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251y0511 2/14/05
ECO251 QBA1
FIRST HOUR EXAM
February 18, 2005
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: Prem S. Mann) The following numbers represent the price earnings ratio of 12 corporations.
7, 16, 18, 18, 22, 20, 20, 19, 31, 34, 38, 58
Compute the following:
a) The Median (1)
b) The Standard Deviation (3)
c) The 61st percentile (2)
d) The Coefficient of variation (1)
Solution: The numbers in order are 7, 16, 18, 18, 19, 20, 20, 22, 31, 34, 38, 58.
n  12.
a) pn  1  .513  65 The middle
x
7
x2
49
16
256
18
324
x4
x5
18
324
19
361
x6
20
400
x7
20
400
x8
22
484
x9
31
961
x10
34
1156
s  179.3579  13.392
c) pn  1  .6113   7.93 . So a  7 and
.b  0.93
x1 p  xa  .b( xa1  xa ) so
x11
x12
38
1444
x1.61  x.39  x7  0.93( x8  x7 )
58
3364
301
9523
 20  0.93(22  20)  21.86
s
13 .392
d) C  
 0.5339 or 53.39%
x 25 .0833
x1
x2
x3
Total
numbers are the 6th and 7th number, which are
x  x8
 20 .
both 20. x.50  7
2
x 301

