11813
0.42875
0.83188
0.42188
-8.58138
If we like to work, we may have used the definitional method instead.
Row
Class
f
1 -.95 to -.76 10
2 -.75 to -.56 11
3 -.55 to -.36
9
4 -.35 to -.16 25
5 -.15 to .04
65
6 .05 to .24
35
7 .25 to .44
10
8 .45 to .64
5
9 .65 to .84
1
Total
171
x
-0.85
-0.65
-0.45
-0.25
-0.05
0.15
0.35
0.55
0.75
fx
-8.50
-7.15
-4.05
-6.25
-3.25
5.25
3.50
2.75
0.75
-16.95
xx
-0.750877
-0.550877
-0.350877
-0.150877
0.049123
0.249123
0.449123
0.649123
0.849123
f x  x 
-7.50877 5.63817
-6.05965 3.33812
-3.15789 1.10803
-3.77193 0.56910
3.19298 0.15685
8.71930 2.17218
4.49123 2.01711
3.24561 2.10680
0.84912 0.72101
0.00000 17.8274
From the f and fx columns of either display, we find n 
mean is x 
171
f x  x 3
-4.23357
-1.83890
-0.38878
-0.08586
0.00770
0.54114
0.90593
1.36757
0.61223
-3.11254
 f  171 and  fx
 fx   16.95  0.0991228 . We also find that
n
f x  x 2
 fx
2
 19 .5075 ,
 16 .95 , so that the
 fx
3
 3.11254 ,
 f x  x   0 (except for a rounding error),  f x  x 2  17.8274, and  f x  x 3  3.11254 .
The width of all these intervals is 0.75  0.95   0.20 . (Note that, to be reasonable, the mean, median
and quartiles must fall between -0.95 and 0.85.)
a. Calculate the Cumulative Frequency (0.5): See the F column above.
fx  16 .95
b. Calculate the Mean (0.5): We have already found x 

 0.0991228 .
n
171
c. Calculate the Median (1): position  pn  1  .5172   86 . This is above F  55 and below F  120 ,

 pN  F 
so the interval is the 5th, -0.15 to 0.04, which has a frequency of 65. x1 p  L p  
 w so
 f p 
 .5171   55 
x1.5  x.5  0.15  
 0.20   0.15  0.4692 0.20   0.0561539
65


d. Calculate the Mode (0.5): The mode is the midpoint of the largest group. Since 65 is the largest
frequency, the modal group is -0.15 to 0.04 and the mode is -0.05.
10
251y0611 2/15/06
e. Calculate the Variance (1.5): s 2 
or s 2 
 f x  x 
n 1
2

 fx
2
 nx 2
n 1

19 .5075  171 0.0991228
170
2

17 .82737
 0.104867
170
17 .8274
 0.104867 . The computer got 0.104867.
170
f. Calculate the Standard Deviation (1): s  0.104867  0.323832
g. Calculate the Interquartile Range (1.5): First Quartile: position  pn  1  .25172   43 . This is above
F  30 and below F  55, so the interval is the 4th, -0.35 to 0.16, which has a frequency of 25.
 pN  F 
 .25171   30 
x1 p  L p  
 w gives us Q1  x1.25  x.75  0.35  
 0.20 
25


 f p 
 0.35  0.5100 0.20   0.248000 .
Third Quartile: position  pn  1  .75172   129 . This is above F  120 and below F  155 , so the
interval is the 6th, 0.05 to 0.24 which has a frequency of 35.
 .75171   120 
x1.75  x.25  0.05  
 0.20   0.05  0.23571 0.20  0.097143 ..
35


IQR  Q3  Q1  0.097143  0.248000   0.345143 .
h. Calculate a Statistic showing Skewness and interpret it (1.5): We had n  171 and
so that the mean is x  0.0991228 . We also found that
 f x  x 
3
k 3

 fx
2
 19 .5075 ,
 fx
3
 fx
 16 .9500 ,
 8.58038 and
 3.11254 . x.5  0.0561539 and mo  0.05.
n
(n  1)( n  2)
 fx
3
 3x
 fx
2
 2nx 3


171
 8.58038  3 0.0991228 19 .5075  2171  0.0991228
170 169 
3 
 0.00595197 8.58038  5.800914  0.333079   0.00595197 3.112545   0.018526 .

