Preview Extract

6
Chapter 2
Chapter
Methods for Describing
Sets of Data
2
2.2
In a bar graph, a bar or rectangle is drawn above each class of the qualitative variable
corresponding to the class frequency or class relative frequency. In a pie chart, each slice of
the pie corresponds to the relative frequency of a class of the qualitative variable.
2.4
First, we find the frequency of the grade A. The sum of the frequencies for all 5 grades must
be 200. Therefore, subtract the sum of the frequencies of the other 4 grades from 200. The
frequency for grade A is:
200 ๏ญ (36 + 90 + 30 + 28) = 200 ๏ญ 184 = 16
To find the relative frequency for each grade, divide the frequency by the total sample size,
200. The relative frequency for the grade B is 36/200 = .18. The rest of the relative
frequencies are found in a similar manner and appear in the table:
Grade on Statistics Exam
A: 90๏ญ100
B: 80๏ญ 89
C: 65๏ญ 79
D: 50๏ญ 64
F: Below 50
Total
Relative Frequency
.08
.18
.45
.15
.14
1.00
a.
The graph shown is a pie chart.
b.
The qualitative variable described in the graph is opinion on library importance.
c.
The most common opinion is more important, with 46.0% of the responders indicating
that they think libraries have become more important.
d.
Using MINITAB, the Pareto diagram is:
Chart of Percent
50
40
Percent
2.6
Frequency
16
36
90
30
28
200
30
20
10
0
More
Same
Less
Importance
Of those who responded to the question, almost half (46%) believe that libraries have
become more important to their community. Only 18% believe that libraries have
become less important.
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
2.8
a.
From the pie chart, 50.4% or 0.504 of adults living in the U.S. use the internet and pay
to download music. From the data, 506 out of 1,003 adults or 506/1,003 = 0.504 of
adults in the U.S. use the internet and pay to download music. These two results agree.
b.
Using MINITAB, a pie chart of the data is:
7
Pie Chart of Download-Music
C ategory
Pay
No Pay
No Pay
33.0%
Pay
67.0%
a.
Data were collected on 3 questions. For questions 1 and 2, the responses were either
โyesโ or โnoโ. Since these are not numbers, the data are qualitative. For question 3, the
responses include โcharacter countsโ, โroots of empathyโ, โteacher designedโ, otherโ, and
โnoneโ. Since these responses are not numbers, the data are qualitative.
b.
Using MINITAB, bar charts for the 3 questions are:
Chart of Classroom Pets
60
50
40
Count
2.10
30
20
10
0
Yes
No
Classroom Pets
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8
Chapter 2
Chart of Pet Visits
40
Count
30
20
10
0
Yes
No
Pet Visits
Chart of Education
30
25
Count
20
15
10
5
0
Character Counts Roots of Empathy Teacher designed
Other
None
Education
c.
2.12
Many different things can be written. Possible answers might be: Most of the classroom
teachers surveyed (61 / 75 ๏ฝ 0.813) keep classroom pets. A little less than half of the
surveyed classroom teachers (35 / 75 ๏ฝ 0.467) allow visits by pets.
Using MINITAB, the pie chart is:
Pie Chart of Loc
Category
Urban
Suburban
Rural
Rural
5.7%
Suburban
32.8%
Urban
61.5%
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
2.14
a.
b.
2.16
The two qualitative variables graphed in the bar charts are the occupational titles of clan
individuals in the continued line and the occupational titles of clan individuals in the
dropout line.
In the Continued Line, about 63% were in either the high or the middle grade. Only
about 20% were in the nonofficial category. In the Dropout Line, only about 22% were
in either the high or middle grade while about 64% were in the nonofficial category.
The percentages in the low grade and provincial official categories were about the same
for the two lines.
Using MINITAB, the Pareto chart is:
Chart of Allocation
.14
Relative frequency
.12
.10
.08
.06
.04
.02
0
#5
#8
#6
#7
#10
#11
#2
#3
#4
#9
#1
Track
Proportion within all data.
From the graph, it appears that tracks #5 and #8 were over-utilized and track #1 is underutilized.
a. Using MINITAB, the Pareto chart of the total annual shootings involving the Boston
street gang is:
Chart of Total Shootings
.5
Proportion of Total Shootings
2.18
9
.4
.3
.2
.1
0
2006
2007
2009
2008
2010
Year
Proportion within all data.
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10
Chapter 2
b. Using MINITAB, the Pareto chart of the annual shootings of the Boston gang members
is:
Chart of Gang Member Shootings
Proportion of Gang Members
.5
.4
.3
.2
.1
0
2006
2007
2009
2008
2010
Year
Proportion within all data.
c.
Using MINITAB, the side-by-side bar graphs showing the distribution of dives for the three
match situations are:
Chart of Team behind, Tied, Team ahead
Left
Team behind
Middle
Tied
Right
.8
.6
.4
Proportion
2.20
Because the proportion of shootings per year dropped drastically after 2007 for both the
total annual shootings and annual shootings of the Boston street gang members, it
appears that Operation Ceasefire was effective.
.2
0
Team ahead
.8
.6
.4
.2
0
Left
Middle
Right
Dive
Proportion within all data.
From the graphs, it appears that when a team is tied or ahead, there is no difference in the
proportion of times the goal-keeper dives right or left. However, if the team is behind, the
goal-keeper tends to dive right much more frequently than left.