 25 .0833 ,
b) x 
n
12
x 2  nx 2 9523  12 25 .0833 2
2
s 

n 1
12  1
1972 .9367

 179 .3579 . So
11


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251x0511 2/16/05
How mean and variance were checked. The numbers were put into c1.
————— 2/15/2005 10:44:12 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > let c2=c1*c1
MTB > sum c1
Sum of x
Sum of x = 301
MTB > sum c2
Sum of xsq
Sum of xsq = 9523
MTB > describe c1;
SUBC> mean;
SUBC> variance;
SUBC> stdev.
Descriptive Statistics: x
Variable
Mean StDev
x
25.08 13.39
Variance
179.36
MTB > print c1 c2
Data Display
Row
x
xsq
1
7
49
2 16
256
3 18
324
4 18
324
5 19
361
6 20
400
7 20
400
8 22
484
9 31
961
10 34 1156
11 38 1444
12 58 3364
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251x0511 2/16/05
Part II. (At least 35 points – 2 points each unless marked - Parentheses give points on individual
questions. Brackets give cumulative point total.)
1.
I have the average time of the first 10 runners in the Boston Marathon.
a) Is this a parameter or a statistic? (Think!)
b) What symbol should you use to indicate this mean?
[2]
Answer: This is a parameter, since the first 10 runners are a population – we have all of them.
The symbol for a population mean is mu   .
2.
The data in question 1 is an example of
a) Ordinal Data
b) Nominal Data
c) Discrete ratio data
d) Continuous interval data
e) *None of the above.
[4]
Answer:
They are none of the above. Both the times of the runners and their average are continuous
ratio data.
3.
Assume now that I have the times of all the runners who finish the Boston Marathon and that
the first ten or 20 runners have times that are far below most of the rest, but that the more
typical runners are relatively close together. Which of the following is most likely?
a) *mean < median < mode
b) mean < mode < median
c) mode < mean < median
d) mode < median < mean
e) none of the above.
[6]
Answer: The description (which might be highly inaccurate) is of a data set that is skewed to
the left. The mean is the least robust of the measures of central tendency, so it will be pulled
furthest to the right. The mode is the most robust and will hold its ground. The median is
generally between them.
4.
Mark the variables below as qualitative (A) or quantitative (B)
a) Celsius Temperature
b) Absolute Temperature
c) Cost of a new thermometer
d) The number of thermometers you have in your house.
[8]
Answer: All of these variables are quantitative (B). The first is continuous interval data, the
second and third are usually considered continuous ratio data and the last is discrete ratio data.
5.
Which of the following is not a dimension – free measurement.
a) *The population variance.
b) Pearson’s measure of skewness
c) g 1
d) The coefficient of variation.
e) The coefficient of excess.
f) All of the above are dimension free
g) None of the above are dimension free.
[10]
Explanation: b – e are dimension-free ratios because they have data that are measured in the
same units in their numerators and denominators.
6.
Classify a deck of cards as follows: Write yes or no in each location.
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251x0511 2/16/05
A1 Hearts; A2 Red cards; A3 Black cards; A4 Face cards (4)
A1 and A2
A2
and A3
A2 , A3 , A4
A1 and A3
Mutually Exclusive?
__no__
Collectively Exhaustive?
__no__
__yes__
__yes_
__no__
__yes_
__yes__
__no__
[14]
Explanation: A1 is inside A2 , and A2 hardly includes all cards. A2 and A3 do not overlap and
between them include all cards. A4 includes parts of both A2 and A3 , but since A2 and A3
already include all cards, the three classes are collectively exhaustive. Hearts cannot be black cards
but A1 and A3 together leave out diamonds.
7.
What characteristic do the variance, standard deviation and Interquartile range have in
common that they do not share with the mean, median, mode or skewness? (1)
[15]
Answer: They are all measures of dispersion.
Exhibit 1. The boxplot, stem-and-leaf display and 5 number summary describe the data set ‘Length’
Boxplot of Length
602
Length
601
600
599
598
597
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Exhibit 1. (ctd.)
Stem-and-Leaf Display: Length (Numbers are in the 2nd and 3rd columns – 1st column is a form of
cumulative count)
1
597 2
4
597 688
17 598 0000222224444
28 598 88888888888
41 599 0000022244444
44 599 666
49 600 00224
(6) 600 668888
45 601 00000000002222222222224444444444444
10 601 666666888
1
602 2
Five number summary: Length
597.20
598.80
600.60
601.20
Descriptive Statistics: Length
602.20
Mean 600.07
StDev 1.34
8. What are the median and the interquartile range for the data set in exhibit 1? [17]
Solution: The mean is the middle number of the 5 number summary, 600.60. The IQR
= Q3 – Q1 = 601.20 – 598.80 = 2.40.
9.
Assume that you were asked to present these data in seven intervals, what class interval would
you use? (Show your work!!!)
Solution: The highest number is 602.20 and the lowest is 597.20. w 
 .7143 . 0.8 is probably the smallest interval one could consider.
10. Show the intervals that you would actually use.
Class
A
B
C
D
E
F
G
From
597.0
597.8
598.6
599.4
600.2
601.0
601.8
to
597.7
598.5
599.3
600.1
600.9
601.7
602.5
602.20  597.20
7
[21]
11. If we take the area between 602.75 and 597.39
a) According to the empirical rule, what percent of the data should be between these points?
Solution: It says: Mean 600.07 StDev 1.34. 597.39 = 600.07 – 2(1.34) . 602.75 = 600.07
+ 2(1.34) This is two standard deviations from the mean and should include about 95% of the
data in a unimodal symmetrical distribution.
b) According to the Tchebyschev inequality, what percent of the data should be between
these points?
Solution: If k  2, the tails should include at most 1 2  1 2  .25 or 25% - so the center
k
2
should have at least 75%.
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251x0511 2/16/05
c) What percent of the data is actually between these points.? Comment.
Solution: 597.2 is the only point that is not between 597.39 and 602.75. So 99% of the 100
points are in the interval. This is above what is predicted by the empirical rule, which is not
surprising, since the distribution is asymmetrical and bimodal. It is above the 75% in the
Tchebyschev inequality, which is actually exactly what the inequality says it will be.
Original Data
Frequency
1
4
17
28
41
44
49
6)
45
10
1
Exhibit 2:
597
597
598
598
599
599
600
600
601
601
602
class
2
688
0000222224444
88888888888
0000022244444
666
00224
668888
00000000002222222222224444444444444
666666888
2
f
0-10
10-20
20-30
30-40
40-50
Total
f rel
F
1
3
13
11
13
3
5
6
35
9
1
100
[27]
Frel
.10
.30
.70
.20
.10
100
xxxx
1.00
xxxx
12. Fill in the missing numbers in Exhibit 2. (4)
[31]
Explanation: n is the sum of the f column and also must be the last number in the F column.
F
f
and f rel  , so the F and f columns can be gotten by multiplying Frel and f rel by
Frel 
n
n
n . Since Frel is the cumulative version of f rel , the missing number in the Frel column can be
gotten by adding .20 to .70 to get .90.
class
f
F
f
F
rel
0-10
10-20
20-30
30-40
40-50
Total
10
20
40
20
10
100
.10
.20
.40
.20
.10
1.00
rel
10
30
70
90
100
xxxx
.10
.30
.70
.90
1.00
xxxx
13. Find the 61st percentile in exhibit 2.(3)
[34]
Solution: position  pn  1  .61101  61.61 . This is above 30 and below 70 in the F column,
 pN  F 
so we look for the 61st percentile in the 20-30 class. x1 p  L p  
 w , so
 f p 
 0.61100   30 
x.39  20  
10  20  0.775 10   27.75.
40