n
171
 3.11254   0.00595197  3.11254   0.018526 .
f x  x 3 
(n  1)( n  2)
170 169 
The computer gets -0.0185257
k
 0.018526
 0.5455
or g 1  33 
s
0.323832 3
or k 3 
or
Pearson's Measure of Skewness SK 
Actually, this should be SK1 
3mean  mode  3 0.0991228   0.05 

 0.4551
std .deviation
0.323832
mean  mode   0.0991228   0.05   0.1517 or
std .deviation
0.323832
3mean  median 3 0.0991228   0.0561539 

 .5547
0.2323832
std .deviation
Because of the negative sign, the measures all imply skewness to the left.
SK 2 
i. Make an ogive of the data (Neatness Counts!)(1): An ogive is a simple line graph with cumulative
frequency on the y-axis and the numbers -0.95 to 0.84 (actually 0 to 1 would be fine) on the x-axis. Each F
or Frel point is plotted against the upper limit of the class, x . In addition 1 empty class must be added at
the beginning. The first point on your graph should be (-0.95, 0); the next point is (-0.75, 10) or (-.75,
0.05848). At x  0.85 the height of the line should be either 171 or 1. After this point the ogive is a
horizontal line.
11
251y0611 2/15/06
j. Extra credit: Put a (horizontal) box plot below the ogive chart using the same horizontal scale (1)
The five-number summary is [-0.95 (or -0.85), -0.248000 , -0.0561539, 0.097143, 0.85(or .75)]
IQR  Q3  Q1  0.097143  0.248000   0.345143 . 1.5IQR   0.517715 . If you use fences, they should
be at 0.248000  .517715  0.7657 and 0.097143  .517715  0.6149 . So the box extends from
(-.0.248000) to .097143, with a median marked by a horizontal line at -0.0561539. The whiskers go from
the box to -.7657 and 0.6149 with dotted lines showing the full range.
12
251y0611 2/15/06
2. The following problems (1 point each) are given in their raw form. Personalize them by using the third
to last and second to last numbers in your student number. Add them as percents to the first two percents in
a) and then add them to the first two numbers in b) and finally in c) divide them by 100 and add them to the
first two numbers. For example, Seymour Butz’s student number is 876509, so he uses 5 and 0. This means
in a) the percents are 5% + 5% =10%, 10% + 0% = 10%, -1%, 3%, and 6%, in b) the speeds are 55 + 5 =
60, 30 + 0 = 30, 70, 35 and 80 and in c) the numbers are 0.01 + 0.05 = 0.06, 0.25 + 0.00 = 0.25, 0.37, 0.26,
0.85 and 0.27
a) Geometric Mean
Over 5 years GDP grew at 5%, 10%, -1%, 3% and 6%. Modify these as was done in the example given in
class and use the geometric mean to find the average rate of growth.
b) Harmonic mean
I drive 300 miles a day, on the first day at (an average of) 55 mph, on the second at 30 mph, on the third at
70 mph, on the 4th at 35 mph and on the last at 80 mph. Use the harmonic mean to find my average speed
over 2 days.
c) Root-mean-square
Find the rms of the following numbers: 0.01, 0.25, 0.37, 0.26, 0.85 and 0.27.
If you wish, d) Compute the geometric mean from a) using natural and/or base 10 logarithms. (1 point extra
credit each).
Solution: Using the original data, before I started, I computed the following table.
(1)
(2)
Row growth growth
Natural
logarithm mph
1/mph
y
y2
rates
rates
as
plus
decimals one
1
2
3
4
5
6
0.05
0.10
-0.01
0.03
0.06
1.05
1.10
0.99
1.03
1.06
logarithm
of
(2)
0.0487902
0.0953102
-0.0100503
0.0295588
0.0582689
0.221878
of (2)
to base
ten
0.0211893
0.0413927
-0.0043648
0.0128372
0.0253059
0.0963603
55
30
70
35
80
0.0181818
0.0333333
0.0142857
0.0285714
0.0125000
0.106872
0.01
0.25
0.37
0.26
0.85
0.27
0.0001
0.0625
0.1369
0.0676
0.7225
0.0729
1.0625
a) The Geometric Mean. For average growth rates add 1 to each rate.
1
x g  x1  x 2  x3  x n  n  n
x 
5
1.05 1.10 0.99 1.031.06 
 5 1.24842  1.24842 
1
5
 1.24842 0.2  1.04537 So the average growth rate was 4.537%.
b) The Harmonic Mean.
1 1

xh n

1 1
1 
 x  5  55  30  70  35  80   5 0.0181818  0.033333  0.014857  0.0285714  0.125000 
1
1
1
1
0.106872   0.0213745 .
5
1
1
So xh 
1
1
n