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
2.22
11
Using MINITAB, a bar graph of the data is:
Chart of Measure
9
8
7
Freq
6
5
4
3
2
1
0
Total Visitors
Paying visitors
Big shows
Funds raised
Members
Measure
The researcher concluded that โthere is a large amount of variation within the museum
community with regards to . . . performance measurement and evaluationโ. From the data,
there are only 5 different performance measures. I would not say that this is a large amount.
Within these 5 categories, the number of times each is used does not vary that much. I would
disagree with the researcher. There is not much variation.
Using MINITAB a bar chart for the Extinct status versus flight capability is:
Chart of Extinct, Flight
80
70
60
50
Count
2.24
40
30
20
10
0
Flight
Extinct
No
Yes
Absent
No
Yes
No
No
Yes
Yes
It appears that extinct status is related to flight capability. For birds that do have flight
capability, most of them are present. For those birds that do not have flight capability, most
are extinct.
Copyright ยฉ 2017 Pearson Education, Inc.
12
Chapter 2
The bar chart for Extinct status versus Nest Density is:
Chart of Extinct, Nest Density
60
50
Count
40
30
20
10
0
Nest Density
H
Extinct
L
H
Absent
L
H
No
L
Yes
It appears that extinct status is not related to nest density. The proportion of birds present,
absent, and extinct appears to be very similar for nest density high and nest density low.
The bar chart for Extinct status versus Habitat is:
Chart of Extinct, Habitat
40
Count
30
20
10
0
Habitat
Extinct
A
TA
Absent
TG
A
TA
No
TG
A
TA
TG
Yes
It appears that the extinct status is related to habitat. For those in aerial terrestrial (TA), most
species are present. For those in ground terrestrial (TG), most species are extinct. For those
in aquatic, most species are present.
2.26
The difference between a bar chart and a histogram is that a bar chart is used for qualitative
data and a histogram is used for quantitative data. For a bar chart, the categories of the
qualitative variable usually appear on the horizontal axis. The frequency or relative
frequency for each category usually appears on the vertical axis. For a histogram, values of
the quantitative variable usually appear on the horizontal axis and either frequency or relative
frequency usually appears on the vertical axis. The quantitative data are grouped into
intervals which appear on the horizontal axis. The number of observations appearing in each
interval is then graphed. Bar charts usually leave spaces between the bars while histograms
do not.
2.28
In a histogram, a class interval is a range of numbers above which the frequency of the
measurements or relative frequency of the measurements is plotted.
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
2.30
2.32
a.
This is a frequency histogram because the number of observations is displayed rather
than the relative frequencies.
b.
There are 14 class intervals used in this histogram.
c.
The total number of measurements in the data set is 49.
13
Using MINITAB, the relative frequency histogram is:
.25
Relative frequency
.20
.15
.10
.05
0
2.34
a.
0.5
2.5
4.5
6.5
8.5
10.5
Class Interval
12.5
14.5
16.5
Using MINITAB, the relative frequency histogram is:
Histogram of RDER
0.25
Relative Frequency
0.20
0.15
0.10
0.05
0
-45
-15
15
45
75
105
135
165
195
225
255
RDER Value
2.36
b.
From the graph, the proportion of subjects with RDER values between 75 and 105 is
about 0.18. The exact proportion is 13 / 71 ๏ฝ 0.183 .
b.
From the graph, the proportion of subjects with RDER values below 15 is about
0.01 ๏ซ 0.08 ๏ฝ 0.09 . The exact proportion is ๏จ1 ๏ซ 6 ๏ฉ / 71 ๏ฝ 0.099 .
a.
Because the label on the vertical axis is โPercentโ , this is a relative frequency histogram.
Copyright ยฉ 2017 Pearson Education, Inc.
14
Chapter 2
b.
2.38
From the graph, the percentage of the 992 senior managers who reported a high level of
support for corporate sustainability is about
3.8 ๏ซ 2.4 ๏ซ 2.1 ๏ซ 1.2 ๏ซ 1.2 ๏ซ 0.5 ๏ซ 0.7 ๏ซ 0.2 ๏ซ 0.1 ๏ซ 0 ๏ซ 0.1 ๏ฝ 12.3% .
Using MINITAB, the stem-and-leaf display is:
Stem-and-Leaf Display: Depth
Stem-and-leaf of Depth
Leaf Unit = 0.10
2
4
8
(3)
7
5
3
13
14
15
16
17
18
19
N
= 18
29
00
7789
125
08
11
347
The data in the stem-and-leaf display are displayed to 1 decimal place while the actual data is
displayed to 2 decimal places. To 1 decimal place, there are 3 numbers that appear twice โ
14.0, 15.7, and 18.1. However, to 2 decimal places, none of these numbers are the same.
Thus, no molar depth occurs more frequently in the data.
2.40
a.
Using MINITAB, the dot plot of the honey dosage data is:
Dotplot of Honey Dosage Group
4
6
8
10
12
14
16
Improvement Score
b.
Both 10 and 12 occurred 6 times in the honey dosage group.
c.
From the graph in part c, 8 of the top 11 scores (72.7%) are from the honey dosage
group. Of the top 30 scores, 18 (60%) are from the honey dosage group. This supports
the conclusions of the researchers that honey may be a preferable treatment for the
cough and sleep difficulty associated with childhood upper respiratory tract infection.
