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ECO251 QBA1
FIRST EXAM
February 21, 2008
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)
The following numbers are to be considered a sample and represent the scores of a class of ten seniors on
the first exam in Dr. Hardnose’s accounting class (Doane and Seward).
60
x1
88
60
71
60
73
74
75
60
99
Compute the following: Show your work!
a) The Median (1)
b) The Standard Deviation (3)
c) The 7th decile (2)
d) The Coefficient of variation (1)
e) (Extra Credit) Dr. Hardnose gives a second exam, but one student is absent. He enters the data from both
exams into Minitab with the following results.
MTB > describe c1 c2
Descriptive Statistics: x1, x2
Variable
N N*
Mean
x1
10
0 ----x2
9
0 72.78
SE Mean
---2.19
StDev
----6.57
Minimum
60.00
65.00
Q1
60.00
65.00
Median
72.00
74.00
Q3
78.25
79.00
Maximum
99.00
79.00
The numbers computed should be self explanatory except for ‘SE Mean,’ (the standard error of the mean)
s
which can be gotten by computing s x 
. I have erased the numbers related to the statistics you should
n
compute. So – given the statistics that we have learned to compute and the facts that the mean and sample
size have changed, can you compare the variability of the results of the first and second exams? (Yes or No
won’t work here, give me some numbers!)
a) The numbers in order are at left. The median
xx
x
x2
x  x 2
is the average of the middle numbers
60 3600 -12
144
x1
71  73
x.50 
 72 .
144
x 2 60 3600 -12
2
More formally, position  pn  1  .511  5.5
60 3600 -12
144
x3
The median is thus
144
x 4 60 3600 -12
x.50  x5  .5x6  x5   71  0.573  71  72
x5
71 5041
-1
1
b) The variance is computed two ways here.
Please do it one way only. We have
x6
73 5329
1
1
n  10,
x  720 ,
x 2  53416 ,
x7
x8
x9
x10
Total
74
5476
2
4
75
5625
3
9
88
7744
16
256
99
9801
27
729
720 53416
0
1576

 x  x   0 and  x  x 2  1576 .
 x  720  72 ,
x

n
10
1
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x
2/11/07

x  x  1576
53416  1172 2

 175 .11111 . So s  175.1111  13.2330

9
n 1
10  1
n 1
c) The 7th decile has 7 10 or 70% below it, so p  .70 and pn  1  .710   7.00 .
s2 
2
 nx 2

2
So a  7 and .b  0.0 . Then x1 p  xa  .b( xa1  xa ) .
So x1.70  x.30  x 7  0.0( x8  x 7 )  74  0.075  74   74
s 13 .2330

 0.1838 or 18.38%
x
72
Note that mean, median and seventh decile must be between 60 and 99. In the variance excess rounding
often will give you a negative variance. s 2 cannot be negative.
e) We have the following Descriptive Statistics for x1 and x2
d) C 
Variable
x1
x2
N
10
9
N*
0
0
For the first exam C1 
C2 
Mean
----72.78
SE Mean
---2.19
StDev
----6.57
Minimum
60.00
65.00
Q1
60.00
65.00
Median
72.00
74.00
Q3
78.25
79.00
Maximum
99.00
79.00
s1 13 .2330