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251x0511 2/16/05
14. An Economics course has students of all classes in it. 10% Freshmen, 46% sophomores,
30%Juniors and 14% Seniors. Make this information into a Pareto chart. (4)
[38]
Solution: The graph consists of 4 labeled bars with heights 46%, 30%, 14% and 10%. The line
starts at 46% and rises to 76%, 90% and 100%.
Extra Credit. I forwarded this to several people a few days ago. The headline appeared
in a Slovak newspaper. The subject line is my comment. From what you know about ordinal data,
write a short essay explaining my comment. I’m looking for your ability to express yourself
persuasively as much as for facts.
Subject: A great example of the uselessness of ordinal data
Bratislava is the 44th most expensive city in the world
Suggested comments: The main thing that we learned about ordinal data is that the differences
between successive items in the listing are not usually equal. What the reader wants to know is how
expensive Bratislava is relative to cities that the reader might know better, perhaps like New York
and Tokyo. Actually nowhere in the article does it say how many cities were ranked, but that
wouldn’t really help. It is still possible that Bratislava, as the 45th city is still not very expensive
relative to the, perhaps, 100 cities below it, and that Bratislava is very cheap compared with the 10
or so most expensive. But it is also possible that there are 40 or so European cities, including
Bratislava, in the first 45 that are all relatively similar in cost, even if Bratislava is the cheapest.
The study probably computed some sort of index number for businesspersons’ expenses in each
city that would be much more useful. One of my companions in a Slovene language class in
Ljubljana was a Greek lawyer, who was proficient at languages and went to Slovenia on a lark. He
felt that Slovenia’s capital was expensive and had to call home for money, while I found it
incredibly cheap. Index values or estimated costs per day compared to Philadelphia and Athens
would have done us much more good than knowing the ranks of Philadelphia, Ljubljana and
Athens.
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251x0511 2/16/05
ECO251 QBA1
FIRST EXAM
February 18, 2005
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 (11 Points) Show your work!
1. The frequency distribution below represents mortgage payments in hundreds of dollars of 85 families to
the Relatively Reliable Bank and Trust Company. Personalize the data below by adding the last digit of
your student number to the last frequency and the second to last digit of your student number to the second
to last digit,. 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 (1, 4, 18, 32, 16, 10, 0, 13).
Payment
frequency
a. Calculate the Cumulative Frequency (0.5)
b. Calculate the Mean (0.5)
0 - 4
1
c. Calculate the Median (1)
4 - 8
4
d. Calculate the Mode (0.5)
8 – 12
18
e. Calculate the Variance (1.5)
12 – 16
32
f. Calculate the Standard Deviation (1)
16 – 20
16
g. Calculate the Interquartile Range (1.5)
20 – 24
10
h. Calculate a Statistic showing Skewness and
24 - 28
0
28 - 32
4
interpret it (1.5)
i. Make a frequency polygon of the data showing
relative or percentage frequency (Neatness
Counts!)(1)
j. Extra credit: Put a (horizontal) box plot below
the relative frequency chart using the same scale
(1)
.
Solution: If we use the Seymour’s numbers and the computational method, we get the following.
Row Class
f F x
fx2
fx3
fx
1
0- 4
1
1
2
2
4
8
x is the class midpoint.
2
3
4
5
6
7
8
4- 8
8-12
12-16
16-20
20-24
24-28
28-32
4
18
32
16
10
0
13
94
5
23
55
71
81
90
94
6
10
14
18
22
26
30
24
144
180 1800
448 6272
288 5184
220 4840
0
0
390 11700
1552 29944
864
18000
87808
93312
106480
0
351000
657472
If we like to work, we may have used the definitional method instead.
Row
1
2
3
4
5
6
7
8
Class
0- 4
4- 8
8-12
12-16
16-20
20-24
24-28
28-32
f
x
fx
1
4
18
32
16
10
0
13
94
2
6
10
14
18
22
26
30
2
24
180
448
288
220
0
390
1552
xx
-14.5106
-10.5106
-6.5106
-2.5106
1.4894
5.4894
9.4894
13.4894
f x  x 
-14.511
-42.043
-117.191
-80.340
23.830
54.894
0.000
175.362
0.000
f x  x 2
f x  x 3
210.56
441.89
762.99
201.71
35.49
301.33
0.00
2365.52
4319.49
-3055.3
-4644.6
-4967.6
-506.4
52.9
1654.1
0.0
31909.3
20442.4
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251x0511 2/16/05
From the f and fx columns of either display, we find n 
 fx  1552  16.5106 . We also find that
 f  94 and  fx
 fx  657472 ,
 f x  x   0 (except for a rounding error),  f x  x 2  3478.47, and  f x  x 3  20442.4
mean is x 
n
94
 fx
 1552 , so that the
2
 29944 ,
3
(Note that, to be reasonable, the mean, median and quartiles must fall between 0 and 32.)
a. Calculate the Cumulative Frequency (1): (See above) The cumulative frequency is the whole F column.
fx 1552