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
2.42
a.
15
Using MINITAB, the stem-and-leaf display is:
Stem-and-Leaf Display: Spider
Stem-and-leaf of Spider
Leaf Unit = 10
1
3
(3)
4
2
1
2.44
0
0
0
0
0
1
N
= 10
0
33
455
67
9
1
b.
The spiders with a contrast value of 70 or higher are in bold type in the stem-and-leaf
display in part a. There are 3 spiders in this group.
c.
The sample proportion of spiders that a bird could detect is 3 / 10 ๏ฝ 0.3 . Thus, we could
infer that a bird could detect a crab-spider sitting on the yellow central part of a daisy
about 30% of the time.
a.
A stem-and-leaf display of the data using MINITAB is:
Stem-and-Leaf Display: FNE
Stem-and-leaf of FNE
Leaf Unit = 1.0
2
3
6
10
12
(2)
11
7
3
2
N
= 25
0 67
0 8
1 001
1 3333
1 45
1 66
1 8999
2 0011
2 3
2 45
b.
The numbers in bold in the stem-and-leaf display represent the bulimic students. Those
numbers tend to be the larger numbers. The larger numbers indicate a greater fear of
negative evaluation. Thus, the bulimic students tend to have a greater fear of negative
evaluation.
c.
A measure of reliability indicates how certain one is that the conclusion drawn is
correct. Without a measure of reliability, anyone could just guess at a conclusion.
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16
Chapter 2
2.46
a.
Using MINITAB, histograms of the two sets of SAT scores are:
Histogram of TOT2011, TOT2014
1400
1500
TOT2011
1600
1700
1800
TOT2014
10
Frequency
8
6
4
2
0
1400
1500
1600
1700
1800
It appears that the distributions of both sets of scores are somewhat skewed to the right.
Although the distributions are not identical for the two years, they are similar.
b.
Using MINITAB, a histogram of the differences of the 2014 and 2011 SAT scores is:
Histogram of Diff
35
30
Frequency
25
20
15
10
5
0
-200
-150
-100
-50
0
50
Diff
c.
It appears that there are more differences above 0 than below 0. Thus, it appears that in
general, the 2014 SAT scores are higher than the 2011 SAT scores. However, there are
many differences that are very close to 0.
d. Wyoming had the largest improvement in SAT scores from 2011 to 2014, with an
increase of 65 points.
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
2.48
17
Using MINITAB, the side-by-side histograms are:
Histogram of ZETA without, ZETA with GYPSUM
-60
ZETA without
-48
-36
-24
-12
ZETA with GYPSUM
Frequency
40
30
20
10
0
-60
-48
-36
-24
-12
The addition of calcium/gypsum increases the values of the zeta potential of silica. All of the
values of zeta potential for the specimens containing calcium/gypsum are greater than all of
the values of zeta potential for the specimens without calcium/gypsum.
2.50
A measure of central tendency measures the โcenterโ of the distribution while measures of
variability measure how spread out the data are.
2.52
A skewed distribution is a distribution that is not symmetric and not centered around the
mean. One tail of the distribution is longer than the other. If the mean is greater than the
median, then the distribution is skewed to the right. If the mean is less than the median, the
distribution is skewed to the left.
2.54
a.
For a distribution that is skewed to the left, the mean is less than the median.
b.
For a distribution that is skewed to the right, the mean is greater than the median.
c.
For a symmetric distribution, the mean and median are equal.
2.56
Assume the data are a sample. The mode is the observation that occurs most frequently. For
this sample, the mode is 15, which occurs 3 times.
The sample mean is:
๏ฅ x ๏ฝ 18 ๏ซ 10 ๏ซ 15 ๏ซ 13 ๏ซ 17 ๏ซ 15 ๏ซ 12 ๏ซ 15 ๏ซ 18 ๏ซ 16 ๏ซ 11 ๏ฝ 160 ๏ฝ 14.545
x๏ญ
n
11
11
The median is the middle number when the data are arranged in order. The data arranged in
order are: 10, 11, 12, 13, 15, 15, 15, 16, 17, 18, 18. The middle number is the 6th number,
which is 15.
2.58
The median is the middle number once the data have been arranged in order. If n is even,
there is not a single middle number. Thus, to compute the median, we take the average of the
middle two numbers. If n is odd, there is a single middle number. The median is this middle
number.
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18
Chapter 2
A data set with 5 measurements arranged in order is 1, 3, 5, 6, 8. The median is the middle
number, which is 5.
A data set with 6 measurements arranged in order is 1, 3, 5, 5, 6, 8. The median is the
5 ๏ซ 5 10
๏ฝ
๏ฝ 5.
average of the middle two numbers which is
2
2
n
๏ฅx
i
2.60
1 ๏ซ 2 ๏ซ 3 ๏ซ … ๏ซ 9 42
๏ฝ
๏ฝ 3.23 . This is the average number of
n
13
13
sword shafts buried at each grave site.
i ๏ฝ1
a.
The mean is x ๏ฝ
๏ฝ
b.
To find the median, the data must be arranged in order. The data arranged in order are:
1 1 1 2 2 2 2 3 4 4 5 6 9
There are a total of 13 observations, which is an odd number. The median is the middle
number which is 2. Half of the grave sites had 2 or fewer sword shafts buried and half
had 2 or more.
c.