 0.1838 or 18.38%. For the second exam
x1
72
s2
6.57

 0.0903 or 9.03%. The second exam scores were much less variable than the first.
x 2 72 .78
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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. Which of the following is not a measure of central tendency (2)
a) The arithmetic mean
b) The geometric mean
c) *The standard deviation
d) The median
e) The mode.
f) All of the above are measures of central tendency
g) None of the above is a measure of central tendency.
2. The smaller the spread of data around the mean (2)
a) The smaller the interquartile range
b) The smaller the standard deviation
c) The smaller the coefficient of variation
d) *All of the above
e) None of the above.
3. Mark the following items N (nominal), O (ordinal), I (interval) or R (ratio) data. If the data is interval or
ratio data, would it be considered C (continuous) or D (discrete)? (4) [8]
a) The weights of Sumo wrestlers _RC_____
b) Your Social Security number __N______
c) Volume of traffic on I-95 (light, medium, heavy) __O________
d) Number of hits in the World Series _RD_____
4. A number that is used to summarize population data is called (2)
a) *A parameter
b) s
c) A statistic
d) None of the above would be used to summarize population data
e) All of the above could be used to summarize population data.
5. If a frequency distribution has a positive coefficient of skewness, we would expect (2) [10]
a) The mean to be between the median and the mode
b) *The mean to exceed the median
c) The median to be larger than the mean or the mode.
d) The standard deviation to exceed the mean
e) The coefficient of excess to be positive
f) None of the above would be likely to be true.
6. Under what circumstances would you expect a population variance be zero? [12]
a) If the mean is zero
b) If every observation above the sample mean is precisely offset by a number the same distance
below the sample mean.
c) If the mean, median and mode were all the same.
d) When there are an equal number of observations above and below the median
e) *None of the above is likely to result in a zero population variance
f) All of the above are likely to result in a zero population variance.
Explanation: A zero population variance can only occur if all numbers are the same.
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7. Batting averages of major league baseball players are thought to follow a symmetrical unimodal
distribution with a mean of .260 and a standard deviation of .03.
a) What proportion of major league players would you expect to have a batting average above
.320? Why? (3)
b) If you did not know that the distribution was symmetrical, what is the largest proportion of
players that you would expect to have a batting average above .320? Why? (3) [18]
.320  .260
 2 a) The empirical rule says that about 96% of data will fall between 2 standard
Solution:
.03
deviations above and below the mean. This means that 4% will be in the tails of the distribution or about
2% in each tail. b) Chebychef’s rule says that, at most 1 2  1 of the data will be more than k  2
4
k
standard deviations from the mean. There is no way to divide it between tails.
Table 1
Given below is a stem-and-leaf display of the heights in inches of 50 Christmas trees being grown for sale.
The smallest tree is 44 inches.
4|44668899
5|00112244555577888899
6|000022223344557799
7|3355
8. In Table 1, what is the median height ? (2)
Solution: There are 50 numbers and the position formula gives us pn  1  .5(51)  25.5
4|44668899
8 items
5|00112244555577888899
20 items
6|000022223344557799
18 items
7|3355
4 items
Both the 25th and the 26th items are 58, so that is the median.
9. In Table 1, what are the first and third quartiles? What do they lead you to think about the skewness of
the distribution? (3)
Solution: There are 50 numbers and the position formula gives us, for Q1, pn  1  .25(51)  12.75 .
x12  51 and x13  52 , so we have a  12 and .b  0.75 . x1 p  xa  .b( xa1  xa ) .
So Q1  x1.25  x.75  x12  0.75( x13  x12 )  51  0.7552  51  51 .75
For Q2, pn  1  .75(51)  38 .25 and x 38  63 and x 39  64 . So we have a  38 and .b  0.25 and
Q1  x1.75  x.25  x38  0.25( x39  x38 )  63  0.2564  63  63 .25 . 58 – 51.75 = 6.25 and 63.25 – 58 =
5.25. The data seems to be more spread out on the left, so it may be skewed to the left.
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10. In Table 1, assume that you were asked to present the data in 5 classes. Show how you would decide
what class interval to use and list the classes below with their frequencies. (4) [27]
Class
A
B
C
D
E
___
___
___
___
___
to
to
to
to
to
under
under
under
under
under
Frequency
___
___
___
___
___
___
___
___
___
___
Solution: The highest number is 75 and the lowest is 44. We calculate
75  44
 6.2 . Use 7 or 8.
5
If you use 10 and start at 40, you will only get 4 classes.
I will repeat the table. Two possible arrangements are below.
4|44668899
8 items
5|00112244555577888899
20 items
6|000022223344557799
18 items
7|3355
4 items
Class
A
B
C
D
E
_40
_48
_56
_64
_72
Table 2
Class
20-35
35-50
50-65
65-80
80-95
to
to
to
to
to
under
under
under
under
under
Frequency
_48
_56
_64
_72
_80
__6_
_14_
_18_
_10_
__2_
f
11
f rel
.122
.278
F
11
36
10
.111
1.000
90
Class
A
B
C
D
E
_42
_49
_56
_63
_70
to
to
to
to
to
under
under
under
under
under
Frequency
_49
_56
_63
_70
_77
__6_
_14_
_16_
_10_
__4_
Frel
.122
.400
.700
.889
11. Fill in the missing numbers in Table 2 (3)
Class
f rel
f
Frel
F
20-35 11
.122
11
.122
35-50 25
.278
36
.400
50-65 27
.300
63
.700
65-80 17
.189
80
.889
80-95 10
.111
90
1.000
12. Find the 1
fractile of the data in Table 2. (3) [33]
3
position  pn  1  13 91  30.333 . This is above F  11 and below F  36, so the interval is the 2nd,
 pN  F 
35 to 50, which has a frequency of 25. Each interval width is 35 - 20 = 15. x1 p  L p  
 w so
 f p 
 1 90   11 
x1 13  x 2 3  35   3
15   35  0.760 15   46 .40
25