1
x

1
 46 .7848 .
0.0213745
Of course some of you decided that 1  1  1  1  1  1  1  1  1   ? 1 
xh
n
x
5  55
30
70
35
80 
1

  ??? .
5  55  30  70  35  80 
This is, of course, an easier way to do the problem, and you will get an A for the course if you can
1 1
1
prove to me that   ? .
2 2
4
13
251y0611 2/15/06
c) The Root-Mean-Square.
1
1
2
x rms

x 2  0.01 2  0.25 2  0.37 2  0.26 2  0.85 2  0.27 2
n
6
1
1
 0.0001  0.0625  0.1369  0.0676  0.7225  0.0729   1.0625   0.177983
6
6


So x rms 
1
n
x

 0.17783  0.4208 .
2
 

2
1
1
1
x 2  ???
x  2.012 . This is, of course, an
n
n
12
easier way to do the problem, but I warned you that it wouldn’t work. It is equivalent to believing that
22  22  42 .
2

Of course some of you decided that x rms
d) (i) Geometric mean using natural logarithms
1
ln( x)   1 ln 1.05   ln 1.10   ln 0.99   ln 1.03   ln 1.06 
ln x g 
n
5
1
1
 0.0487902  0.0953102  0.0100503  0.0295588  0.0582689   0.221878   0.0443755
5
5
 

x g  e 0.0443755  1.04537 .
So
(ii) Geometric mean using logarithms to the base 10
1
log( x)  1 log1.05   log1.10   log0.99   log1.03   log1.06 
log x g 
n
5
1
1
 0.0211893  0.0413927  0.0043648  0.0.0128372  0.0.0253059   0.0963603   0.0192721
5
5
 

So x g  10 0.0192721  1.04537
Solution: Using the original data with nines added to the first two numbers, before I started, I
computed the following table.
(1)
(2)
Row growth growth
Natural
logarithm mph
1/mph
y
y2
1
2
3
4
5
6
rates
as
decimals
0.14
0.19
-0.01
0.03
0.06
rates
logarithm
of (2)
plus
of
to base
one
(2)
ten
1.14
0.131028
0.0569049
1.19
0.173953
0.0755470
0.99 -0.010050 -0.0043648
1.03
0.029559
0.0128372
1.06
0.058269
0.0253059
0.382759
0.166230
64
39
70
35
80
0.0156250
0.0256410
0.0142857
0.0285714
0.0125000
0.0966232
0.10
0.34
0.37
0.26
0.85
0.27
0.0100
0.1156
0.1369
0.0676
0.7225
0.0729
1.1255
a) The Geometric Mean. For average growth rates add 1 to each rate.
1
x g  x1  x 2  x3  x n  n  n
x 
5
1.14 1.19 0.99 1.031.06 
 5 1.46632  1.46632 
1
5
 1.46632 0.2  1.07956 . So the average growth rate was 7.956%.
b) The Harmonic Mean.
1 1

xh n

1 1
1 
 x  5  64  39  70  35  80   5 0.015625  0.0256410  0.014857  0.0285714  0.125000 
1
1
0.0.0966232
5
1
1
1
  0.0193246
.
1
So xh 
1
1
n

1
x

1
 51 .7474 .
0.0193246
14
251y0611 2/15/06
c) The Root-Mean-Square.
1
1
2
x rms

x 2  0.10 2  0.34 2  0.37 2  0.26 2  0.85 2  0.27 2
n
6
1
1
 0.0100  0.1156  0.1369  0.0676  0.7225  0.0729   1.1255   0.187583
6
6


. So x rms 
1
n
x

2
 0.187583  0.43311 .
d) (i) Geometric mean using natural logarithms
1
ln( x)   1 ln 1.14   ln 1.19   ln 0.99   ln 1.03   ln 1.06 
ln x g 
n
5
1
1
 0.131028  0.173953  0.010050  0.0295588  0.0582689   0.382759   0.0765518
5
5
 