The mode is the number that occurs most frequently. In this case, the mode is 2.
n
๏ฅx
i
2.62
a.
b.
54 ๏ซ 42 ๏ซ 51 ๏ซ … ๏ซ 40 365
๏ฝ
๏ฝ 45.625 . This is the Performance
The mean is x ๏ฝ i ๏ฝ1 ๏ฝ
n
8
8
Anxiety Inventory scale for participants in 8 different studies.
To find the median, the data must be arranged in order. The data arranged in order are:
39 40 41 42 43 51 54 55
There are a total of 8 observations, which is an even number. The median is the average
42 ๏ซ 43 85
๏ฝ
๏ฝ 42.5 . Half
of the middle 2 numbers which are 42 and 43. The median is
2
2
of the studies had a PAI scale less than 42.5 and half had a value greater than 42.5.
c.
If 39 were eliminated, the mean becomes
n
๏ฅx
i
54 ๏ซ 42 ๏ซ 51 ๏ซ … ๏ซ 40 326
x ๏ฝ i ๏ฝ1 ๏ฝ
๏ฝ
๏ฝ 46.571 . The data arranged in order are now:
n
7
7
40 41 42 43 51 54 55. The median is the middle number which is 43. The mean
increased by 0.946 while the median only increased by 0.5.
2.64
a.
There are 35 observations in the honey dosage group. Thus, the median is the middle
number, once the data have been arranged in order from the smallest to the largest. The
middle number is the 18th observation which is 11.
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
b.
There are 33 observations in the DM dosage group. Thus, the median is the middle
number, once the data have been arranged in order from the smallest to the largest. The
middle number is the 17th observation which is 9.
c.
There are 37 observations in the control group. Thus, the median is the middle number,
once the data have been arranged in order from the smallest to the largest. The middle
number is the 19th observation which is 7.
d.
Since the median of the honey dosage group is the highest, the median of the DM groups
is the next highest, and the median of the control group is the smallest, we can conclude
that the honey dosage is the most effective, the DM dosage is the next most effective,
and nothing (control) is the least effective.
๏ฅ x ๏ฝ 77.07 ๏ฝ 1.927
The mean of the driving performance index values is: x ๏ฝ
40
n
a.
The median is the average of the middle two numbers once the data have been arranged
in order. After arranging the numbers in order, the 20th and 21st numbers are 1.75 and
1.75 ๏ซ 1.76
๏ฝ 1.755
1.76. The median is:
2
The mode is the number that occurs the most frequently and is 1.4.
b.
The average driving performance index is 1.927. The median is 1.755. Half of the
players have driving performance index values less than 1.755 and half have values
greater than 1.755. Three of the players have the same index value of 1.4.
c.
Since the mean is greater than the median, the data are skewed to the right. Using
MINITAB, a histogram of the data is:
Histogram of Performance
10
8
Fr equency
2.66
19
6
4
2
0
1.5
2.0
2.5
P er for mance
3.0
Copyright ยฉ 2017 Pearson Education, Inc.
3.5
20
Chapter 2
2.68
a.
The mean number of ant species discovered is:
x๏ฝ
๏ฅ x ๏ฝ 3 ๏ซ 3 ๏ซ … ๏ซ 4 ๏ฝ 141 ๏ฝ 12.82
n
11
11
The median is the middle number once the data have been arranged in order:
3, 3, 4, 4, 4, 5, 5, 5, 7, 49, 52.
The median is 5.
The mode is the value with the highest frequency. Since both 4 and 5 occur 3 times,
both 4 and 5 are modes.
b.
For this case, we would recommend that the median is a better measure of central
tendency than the mean. There are 2 very large numbers compared to the rest. The
mean is greatly affected by these 2 numbers, while the median is not.
c.
The mean total plant cover percentage for the Dry Steppe region is:
x๏ฝ
๏ฅ x ๏ฝ 40 ๏ซ 52 ๏ซ … ๏ซ 27 ๏ฝ 202 ๏ฝ 40.4
n
5
5
The median is the middle number once the data have been arranged in order:
27, 40, 40, 43, 52.
The median is 40.
The mode is the value with the highest frequency. Since 40 occurs 2 times, 40 is the
mode.
d.
The mean total plant cover percentage for the Gobi Desert region is:
x๏ฝ
๏ฅ x ๏ฝ 30 ๏ซ 16 ๏ซ … ๏ซ 14 ๏ฝ 168 ๏ฝ 28
n
6
6
The median is the mean of the middle 2 numbers once the data have been arranged in
order: 14, 16, 22, 30, 30, 56.
The median is
22 ๏ซ 30 52
๏ฝ
๏ฝ 26 .
2
2
The mode is the value with the highest frequency. Since 30 occurs 2 times, 30 is the
mode.
e.
Yes, the total plant cover percentage distributions appear to be different for the 2
regions. The percentage of plant coverage in the Dry Steppe region is much greater
than that in the Gobi Desert region.
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
2.70
a.
21
Using MINITAB, the simple statistics are:
Descriptive Statistics: ZETA without, ZETA with GYPSUM
Variable
ZETA without
ZETA with GYPSUM
N
50
50
Mean
-52.070
-10.958
Median
-52.250
-11.300
Mode
-50.2
-11.3
N for
Mode
3
5
For the liquid solutions prepared without calcium/gypsum, the mean zeta potential
measurement is -52.070, the median is -52.250 and the mode is -50.2. The average zeta
potential measurement for liquid solutions prepared without calcium/gypsum is -52.070.