5
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13. The Des Moines Metropolitan area consists of Dallas, Polk and Warren Counties. In the 1990 census
Dallas County had 30700 people and a mean income of $9673, Polk County had 309000 people and a mean
income of $11779 and Warren County had 37300 people with a mean income of $9749.
a) What is the mean income of residents of the Des Moines area? (Note: I am much more interested in
how you do this that whether you get the right answer, so show your work!) (2)
x
w
9673
30.7
296961.1
xw 4300309 .8
11779 309.0 3639711.0
xw 

 11406 .7
377 .0
w
9749
37.3
363637.7
377.0 4300309.8


b) Suppose that in 1991 Census officials believe that the population and mean income in Polk county and
Warren County stayed the same, but the mean income in Dallas county fell while the population stayed
constant. For the Des Moines area, which of the following is likely to have occurred? (2) [37]
a) The median income fell most
b) The modal income fell most
c) *The mean income fell most
d) The mean, median and modal income all fell to the same degree.
Explanation: In general, the mean income is more influenced by movements of individual numbers that
the median or mode. The mode, especially will be unlikely to be influenced since the overwhelming
majority were Polk County and poorer people seem to be more concentrated in Dallas County.
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ECO251 QBA1
FIRST EXAM
February 21, 2008
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 III. Do all the Following (12+ Points). The problems are based on problems by Doane and Seward.
Show your work!
1. The table below represents the distribution of winning times in seconds for horses in the Kentucky
Derby. Treat these data as a sample. Personalize the data below by adding the last two digits of your
student number to the last 2 frequencies. .Add one digit per frequency For example, Seymour Butz’s
student number is 876509 so he adds 0 to the second-to-last frequency and 9 to the last frequency and uses
{2, 5, 16, 22, 12, 8, 5, 3 and 11} (adding to 85). You may check your work on the computer, but what is
turned in should look as if it had all been done by hand.
Row
Time in seconds
Frequency
a. Calculate the Cumulative Frequency (0.5)
b. Calculate the Mean (0.5)
c. Calculate the Median (1)
d. Calculate the Mode (0.5)
e. Calculate the Variance (1.5)
f. Calculate the Standard Deviation (1)
g. Calculate the Interquartile Range (1.5)
h. Calculate a Statistic showing Skewness and
interpret it (1.5)
i. Make an ogive of the data (Neatness
Counts!)(1)
j. Extra credit: Put a (horizontal) box plot below the ogive using the same horizontal scale (1)
k. (Extra, extra credit) the trimmed mean is a measure of the data that mitigates the effect of extreme
values. A 5% trimmed mean is a mean calculated after 5% of the data is removed from both the top and
bottom of the data. (For example, if there are 100 points, the top 5 and the bottom 5 are removed and the
mean of the middle 90 points is calculated.) Try to calculate a 5% trimmed mean of your Kentucky Derby
numbers. (1)
l. (More extra credit) Davies’ test: In 1929 Professor George Davies tried to find a method that would tell
us whether an arithmetic mean or a geometric mean was a better way to characterize a set of data. Davies
logQ1  logQ2  2 logx.50  . We should decide to use the geometric mean if
recommended using D 
log Q3  log Q1
(α) D  0.20 . (β) The data seems to be convincingly skewed to the right. (γ) There are at least 50
observations.
(i) Using the Derby data compute D and make a recommendation with an explanation.
(ii) See if you can compute a geometric mean for this grouped data. Be sure that you explain what
you do so that I can follow it.
1
2
3
4
5
6
7
8
9
119
120
121
122
123
124
125
126
127
to
to
to
to
to
to
to
to
to
under
under
under
under
under
under
under
under
under
120
121
122
123
124
125
126
127
128
1
5
16
22
12
8
5
3
2
Note that unreasonable answers are answers where the mean, median, mode, first quartile and third
quartile do not fall between 119 and 128.
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Solution using the original numbers with 5 added to the last two frequencies: If we use the original
numbers and the computational method or the definitional method (Columns 1-5, 8-11), we get the
following. x is the midpoint of the class.
Row
1
2
3
4
5
6
7
8
9
Class
119-120
120-121
121-122
122-123
123-124
124-125
125-126
126-127
127-128
f
1
5
16
22
12
8
5
8
7
84
fx2
fx
F
x
1
6
22
44
56
64
69
77
84
119.5
120.5
121.5
122.5
123.5
124.5
125.5
126.5
127.5
fx3
119.5
14280
1706490
602.5
72601
8748451
1944.0 236196 28697814
2695.0 330138 40441844
1482.0 183027 22603835
996.0 124002 15438249
627.5
78751
9883282
1012.0 128018 16194277
892.5 113794 14508703
10371.0 1280807 158222944
If we use the original numbers and the definitional method, we get the following for the frequencies. x is
the midpoint of the class. I usually tell people that they are wasting their time if they use the definitional
method. Because of the large numbers here that may not be true.
Row Class
1 119-120
2 120-121
3 121-122
4 122-123
5 123-124
6 124-125
7 125-126
8 126-127
9 127-128
f
F
1
5
16
22
12
8
5
8
7
84
1
6
22
44
56
64
69
77
84
119.5
120.5
121.5
122.5
123.5
124.5
125.5
126.5
127.5
119.5
602.5
1944.0
2695.0
1482.0
996.0
627.5
1012.0
892.5
10371.0
-3.96429
-2.96429
-1.96429
-0.96429
0.03571
1.03571
2.03571
3.03571
4.03571
-3.9643
-14.8214
-31.4286
-21.2143
0.4286
8.2857
10.1786
24.2857
28.2500
0.0000
If you used the computational method, you would have gotten n 
that the mean is x 
 f  84
 fx  10371  123 .4643 . You would also find
n
84
158222944..
If you used the definitional method, you would have and gotten n 
that the mean is x 
check),
 f x  x 
f x  x 2
f x  x 
xx
fx
x
 fx
15.716
43.935
61.735
20.457
0.015
8.582
20.721
73.724
114.009
358.893
and
2
 f  84
 fx
2
84
 358.893 and
 f x  x 
3
-62.301
-130.236
-121.265
-19.726
0.001
8.888
42.181
223.806
460.107
401.457
 10371 .0 , so
 1280807 and
and
 fx
 fx  10371  123 .4643 . You would have followed by getting
n
f x  x 3
 fx
3