So

x g  e 0.0765518  1.07956 .
(ii) Geometric mean using logarithms to the base 10
1
log( x)  1 log 1.14   log1.19   log0.99   log 1.03   log1.06 
log x g 
n
5
1
1
 0.0569049  0.0755470  0.0043648  0.0.0128372  0.0.0253059   0.166230   0.0332460
5
5
 

So x g  10 0.0332460  1.07956
15
251y0611 2/15/06
Programs used to Analyze Grouped Data
#grp.mtb
#Put frequencies in column 1 ‘f’
#and midpoints on column 2 ‘x’
let c3=c1*c2
let c4=c3*c2
let c5=c4*c2
name k1 'n'
name k2 'mean'
name k3 'Sfx'
name k4 'Sfx2'
name k5 'Sfx3'
name k7 'Sfx^'
name k8 'Sfx^2'
name k9 'Sfx^3'
let k1=sum(c1)
let k3=sum(c3)
let k4=sum(c4)
let k5=sum(c5)
let k2=k3/k1
print c10, c1-c5
print k1-k5
let c6=c2-k2
let c7=c1*c6
let c8=c7*c6
let c9=c8*c6
let k7 =sum(c7)
let k8 = sum(c8)
let k9 = sum(c9)
print c10, c6-c9
print k7-k9
end
#grpv.mtb
let k6=k1-1
name k10 'var1'
name k11 'var2'
name k17 'stdev'
let k10=k8/k6
let k11=k1*k2*k2
let k11=k4-k11
let k11=k11/k6
let k17=sqrt(k11)
end
#groups.mtb
let k12=k6-1
let k13=k1/k6
let k13=k13/k12
let k14=k13*k9
let k15=2*k1*k2*k2*k2
let k16=3*k2*k4
let k15=k5-k16+k15
let k16=k13*k15
let k18=k16/k11
let k18=k18/k17
let k19=k14/k11
let k19=k19/k17
print k1-k19
end
16
251y0611 2/10/06
Processing of Original Data
————— 2/15/2006 4:29:58 PM ————————————————————
Welcome to Minitab, press F1 for help.
Results for: 251x0601-1.MTW
MTB > sum c1
Sum of f
Sum of f = 153
MTB > print c10 c1 c11 c2
Data Display
Row
1
2
3
4
5
6
7
8
9
Class
-.95 to -.76
-.75 to -.56
-.55 to -.36
-.35 to -.16
-.15 to .04
.05 to .24
.25 to .44
.45 to .64
.65 to .84
f
1
2
9
25
65
35
10
5
1
bigF
1
3
12
37
102
137
147
152
153
#Original Numbers
x
-0.85
-0.65
-0.45
-0.25
-0.05
0.15
0.35
0.55
0.75
MTB > Save "C:\Documents and Settings\rbove\My Documents\Minitab\251x06011.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\251x0601-1.MTW'
Existing file replaced.
MTB > let c12 = c11/153
MTB > let c13 = c1/153
MTB > print c10 c1 c13 c11 c12
Data Display
Row
1
2
3
4
5
6
7
8
9
Class
-.95 to -.76
-.75 to -.56
-.55 to -.36
-.35 to -.16
-.15 to .04
.05 to .24
.25 to .44
.45 to .64
.65 to .84
#Relative Frequencies
f
1
2
9
25
65
35
10
5
1
frel
0.006536
0.013072
0.058824
0.163399
0.424837
0.228758
0.065359
0.032680
0.006536
bigF
1
3
12
37
102
137
147
152
153
bigFrel
0.00654
0.01961
0.07843
0.24183
0.66667
0.89542
0.96078
0.99346
1.00000
MTB > Execute "C:\Documents and Settings\rbove\My Documents\Minitab\grp.mtb" 1.
Executing from file: C:\Documents and Settings\rbove\My
Documents\Minitab\grp.mtb
17
251y0611 2/10/06
Data Display
Row
1
2
3
4
5
6
7
8
9
Class
-.95 to -.76
-.75 to -.56
-.55 to -.36
-.35 to -.16
-.15 to .04
.05 to .24
.25 to .44
.45 to .64
.65 to .84
f
1
2
9
25
65
35
10
5
1
x
-0.85
-0.65
-0.45
-0.25