 16 .5106
b. Calculate the Mean (1): We already found that x 
n
94
c. Calculate the Median (2): position  pn  1  .595   47.5 . This is above F  23 and below

 pN  F 
 w so
 f p 
F  55, so the interval is the 4th, 12 -16, which has a frequency of 32. x1 p  L p  
 .594   23 
x1.5  x.5  12  
4  12  0.754  15
32


d. Calculate the Mode (1) The mode is the midpoint of the largest group. Since 32 is the largest frequency,
the modal group is 12 to 16 and the mode is 14.
fx 2  nx 2 29944  94 16 .5106 2 4319 .6082


 46 .4474 or
e. Calculate the Variance (3): s 2 
n 1
93
93

 f x  x 
2
4319 .49

 46 .4509 . The computer got 46.4461.
n 1
93
f. Calculate the Standard Deviation (2): s  46.4461  6.8151 .
g. Calculate the Interquartile Range (3): First Quartile: position  pn  1  .2595   23.75 . This is above
s2 
 pN  F 
 w gives us,
 f p 
F  23 and below F  55, so the interval is the 4th, 12-16. x1 p  L p  
 .2594   23 
Q1  x1.25  x.75  12  
4  12  .015625 4  12 .0625 .
32


Third Quartile: position  pn  1  .7595   71.25 . This is above F  71 and below F  81, so the
 .7594   71 
interval is the 6th, 20-24. x1.75  x.25  20  
4  20   0.05 4  19.8 . The formula isn’t
10


working as well as it should, so lets try the 5th interval, 16-20.
 .7594   55 
x1.75  x.25  16  
4  16  0.96875 4  19.875 It doesn’t seem to matter which you use.
16


IQR  Q3  Q1  19.8  12.0625  7.7375 .
9
251x0511 2/16/05
h. Calculate a Statistic showing Skewness and interpret it (3):
fx  1552 , so that the mean is x  16.5106 . We also found that
We had n  94 and

 fx  29944 ,  fx
n
 fx
k 
(n  1)( n  2) 
2
3
 657472 and
3
 3x
3
 fx
2
 f x  x 
3

 2nx 3  94
 20442.4 .
93 92 
657472  316.5106 29944  294 16.5106  
3
 0.0109864 657472  1483180 .2  846148 .2  0.0109864 20440   224 .562 .
n
94
or k 3
20442 .4  224 .589 The computer gets 224.589
f x  x 3 
(n  1)( n  2)
9392 