 10371 .0 , so
 f x  x   0 (a
 401.457 .
If you used one of Pearson’s measures of skewness, you would not have bothered with the f x  x 3 or the
fx3 columns. In any case only an Adrian Munk personality would have computed everything here.
a. Calculate the Cumulative Frequency (0.5): See the F column above.
b. Calculate the Mean (0.5): We have already found x 
 fx  10371  123 .4643 .
n
84
c. Calculate the Median (1): position  pn  1  .585   42.5 . This is above F  22 and below F  44,
so the interval is the 4th, 122 to 123, which has a frequency of 22. Each interval width is 123 - 122 = 1.
 pN  F 
 .584   22 
x1 p  L p  
 w so x1.5  x.5  122  
 1  122  0.90909 1  122 .9091 .
22


 f p 
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d. Calculate the Mode (0.5): The largest group is 122 to 123, which has a frequency of 22, so by convention
the mode is its midpoint, which is mo  122.5. It is possible that you will have two modes. Note that to be
reasonable, Q1  x50  Q3 and that Q1, x50 , Q3, x and the mode must be between 119 and 128.
e. Calculate the Variance (1.5): We have

f x  x 2  358.893 . s 2 
s2 
 f x  x 
n 1
2

 fx
2
 fx
 nx 2
n 1
2

 1280807 and x  123 .4643 or
1280807  84 123 .4643 2 358 .597

 4.32045 or
83
83
358 .893
 4.32401 . The computer got 346637 too.
83
f. Calculate the Standard Deviation (1): s  4.32401  2.07943 or s  4.32045  2.0786
g. Calculate the Interquartile Range (1.5) First Quartile: position  pn  1  .2585   21 .25 . This is above
F  6 and below F  22, so the interval is the 3rd, 121 to 122, which has a frequency of 16.
 pN  F 
 .2584   6 
x1 p  L p  
 w gives us Q1  x1.25  x.75  121  
 1  121  0.9375 1  121 .9375 .
16


 f p 
Third Quartile: position  pn  1  .7585   63.75 . This is above F  56 and below F  64, so the
interval is the 6th, 124 to 125 which has a frequency of 8.
 .7584   56 
Q3  x1.75  x.25  124  
 1  124  0.875 1  124 .875 .
8