Half of the zeta potential measurements for liquid solutions prepared without
calcium/gypsum are less than or equal to -52.250 and half are greater than -52.250. The
most common zeta potential measurement for liquid solutions prepared without
calcium/gypsum is -50.2.
2.72
b.
For the liquid solutions prepared with calcium/gypsum, the mean zeta potential
measurement is -10.958, the median is -11.300 and the mode is -11.3. The average zeta
potential measurement for liquid solutions prepared with calcium/gypsum is -10.958.
Half of the zeta potential measurements for liquid solutions prepared with
calcium/gypsum are less than or equal to -11.300 and half are greater than -11.300. The
most common zeta potential measurement for liquid solutions prepared with
calcium/gypsum is -11.3.
c.
The interpretation remains the same as in Exercise 2.48. The addition of
calcium/gypsum increases the values of the zeta potential of silica. The mean, median
and mode of the values of zeta potential for the specimens containing calcium/gypsum
are greater than the mean, median and mode of the values of zeta potential for the
specimens without calcium/gypsum.
a.
The mean number of power plants is:
n
๏ฅx
i
x๏ฝ
i ๏ฝ1
n
๏ฝ
5 ๏ซ 3 ๏ซ 4 ๏ซ … ๏ซ 3 79
๏ฝ
๏ฝ 3.95
20
20
The median is the mean of the middle 2 numbers once the data have been arranged in
order: 1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 7, 9, 11
The median is
3๏ซ 4 7
๏ฝ ๏ฝ 3.5 .
2
2
The number 1 occurs 5 times. The mode is 1.
b. Deleting the largest number, 11, the new mean is:
n
๏ฅx
i
x ๏ฝ i ๏ฝ1
n
๏ฝ
5 ๏ซ 3 ๏ซ 4 ๏ซ … ๏ซ 3 68
๏ฝ
๏ฝ 3.58
19
19
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22
Chapter 2
The median is the middle number once the data have been arranged in order:
1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 7, 9
The median is 3.
The number 1 occurs 5 times. The mode is 1.
By dropping the largest measurement from the data set, the mean drops from 3.95 to
3.58. The median drops from 3.5 to 3 and the mode stays the same.
c.
Deleting the lowest 2 and highest 2 measurements leaves the following:
1, 1, 1, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 7
The new mean is:
n
๏ฅx
i
x ๏ฝ i ๏ฝ1
n
๏ฝ
1 ๏ซ 1 ๏ซ 1 ๏ซ … ๏ซ 7 57
๏ฝ
๏ฝ 3.56
16
16
The trimmed mean has the advantage that some possible outliers have been eliminated.
2.74
The primary disadvantage of using the range to compare variability of data sets is that the two
data sets can have the same range and be vastly different with respect to data variation. Also,
the range is greatly affected by extreme measures.
2.76
The variance of a data set can never be negative. The variance of a sample is the sum of the
squared deviations from the mean divided by n ๏ญ 1. The square of any number, positive or
negative, is always positive. Thus, the variance will be positive.
The variance is usually greater than the standard deviation. However, it is possible for the
variance to be smaller than the standard deviation. If the data are between 0 and 1, the
variance will be smaller than the standard deviation. For example, suppose the data set is
0.8, 0.7, 0.9, 0.5, and 0.3. The sample mean is:
x๏ฝ
๏ฅ x ๏ฝ 0.8 ๏ซ 0.7 ๏ซ 0.9 ๏ซ 0.5 ๏ซ 0.3 ๏ฝ 3.2 ๏ฝ 0.64
n
.5
5
The sample variance is:
s2 ๏ฝ
๏ฅ
๏จ๏ฅ x๏ฉ
x ๏ญ
2
2
n ๏ญ1
n
๏ฝ
3.22
5 ๏ฝ 2.28 ๏ญ 2.048 ๏ฝ 0.058
5 ๏ญ1
4
2.28 ๏ญ
The standard deviation is s ๏ฝ 0.058 ๏ฝ 0.241
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
2.78
a.
b.
2.80
s2 ๏ฝ
s2 ๏ฝ
๏จ๏ฅ x๏ฉ
x ๏ญ
n
๏จ๏ฅ x๏ฉ
x ๏ญ
Range ๏ฝ 4 ๏ญ 0 ๏ฝ 4
s ๏ฝ 3.3333 ๏ฝ 1.826
17 2
20 ๏ฝ 0.1868
๏ฝ
20 ๏ญ 1
18 ๏ญ
๏จ๏ฅ x๏ฉ
x ๏ญ
2
2
n ๏ญ1
1002
40 ๏ฝ 3.3333
40 ๏ญ 1
380 ๏ญ
82
5 ๏ฝ 2.3
๏ฝ
4 ๏ญ1
22 ๏ญ
n
s ๏ฝ 0.1868 ๏ฝ 0.432
s ๏ฝ 2.3 ๏ฝ 1.52
๏ฅ
๏จ๏ฅ x๏ฉ
x ๏ญ
2
2
n ๏ญ1
n
๏ฝ
17 2
7 ๏ฝ 3.619
7 ๏ญ1
s ๏ฝ 3.619 ๏ฝ 1.90
27 2
9 ๏ฝ8
9 ๏ญ1
s ๏ฝ 8 ๏ฝ 2.828
(๏ญ5) 2
18 ๏ฝ 1.624
18 ๏ญ 1
s ๏ฝ 1.624 ๏ฝ 1.274
63 ๏ญ
Range ๏ฝ 8 ๏ญ (๏ญ2) ๏ฝ 10
s2 ๏ฝ
d.
n ๏ญ1
s ๏ฝ 4.8889 ๏ฝ 2.211
Range ๏ฝ 6 ๏ญ 0 ๏ฝ 6
s2 ๏ฝ
c.