-0.05
0.15
0.35
0.55
0.75
#Table using computational method
fx
-0.85
-1.30
-4.05
-6.25
-3.25
5.25
3.50
2.75
0.75
fxsq
0.7225
0.8450
1.8225
1.5625
0.1625
0.7875
1.2250
1.5125
0.5625
fxcu
-0.614125
-0.549250
-0.820125
-0.390625
-0.008125
0.118125
0.428750
0.831875
0.421875
Data Display
n
mean
Sfx
Sfx2
Sfx3
153.000
-0.0225490
-3.45000
9.20250
-0.581625
Data Display
Row
1
2
3
4
5
6
7
8
9
Class
-.95 to -.76
-.75 to -.56
-.55 to -.36
-.35 to -.16
-.15 to .04
.05 to .24
.25 to .44
.45 to .64
.65 to .84
#Column Sums
x^
-0.827451
-0.627451
-0.427451
-0.227451
-0.027451
0.172549
0.372549
0.572549
0.772549
Data Display
Sfx^
Sfx^2
Sfx^3
-0.000000000
9.12471
0.0373887
#Table using definitional method
fx^
-0.82745
-1.25490
-3.84706
-5.68627
-1.78431
6.03922
3.72549
2.86275
0.77255
fx^sq
0.68468
0.78739
1.64443
1.29335
0.04898
1.04206
1.38793
1.63906
0.59683
fx^cu
-0.566535
-0.494048
-0.702913
-0.294173
-0.001345
0.179807
0.517071
0.938443
0.461082
#Column sums
MTB > exec 'grpv'
Executing from file: grpv.MTB
MTB > exec 'grps'
Executing from file: grps.MTB
Data Display
n
mean
Sfx
Sfx2
Sfx3
K6
Sfx^
Sfx^2
Sfx^3
var1
var2
K12
K13
k31
K15
k32
stdev
g11
g12
153.000
-0.0225490
-3.45000
9.20250
-0.581625
152.000
-0.000000000
9.12471
0.0373887
0.0600310
0.0600310
151.000
0.00666609
0.000249236
0.0373887
0.000249236
0.245012
0.0169453
0.0169453
#Final results
MTB > Save "C:\Documents and Settings\rbove\My Documents\Minitab\251x06011.MTW";
SUBC>
Replace.
18
251y0611 2/10/06
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\251x0601-1.MTW'
Existing file replaced.
MTB > print gr
Data Display
#Percent Data for Geometric mean.
gr
5
MTB
MTB
MTB
MTB
>
>
>
>
10
let
let
let
let
-1
c15
c16
c17
c18
=
=
=
=
3
6
c15/100
c15+1
loge(c16)
logten(c16)
MTB > sum c17
Sum of lngrf
Sum of lngrf = 0.221878
MTB > mean c17
Mean of lngrf
Mean of lngrf = 0.0443755
MTB > let k17 = mean(c17)
MTB > let k17 = expon(k17)
MTB > print k17
Data Display
k17
1.04537
MTB > sum c18
Sum of loggrf
Sum of loggrf = 0.0963603
MTB > mean c18
Mean of loggrf
Mean of loggrf = 0.0192721
MTB > let k18 = mean(c18)
MTB > let k18 = 10**k18
MTB > print k18
Data Display
k18
1.04537
MTB > let k16 = c16(1)*c16(2)*c16(3)*c16(4)*c16(5)
MTB > print k16
Data Display
k32
1.24842
MTB > let k16 = k16**0.2
MTB > print k16
Data Display
k32
1.04537
19
251y0611 2/10/06
MTB > print c19
Data Display
mph
55
30
#Data for Harmonic mean
70
35
80
MTB > let c20 = 1/c19
MTB > sum c20
Sum of 1/mph
Sum of 1/mph = 0.106872
MTB > mean c20
Mean of 1/mph
Mean of 1/mph = 0.0213745
MTB > let k20 = mean(c20)
MTB > let k20 = 1/k20
MTB > print k20
Data Display
K20
46.7848
MTB > print c21
Data Display
#Data for RMS
y
0.01
0.25
0.37
0.26
0.85
0.27
MTB > let c22 = c21*c21
MTB > sum c22