IQR  Q3  Q1  124 .875  121 .9375   2.945
h. Calculate a Statistic showing Skewness and interpret it (1.5)
fx  10371 .0 , so that the mean is x  123 .4643 . We also found that, for the
We had n  84 and

computational method find
 fx
2
 1280807 and
 fx
3
 158222944. If we use the definitional
 f x  x   401.457 . We also have s  2.07943 , x  122 .9091 and mo  122.5.
n
 fx  3x  fx  2nx   838482 158222944  3123 .4643 1280807  284123 .4643  
k 
(n  1)( n  2) 
3
method
.5
3
2
3
3
3
 0.012342 158222944  474401819 .1  316179331 .6  0.012342 456 .5  5.6341 .
(The computer got the same answer as for the definitional formula)
or k 3 
n
(n  1)( n  2)
or g 1 
k3
or
s
3

 f x  x 
4.9548
2.07943 3
3

84
401 .457   4.9548 .
8382 
 0.5579
Pearson's Measure of Skewness SK1 
mean  mode  123 .4643  122 .5  0.4637
std .deviation
2.07943
or
3mean  median 3123 .4643  122 .9091 

 0.5961
2.07943
std .deviation
Because of the positive sign, the measures all imply (slight) skewness to the right..
SK 2 
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i. Make an ogive of the data (Neatness Counts!)(1)
x
Row Class
f
fx
F
1
2
3
4
5
6
7
8
9
119-120
120-121
121-122
122-123
123-124
124-125
125-126
126-127
127-128
1
5
16
22
12
8
5
8
7
84
1
6
22
44
56
64
69
77
84
119.5
120.5
121.5
122.5
123.5
124.5
125.5
126.5
127.5
fx2
fx3
119.5
14280
1706490
602.5
72601
8748451
1944.0 236196 28697814
2695.0 330138 40441844
1482.0 183027 22603835
996.0 124002 15438249
627.5
78751
9883282
1012.0 128018 16194277
892.5 113794 14508703
10371.0 1280807 158222944
Solution: An ogive is a graph of the cumulative distribution. The first point will be (119, 0), the second
point will be (120, 1), the third point will be (121, 6) etc.
j. Extra credit: Put a (horizontal) box plot below the ogive using the same horizontal scale (1)
Solution: The box plot would have a box beginning at the first quartile and ending at the third quartile. A
band across the box will indicate the median. Whiskers will indicate the range of the data.
The five-number summary is (119, 121.9375, 122.9091, 124.8750, 128).
IQR  Q3  Q1  124 .875  121 .9375   2.945 . 1.5( IQR )  4.4175 If you use fences, they should be at
121 .9375  4.4175  117 .52 and 124 .875  4.4175  129 .29 . But these are beyond the range of the data,
which makes them irrelevant. So the box extends from 121.94 to 124.88, with a median marked by a
vertical line at 122.91. The whiskers go from the box to 119 and 128 with dotted lines showing the full
range unnecessary. A rough picture is below.
119
121
123
127
129
k. (Extra, extra credit) the trimmed mean is a measure of the data that mitigates the effect of extreme
values. A 5% trimmed mean is a mean calculated after 5% of the data is removed from both the top and
bottom of the data. (For example, if there are 100 points, the top 5 and the bottom 5 are removed and the
mean of the middle 90 points is calculated.) Try to calculate a 5% trimmed mean of your Kentucky Derby
numbers. (1)
There are 84 points here, so we remove 4 points ar each end.
x
Row Class
f
fx
fx2
fx3
F
1
2
3
4
5
6
7
8
9
119-120
120-121
121-122
122-123
123-124
124-125
125-126
126-127
127-128
0
2
16
22
12
8
5
8
3
76
1
6
22
44
56
64
69
77
84
The trimmed mean would be
119.5
120.5
121.5
122.5
123.5
124.5
125.5
126.5
127.5
0.0
241.0
1944.0
2695.0
1482.0
996.0
627.5
1012.0
382.5
9380.0
9380 .0
 123 .4211 . There weren’t any real extreme points, so the mean
76
wasn’t changed much.
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l. (More extra credit) Davies’ test: In 1929 Professor George Davies tried to find a method that would tell
us whether an arithmetic mean or a geometric mean was a better way to characterize a set of data. Davies
logQ1  logQ3  2 logx.50 
recommended using D 
. We should decide to use the geometric mean if
log Q3  logQ1
(α) D  0.20 . (β) The data seems to be convincingly skewed to the right. (γ) There are at least 50
observations.
(i) Using the Derby data compute D and make a recommendation with an explanation.
The five-number summary is (119, 121.9375, 122.9091, 124.8750, 128). As logs to base 10, the three
middle numbers would be (2.08614, 2.08958, 2.09648) .. So
(2.08614  2.09648 )  22.08958  .00346
D