or g1 
k3
s
3

224 .589
6.81514 3
 0.70952
3mean  mode 316 .5106  14 

 .1228
std.deviation
6.81514
Because of the positive sign, the measures imply skewness to the right..
Pearson's Measure of Skewness SK 
or
i. A frequency polygon is a simple line graph with frequency on the y-axis and the numbers 0 to 36 on the xaxis. Each f point is plotted against the midpoint of the class, x . In addition 2 empty classes must be
added. Your graph should show the point (32, 0) and show a line segment rising from (-2, 0) to (2, 0.0106).
Of course, showing the half of the line segment to the left of the y-axis is undesirable. The data Seymour
had is:
x
Row Class
f
f rel
fx
0
1
2
3
4
5
6
7
8
9
-4-0
0- 4
4- 8
8-12
12-16
16-20
20-24
24-28
28-32
32-36
0
1
4
18
32
16
10
0
13
0
94
-2
2
6
10
14
18
22
26
30
34
0
2
24
180
448
288
220
0
390
0
1552
0.0000
.0106
.0426
.1915
.3404
.1702
.1064
0.0000
.1383
0.0000
1.0000
Each number in the f rel column is the corresponding number in the f column divided by
axis could be marked from zero to .35.
n  94. The y
j. The box plot should show the median and the quartiles and use the same x axis as the histogram.
How this was checked.
Three Minitab routines are available to you in PROGRAMS on the website. Click on 'Programs for
grouped data computation.' The three programs are 'grp', 'grpv', and 'grps.' These were run in succession
with the following results.
————— 2/16/2005 7:54:17 PM ————————————————————
Welcome to Minitab, press F1 for help.
Results for: 251x0501-1.MTW
MTB > Save "C:\Documents and Settings\rbove\My Documents\Minitab\251x0501-1.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
10
251x0511 2/16/05
Documents\Minitab\251x0501-1.MTW'
Existing file replaced.
MTB > exec 'grp'
Executing from file: grp.MTB
Data Display
Row Class
1
0- 4
2
4- 8
3
8-12
4 12-16
5 16-20
6 20-24
7 24-28
8 28-32
f
1
4
18
32
16
10
0
13
x
2
6
10
14
18
22
26
30
fx
2
24
180
448
288
220
0
390
fxsq
4
144
1800
6272
5184
4840
0
11700
fxcu
8
864
18000
87808
93312
106480
0
351000
Data Display
n
94.0000
mean
16.5106
Sfx
1552.00
Sfx2
29944.0
Sfx3
657472
Data Display
Row Class
1
0- 4
2
4- 8
3
8-12
4 12-16
5 16-20
6 20-24
7 24-28
8 28-32
x^
-14.5106
-10.5106
-6.5106
-2.5106
1.4894
5.4894
9.4894
13.4894
fx^
-14.511
-42.043
-117.191
-80.340
23.830
54.894
0.000
175.362
fx^sq
210.56
441.89
762.99
201.71
35.49
301.33
0.00
2365.52
fx^cu
-3055.3
-4644.6
-4967.6
-506.4
52.9
1654.1
0.0
31909.3
Data Display
Sfx^
0.000000000
Sfx^2
4319.49
Sfx^3
20442.4
MTB > exec 'grpv'
Executing from file: grpv.MTB
MTB > exec 'grps'
Executing from file: grps.MTB
Data Display
n
94.0000
mean
16.5106
Sfx
1552.00
Sfx2
29944.0
Sfx3
657472
K6
93.0000
Sfx^
0.000000000
Sfx^2
4319.49
Sfx^3
20442.4
var1
46.4461
var2
46.4461
K12
92.0000
K13
0.0109864
k31
224.589
K15
20442.4
k32
224.589
stdev
6.81514
g11
0.709520
g12
0.709520
MTB > Save "C:\Documents and Settings\rbove\My Documents\Minitab\251x0501-1.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My Documents\Minitab\251x0501-1.MTW'
11
251x0511 2/16/05
2. Take your student number followed by 10, 14, 16, 18, 20, 22 as 12 values of x . Change any zeros to
ones. For example, Seymour Butz’s student number is 876509, so he uses 8, 7, 6, 5, 1, 9, 10, 14, 16, 18, 20,
22.
For these twelve numbers, compute the a) Geometric Mean b) Harmonic mean, c) Root-mean-square
(1point each). Label each clearly. If you wish, d) Compute the geometric mean using natural or base 10
logarithms. (1 point extra credit each).
a) The Geometric Mean.
1
x g  x1  x 2  x3  x n  n  n
x 
12
87651910 14 16 1820 22 
 2682408960 00 0.0833333  8.96143 . At least, not many of you tried to
get the answer by dividing 582112000 by 12, but a number of you seem to have convinced yourselves that
you could take a square root instead of an 11th root.
 12 2682408960 00  2682408960 00  12
1
b) The Harmonic Mean.
1
1