Sum of ysq
Sum of ysq = 1.0625
MTB > mean c22
Mean of ysq
Mean of ysq = 0.177083
MTB > let k22 = mean(c22)
MTB > let k22 = sqrt(k22)
MTB > print k22
Data Display
K22
0.420813
MTB > Print c15-c22
Data Display
Row
1
2
3
4
5
6
gr
0.05
0.10
-0.01
0.03
0.06
grf
1.05
1.10
0.99
1.03
1.06
lngrf
0.0487902
0.0953102
-0.0100503
0.0295588
0.0582689
loggrf
0.0211893
0.0413927
-0.0043648
0.0128372
0.0253059
mph
55
30
70
35
80
1/mph
0.0181818
0.0333333
0.0142857
0.0285714
0.0125000
y
0.01
0.25
0.37
0.26
0.85
0.27
ysq
0.0001
0.0625
0.1369
0.0676
0.7225
0.0729
20
251y0611 2/10/06
Processing of Original Data with nines added to first two numbers.
————— 2/15/2006 6:34:07 PM ————————————————————
Welcome to Minitab, press F1 for help.
Results for: 251x0601-2.MTW
MTB > sum c1
Sum of f
Sum of f = 171
MTB > print c10 c1 c11 c2
Data Display
Row
1
2
3
4
5
6
7
8
9
Class
-.95 to -.76
-.75 to -.56
-.55 to -.36
-.35 to -.16
-.15 to .04
.05 to .24
.25 to .44
.45 to .64
.65 to .84
#Original Numbers with nines added to frequencies
f
10
11
9
25
65
35
10
5
1
bigF
10
21
30
55
120
155
165
170
171
x
-0.85
-0.65
-0.45
-0.25
-0.05
0.15
0.35
0.55
0.75
MTB > Save "C:\Documents and Settings\rbove\My Documents\Minitab\251x06012.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\251x0601-2.MTW'
Existing file replaced.
MTB > let c12 = c11/171
MTB > let c13 = c1/171
MTB > sum c13
Sum of frel
Sum of frel = 1
MTB > print c10 c1 c13 c11 c12
Data Display
Row
1
2
3
4
5
6
7
8
9
Class
-.95 to -.76
-.75 to -.56
-.55 to -.36
-.35 to -.16
-.15 to .04
.05 to .24
.25 to .44
.45 to .64
.65 to .84
f
10
11
9
25
65
35
10
5
1
frel
0.058480
0.064327
0.052632
0.146199
0.380117
0.204678
0.058480
0.029240
0.005848
bigF
10
21
30
55
120
155
165
170
171
#Relative Frequencies
bigFrel
0.05848
0.12281
0.17544
0.32164
0.70175
0.90643
0.96491
0.99415
1.00000
MTB > Execute "C:\Documents and Settings\rbove\My Documents\Minitab\grp.mtb" 1.
Executing from file: C:\Documents and Settings\rbove\My
Documents\Minitab\grp.mtb
21
251y0611 2/10/06
Data Display
Row
1
2
3
4
5
6
7
8
9
Class
-.95 to -.76
-.75 to -.56
-.55 to -.36
-.35 to -.16
-.15 to .04
.05 to .24
.25 to .44
.45 to .64
.65 to .84
f
10
11
9
25
65
35
10
5
1
x
-0.85
-0.65
-0.45
-0.25
-0.05
0.15
0.35
0.55
0.75
fx
-8.50
-7.15
-4.05
-6.25
-3.25
5.25
3.50
2.75
0.75
#Table using computational method
fxsq
7.2250
4.6475
1.8225
1.5625
0.1625
0.7875
1.2250
1.5125
0.5625
fxcu
-6.14125
-3.02088
-0.82013
-0.39063
-0.00813
0.11813
0.42875
0.83188
0.42188
Data Display
n
mean
Sfx
Sfx2
Sfx3
171.000
-0.0991228
-16.9500
19.5075
-8.58038
Data Display
Row
1
2
3
4
5
6
7
8
9
Class
-.95 to -.76
-.75 to -.56
-.55 to -.36
-.35 to -.16
-.15 to .04
.05 to .24
.25 to .44
.45 to .64
.65 to .84
#Column Sums
x^