 .3346 . There is no way we would use the geometric
2.09648  2.08614
.01034
mean here since D is above .20 and there is little skewness anyway.
(ii) See if you can compute a geometric mean for these grouped data. Be sure that you explain
what you do so that I can follow it.
1
ln( x)  and make it into
The easiest way to think about this is to look at the formula ln x g 
n
 
 



f ln( x)  This is equivalent to doing x g  x1f1  x 2f 2  x3 3  x nf n
ln x g 
1
n
Here n 
 f . I’ll leave the work to you.
f

1
n
n
x
f
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2. The sales of your firm over the last 5 years are as follows.
Year
1
2
3
4
5
Sales ($millions)
131
227
311
354
403
Personalize these data by subtracting the last two digits of your student number from the first sales figure.
For example, Ima Badrisk’s student number is 876519, so she subtracts 19 from 131 to get 112. The effect
of this will be to raise the growth rates in a). Do the following (3)
a) Find the average growth rate of sales by taking a geometric mean using the four year-to-year growth
rates.
b) Find the harmonic mean of your sales numbers.
c) Find the root-mean-square of your sales numbers.
d) (Extra credit) Compute the geometric mean from a) using natural and/or base 10 logarithms. (1 point
extra credit each).
 
a) The Geometric Mean. The formula table says ln x g 
1
n
 ln( x) , but I said in class that this could
be either natural logs or logs to the base 10.
227 is 1.7328 times 131, 311 is 1.3700 times 227, 354 is 1.1382 times 311 and 403 is 1.1384 times 354.
The relevant geometric mean is 1  r  4 1.7328 1.3700 1.1382 1.1384   4 3.0760  1.3243 . The average
growth rate is 32.43%.
b) The Harmonic Mean. The formula table says
1
1

xh n
 x or x1  15  1311  2271  3111  3541  4031  
1
h
1
1
1
 0.0205606   0.0041121 . So xh 

 243 .1836 .
1
1 0.0041121
5
n
x
Of course some of you decided that

1 1

xh n
1 1
1 
1

 x  5  131  227  311  354  403   ? 5  131  227  311  354  403   ??? .
1
1
1
1
1
This is, of course, an easier way to do the problem. It is also wrong, and you will get an A for the
course if you can prove to me that it is not wrong! And please don’t try any math if you get on “Are
you smarter than a fifth-grader.”
1
1
x 2 or x rms 2 
x2
c) The root-mean-square. The formula table says x rms 
n
n
1
453136
x rms 2  131 2  227 2  311 2  354 2  403 2 
 90627 .2 . So x rms  90627 .2  301 .044 .
5
5
Sums for Harmonic mean and RMS
1
Row x
x2
x


1
2
3
4
5


131
227
311
354
403
17161
51529
96721
125316
162409
453136
0.0076336
0.0044053
0.0032154
0.0028249
0.0024814
0.0205606
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d) (Extra credit) Compute the geometric mean from a) using natural and/or base 10 logarithms. (1 point
extra credit each).
1
ln( x)  . So we have ln 1  r   1 ln(1  r ) 
Natural Logarithms. ln x g 
n
n
1
ln 1  r   1.12362   0.2809 and 1  r  e 0.2809  1.3243 .
4
1
log( x) . So we have log1  r   1 log(1  r )
Logarithms to the base 10. log x g 
n
n
1
log 1  r   0.487982   0.1220 and 1  r  10 0.1220  1.3243
4
 

 



Sums for geometric mean using natural and base 10 logs.
Row 1  r 
ln 1  r 
log1  r 
1
2
3
4
1.7328
1.3700
1.1382
1.1384
0.549739
0.314811
0.129448
0.129624
1.12362
0.238748
0.136721
0.056219
0.056295
0.487982
13