xh n
1 1
1 
 x  12  8  7  6  5  1  9  10  14  16  18  20  22 
1
1
1
1 1 1
1
1
1
1
1

1
0.12500  0.14286  0.16667  0.20000  1.0000  0.11111  0.10000  0.07143  0.06250  0.05556  0.05000  0.04545 
12

1
2.13057   0.177548 .
12
So x h 
1
1
n
x
1

1
 5.63229 .
0.177548
Of course some of you decided that 1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1 
xh
n
x
12  8
7
6
5
1
9
10
14
16
18
20
22 
1  1  1  1 
 ??? 
 
. . This is, of course, an easier way to do the problem, but I warned you that
n
x  12  136 


1 1 1
it wouldn’t work. . It is equivalent to believing that  
2 2 4

c) The Root-Mean-Square.
1
1 2
2
x rms

x2 
8  7 2  6 2  5 2  12  9 2  10 2  14 2  16 2  18 2  20 2  22 2
n
12
1
 64  49  36  25  1  81  100  196  256  324  400  484 
12
1
1
x 2  168 .00  12 .9615 .
 2016   168 .000 . So x rms 
n
12
2
1
1
1
2
Of course some of you decided that xrms

x 2  ???
x  136 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 .





 
d) (i) Geometric mean using natural logarithms
ln 8  ln 7   ln 6  ln 5  ln 1  ln 9   ln 10   ln 14   ln 16   ln 18 
1

ln( x)  1 
ln x g 
n
12   ln 20   ln 22 

 


1  2.07944  1.94591  1.79176  1.60944  0.00000  2.19722  2.30259  2.63906 


11   2.77259  2.89037  2.99573  3.09104


1
26 .3152   2.19293
12
12
251x0511 2/16/05
x g  e 2.19293  8.96143 .
So
(ii) Geometric mean using logarithms to the base 10
 
log x g 
1
n
1  log 8  log 7   log 6  log 5  log 1  log 9  log 10   log 14   log 16 


 log( x)  12   log18   log20   log22 

1  0.90309  0.84510  0.77815  0.69897  0.00000  0.95424  1.00000  1.14613 


12   1.20412  1.25527  1.30103  1.34242


1
11.4285   0.952377
12
So x g  10 0.952377  8.96143 .
Notice that the original numbers and all the means are between 1 and 22.
It’s probably more efficient to handle a problem this large in columns. The arithmetic mean is also
computed below.
1
x
Row
x2
logx 
ln x 
x
1
2
3
4
5
6
7
8
9
10
11
12
Total
Total
n
8
7
6
5
1
9
10
14
16
18
20
22
136
0.12500
0.14286
0.16667
0.20000
1.00000
0.11111
0.10000
0.07143
0.06250
0.05556
0.05000
0.04545
2.13057
11.3333 0.177548
So, as before x  11 .3333 , xh 
64
49
36
25
1
81
100
196
256
324
400
484
2016
168
0.90309
0.84510
0.77815
0.69897
0.00000
0.95424
1.00000
1.14613
1.20412
1.25527
1.30103
1.34242
11.4285
0.952377
2.07944
1.94591
1.79176
1.60944
0.00000
2.19722
2.30259
2.63906
2.77259
2.89037
2.99573
3.09104
26.3152
2.19293
1
 5.63229 , xrms  168  12 .9615 ,
0.177548
x g  10 0.952377  8.96143
and x g  e 2.19293  8.96143 . .
13