-0.750877
-0.550877
-0.350877
-0.150877
0.049123
0.249123
0.449123
0.649123
0.849123
Data Display
Sfx^
Sfx^2
Sfx^3
-0.000000000
17.8274
-3.11254
fx^
-7.50877
-6.05965
-3.15789
-3.77193
3.19298
8.71930
4.49123
3.24561
0.84912
#Table using definitional method
fx^sq
5.63817
3.33812
1.10803
0.56910
0.15685
2.17218
2.01711
2.10680
0.72101
fx^cu
-4.23357
-1.83890
-0.38878
-0.08586
0.00770
0.54114
0.90593
1.36757
0.61223
#Column sums
MTB > exec 'grpv'
Executing from file: grpv.MTB
MTB > exec 'grps'
Executing from file: grps.MTB
Data Display
n
mean
Sfx
Sfx2
Sfx3
K6
Sfx^
Sfx^2
Sfx^3
var1
var2
K12
K13
k31
K15
k32
stdev
g11
g12
171.000
-0.0991228
-16.9500
19.5075
-8.58038
170.000
-0.000000000
17.8274
-3.11254
0.104867
0.104867
169.000
0.00595197
-0.0185257
-3.11254
-0.0185257
0.323832
-0.545529
-0.545529
#Final results
22
251y0611 2/10/06
MTB > print gr
#Percent Data for Geometric mean divided by 100
Data Display
gr
0.14
MTB
MTB
MTB
MTB
>
>
>
>
0.19
let
let
let
sum
-0.01
0.03
0.06
c16 = c15+1
c17 = loge(c16)
c18 = logten(c16)
c17
Sum of lngrf
Sum of lngrf = 0.382759
MTB > mean c17
Mean of lngrf
Mean of lngrf = 0.0765518
MTB > let k17 = mean(c17)
MTB > let k17 = expon(k17)
MTB > print k17
Data Display
K17
1.07956
MTB > sum c18
Sum of loggrf
Sum of loggrf = 0.166230
MTB > mean c18
Mean of loggrf
Mean of loggrf = 0.0332460
MTB > let k18 = mean(c18)
MTB > let k18 = 10**k18
MTB > print k18
Data Display
K18
1.07956
MTB > let k16 = c16(1)*c16(2)*c16(3)*c16(4)*c16(5)
MTB > print k16
Data Display
K18
1.46632
MTB > let k16 = k16**0.2
MTB > print k16
Data Display
K16
1.07956
MTB > print c19
Data Display
mph
64
39
#Data for Harmonic mean
70
35
80
MTB > let c20 = 1/c19
MTB > sum c20
Sum of 1/mph
Sum of 1/mph = 0.0966232
MTB > mean c20
Mean of 1/mph
Mean of 1/mph = 0.0193246
MTB > let k20 = mean(c20)
MTB > let k20 = 1/k20
23
251y0611 2/10/06
MTB > print k20
Data Display
K20
51.7474
MTB > print c21
Data Display
#Data for RMS
y
0.10
0.34
0.37
0.26
0.85
0.27
MTB > let c22 = c21*c21
MTB > sum c22
Sum of ysq
Sum of ysq = 1.1255
MTB > mean c22
Mean of ysq
Mean of ysq = 0.187583
MTB > let k22 = mean(c22)
MTB > let k22 = sqrt(k22)
MTB > print k22
Data Display
K22
0.433109
MTB > print c15 - c22
Data Display
Row
1
2
3
4
5
6
gr
0.14
0.19
-0.01
0.03
0.06
grf
1.14
1.19
0.99
1.03
1.06
lngrf
0.131028
0.173953
-0.010050
0.029559
0.058269
loggrf
0.0569049
0.0755470
-0.0043648
0.0128372
0.0253059
mph
64
39
70
35
80
1/mph
0.0156250
0.0256410
0.0142857
0.0285714
0.0125000
y
0.10
0.34
0.37
0.26
0.85
0.27
ysq
0.0100
0.1156
0.1369
0.0676
0.7225
0.0729
MTB > Save "C:\Documents and Settings\rbove\My Documents\Minitab\251x06012.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\251x0601-2.MTW'
Existing file replaced.
24