2
n
a.
b.
๏ฝ
2
s2 ๏ฝ
๏ฅ
2
2
n ๏ญ1
๏ฅ
202
10 ๏ฝ 4.8889
๏ฝ
10 ๏ญ 1
84 ๏ญ
n
n ๏ญ1
๏ฅ
2
2
c.
s2 ๏ฝ
๏ฅ
๏จ๏ฅ x๏ฉ
x ๏ญ
2
2
n ๏ญ1
n
๏ฝ
145 ๏ญ
Range ๏ฝ 2 ๏ญ (๏ญ3) ๏ฝ 5
s2 ๏ฝ
2.82
๏ฅ
๏จ๏ฅ x๏ฉ
x ๏ญ
๏ฅ
๏จ๏ฅ x๏ฉ
x ๏ญ
2
2
n ๏ญ1
n
๏ฝ
29 ๏ญ
This is one possibility for the two data sets.
Data Set 1: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Data Set 2: 0, 0, 1, 1, 2, 2, 3, 3, 9, 9
The two sets of data above have the same range = largest measurement ๏ญ smallest
measurement ๏ฝ 9 ๏ญ 0 ๏ฝ 9 .
The means for the two data sets are:
๏ฝ ๏ฅ x 0 ๏ซ 1 ๏ซ 2 ๏ซ 3 ๏ซ 4 ๏ซ 5 ๏ซ 6 ๏ซ 7 ๏ซ 8 ๏ซ 9 45
x1
๏ฝ
๏ฝ
๏ฝ 4.5
n
10
10
Copyright ยฉ 2017 Pearson Education, Inc.
23
24
Chapter 2
x2 ๏ฝ
๏ฅ x ๏ฝ 0 ๏ซ 0 ๏ซ 1 ๏ซ 1 ๏ซ 2 ๏ซ 2 ๏ซ 3 ๏ซ 3 ๏ซ 9 ๏ซ 9 ๏ฝ 30 ๏ฝ 3
n
10
10
The dot diagrams for the two data sets are shown below.
Group
Dotplot of Data
2.84
a.
b.
c.
d.
2.86
a.
x-bar1
1
2
0
s2 ๏ฝ
2
๏ฅ
x-bar2
๏จ๏ฅ x๏ฉ
x ๏ญ
4
2
2
n ๏ญ1
6
n
๏ฝ
๏ฅ
๏จ๏ฅ x๏ฉ
x ๏ญ
๏ฅ
๏จ๏ฅ x๏ฉ
x ๏ญ
282
5 ๏ฝ 69.2 ๏ฝ 17.3
5 ๏ญ1
4
226 ๏ญ
2
2
s2 ๏ฝ
2
2
n ๏ญ1
n
๏จ๏ฅ x๏ฉ
x ๏ญ
s ๏ฝ 17.3 ๏ฝ 4.1593
552
4 ๏ฝ 456.75 ๏ฝ 152.25 square feet
4 ๏ญ1
3
1213 ๏ญ
n
๏ฝ
n ๏ญ1
s ๏ฝ 152.25 ๏ฝ 12.339 feet
s2 ๏ฝ
8
Data
๏ฝ
(๏ญ15) 2
6 ๏ฝ 21.5 ๏ฝ 4.3
6 ๏ญ1
5
59 ๏ญ
s ๏ฝ 4.3 ๏ฝ 2.0736
2
24 22
๏ญ
0.2933
n
s2 ๏ฝ
๏ฝ 25 6 ๏ฝ
๏ฝ 0.0587 square ounces
n ๏ญ1
6 ๏ญ1
5
s ๏ฝ 0.0587 ๏ฝ 0.2422 ounce
๏ฅ
2
The range is the difference between the largest and smallest numbers. For this data set,
the range is 9 ๏ญ 1 ๏ฝ 8 .
๏ฅ
๏จ๏ฅ x๏ฉ
x ๏ญ
2
2
n
b.
The sample variance is s 2 ๏ฝ
c.
The sample standard deviation is s ๏ฝ 5.526 ๏ฝ 2.351 .
n ๏ญ1
๏ฝ
422
13 ๏ฝ 5.526 .
13 ๏ญ 1
202 ๏ญ
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
2.88
d.
Both the range and the standard deviation have the same units of measure as the original
variable.
a.
The range is the difference between the largest and smallest observations and is
17.83 ๏ญ 4.90 ๏ฝ 12.93 meters.
b.
The variance is:
s2 ๏ฝ
2.90
25
๏ฅ
๏จ๏ฅ x๏ฉ
x ๏ญ
2
2
n ๏ญ1
n
๏ฝ
126.322
13 ๏ฝ 16.767 square meters
13 ๏ญ 1
1428.64 ๏ญ
c.
The standard deviation is s ๏ฝ 16.767 ๏ฝ 4.095 meters.
a.
For Group A, the range is 67.20. From the printout, the range is 122.40 ๏ญ 55.20 ๏ฝ 67.2 .
b.
For Group A, the standard deviation is 14.48. From the printout,
s ๏ฝ s 2 ๏ฝ 209.53 ๏ฝ 14.48 .
2.92
c.
Group B has the more variable permeability data. Group B has the largest range, the
largest variance and the largest standard deviation.
a.
Range ๏ฝ 11 ๏ญ 1 ๏ฝ 10
s2 ๏ฝ
b.
๏ฅ
๏จ ๏ฅ x๏ฉ
x ๏ญ
2
2
n ๏ญ1
n
๏ฝ
792
20 ๏ฝ 7.524
20 ๏ญ 1
455 ๏ญ
s ๏ฝ s 2 ๏ฝ 7.524 ๏ฝ 2.743
Dropping the largest measurement:
Range ๏ฝ 9 ๏ญ 1 ๏ฝ 8
๏จ ๏ฅ x๏ฉ
x ๏ญ
2
682
n
19 ๏ฝ 5.035
s2 ๏ฝ
๏ฝ
s ๏ฝ s 2 ๏ฝ 5.035 ๏ฝ 2.244
n ๏ญ1
19 ๏ญ 1
By dropping the largest observation from the data set, the range decreased from 10 to 8,
the variance decreased from 7.524 to 5.035 and the standard deviation decreased from
2.743 to 2.244.
๏ฅ
c.
2
334 ๏ญ
Dropping the largest and smallest measurements:
Range ๏ฝ 9 ๏ญ 1 ๏ฝ 8
๏จ ๏ฅ x๏ฉ
x ๏ญ
2
67 2
n
18 ๏ฝ 4.918
s2 ๏ฝ
๏ฝ
s ๏ฝ s 2 ๏ฝ 4.918 ๏ฝ 2.218
n ๏ญ1
18 ๏ญ 1
By dropping the largest and smallest observations from the data set, the range decreased
from 10 to 8, the variance decreased from 7.524 to 4.918 and the standard deviation
decreased from 2.743 to 2.218.
๏ฅ
2
333 ๏ญ
Copyright ยฉ 2017 Pearson Education, Inc.
26
Chapter 2
2.94
The Empirical Rule applies only to data sets that are mound-shapedโthat are approximately
symmetric, with a clustering of measurements about the midpoint of the distribution and that
tail off as one moves away from the center of the distribution.
2.96
Since no information is given about the data set, we can only use Chebyshev’s rule.
2.98
a.
At least 0 of the measurements will fall between x ๏ญ s and x ๏ซ s .
b.
At least 3/4 or 75% of the measurements will fall between x ๏ญ 2s and x ๏ซ 2s .
c.
At least 8/9 or 89% of the measurements will fall between x ๏ญ 3s and x ๏ซ 3s .
a.
x๏ฝ
s2 ๏ฝ
๏ฅ x ๏ฝ 206 ๏ฝ 8.24
n
๏ฅ
25
๏จ๏ฅ x๏ฉ
x ๏ญ
2
2
n ๏ญ1
n
๏ฝ
2062
25 ๏ฝ 3.357
25 ๏ญ 1
1778 ๏ญ
s ๏ฝ s 2 ๏ฝ 1.83
b.
Number of Measurements
in Interval
Interval
Percentage
x ๏ฑ s , or (6.41, 10.17)
18
18 / 25 ๏ฝ 0.72 or 72%
x ๏ฑ 2s , or (4.58, 11.90)
24
24 / 25 ๏ฝ 0.96 or 96%
x ๏ฑ 3s , or (2.75, 13.73)
25
25 / 25 ๏ฝ 1
c.
The percentages in part b are in agreement with Chebyshev’s rule and agree fairly well
with the percentages given by the Empirical Rule.
d.
Range ๏ฝ 12 ๏ญ 5 ๏ฝ 7
s ๏ป range / 4 ๏ฝ 7 / 4 ๏ฝ 1.75
The range approximation provides a satisfactory estimate of s.
a.
Using MINITAB, the histogram is:
Histogram of Wheels
12
10
8
Frequency
2.100
or 100%
6
4
2
0
1
2
3
4
5
6
7
8
Wheels
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
27
Although the distribution is somewhat mound-shaped, the distribution is skewed to the
right.
b.
Using MINITAB, the mean and standard deviation are:
Descriptive Statistics: Wheels
Variable
Wheels
c.
d.
N
28
Mean
3.214
StDev
1.371
x ๏ฑ 2 s ๏ 3.214 ๏ฑ 2(1.371) ๏ 3.214 ๏ฑ 2.742 ๏ (0.472,5.956)
3
According to Chebyshevโs rule, at least of the measurements will fall within 2
4
standard deviations of the mean.
e. According to the Empirical Rule, approximately 95% of the measurements will fall
within 2 standard deviations of the mean.
f.
2.102
Twenty-six of the twenty-eight observations fall within the interval. The proportion is
26
๏ฝ 0.929 . The Empirical Rule does provide a good estimate of the proportion. The
28
actual percentage is 92.9% which is close to 95%.
a. If the data are symmetric and mound shaped, then the Empirical Rule will describe the
data. About 95% of the observations will fall within 2 standard deviation of the mean.
The interval two standard deviations below and above the mean is
x ๏ฑ 2s ๏ 39 ๏ฑ 2(6) ๏ 39 ๏ฑ 12 ๏ (27, 51) . This range would be 27 to 51.
b. To find the number of standard deviations above the mean a score of 51 would be, we
subtract the mean from 51 and divide by the standard deviation. Thus, a score of 51 is
51 ๏ญ 39
๏ฝ 2 standard deviations above the mean. From the Empirical Rule, about .025 of
6
the drug dealers will have WR scores above 51.
c. By the Empirical Rule, about 99.7% of the observations will fall within 3 standard
deviations of the mean. Thus, nearly all the scores will fall within 3 standard deviations
of the mean. The interval three standard deviations below and above the mean is
x ๏ฑ 3s ๏ 39 ๏ฑ 3(6) ๏ 39 ๏ฑ 18 ๏ (21, 57) . This range would be 21 to 57.
2.104
2.106
a.
Because the histogram in Exercise 2.34 is skewed to the right, Chebyshevโs rule is more
appropriate for describing the distribution of the RDER values.
b.
The interval x ๏ฑ 2 s is 78.19 ๏ฑ 2(63.24) ๏ 78.19 ๏ฑ 126.48 ๏ (๏ญ48.29, 204.67) . At least ยพ
or 75% of the observations will be between -48.29 and 204.67.
a.
By Chebyshevโs rule, at least 75% of the observations will fall within 2 standard
deviations of the mean. This interval is
x ๏ฑ 2 s ๏ 0.90 ๏ฑ 2(1.10) ๏ 0.90 ๏ฑ 2.20 ๏ (๏ญ1.30, 3.10) .
Copyright ยฉ 2017 Pearson Education, Inc.
28
Chapter 2
2.108
b.
No. A value of 7 would be (7 ๏ญ 0.90) / 1.10 ๏ฝ 5.55 standard deviations above the mean.
This would be very unusual.
a.
There are 2 observations with missing values for egg length, so there are only 130
useable observations.
x๏ฝ
s2 ๏ฝ
๏ฅ x 7,885
๏ฝ
๏ฝ 60.65
130
n
๏ฅ
๏จ๏ฅ x๏ฉ
x ๏ญ
2
2
n ๏ญ1
n
๏ฝ
(7,885) 2
130 ๏ฝ 249,586.4231 ๏ฝ 1,934.7785
130 ๏ญ 1
129
727,842 ๏ญ
s ๏ฝ s 2 ๏ฝ 1,934.7785 ๏ฝ 43.99
b.
The data are not symmetrical or mound-shaped. Thus, we will use Chebyshevโs Rule.
We know that there are at least 8/9 or 88.9% of the observations within 3 standard
deviations of the mean. Thus, at least 88.9% of the observations will fall in the interval:
x ๏ฑ 3s ๏ 60.65 ๏ฑ 3(43.99) ๏ 60.65 ๏ฑ 131.97 ๏ ( ๏ญ71.32, 192.69)
Since it is impossible to have negative egg lengths, at least 88.9% of the egg lengths
will be between 0 and 192.69.
2.110
2.112
a.
The mean and standard deviation for Group A are 73.62 and 14.48. The histogram of
the data from Group A is skewed to the right so Chebyshevโs rule is more appropriate.
We know at least 8/9 or 88.9% of the observations will fall within 3 standard deviations
of the mean. This interval is
x ๏ฑ 3s ๏ 73.62 ๏ฑ 3(14.48) ๏ 73.62 ๏ฑ 43.44 ๏ (30.18, 117.06) .
b.
The mean and standard deviation for Group B are 128.54 and 21.97. The histogram of
the data from Group B is skewed to the left so Chebyshevโs rule is more appropriate.
We know at least 8/9 or 88.9% of the observations will fall within 3 standard deviations
of the mean. This interval is
x ๏ฑ 3s ๏ 128.54 ๏ฑ 3(21.97) ๏ 128.54 ๏ฑ 65.91 ๏ (62.63, 194.45) .
c.
The mean and standard deviation for Group C are 83.07 and 20.05. The histogram of
the data from Group C is skewed to the right so Chebyshevโs rule is more appropriate.
We know at least 8/9 or 88.9% of the observations will fall within 3 standard deviations
of the mean. This interval is
x ๏ฑ 3s ๏ 83.07 ๏ฑ 3(20.05) ๏ 83.07 ๏ฑ 60.15 ๏ (22.92, 143.22) .
d.
It appears that weathering type B results in faster decay.
To decide which group the patient is most likely to come from, we will compute the z-score
for each group.
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
Group T: z ๏ฝ
Group V: z ๏ฝ
Group C: z ๏ฝ
x๏ญ๏ญ
๏ณ
x๏ญ๏ญ
๏ณ
x๏ญ๏ญ
๏ณ
๏ฝ
22.5 ๏ญ 10.5
๏ฝ 1.58
7.6
๏ฝ
22.5 ๏ญ 3.9
๏ฝ 2.48
7.5
๏ฝ
22.5 ๏ญ 1.4
๏ฝ 2.81
7.5
29
The patient is most likely to have come from Group T. The z-score for Group T is z ๏ฝ 1.58 .
This would not be an unusual z-score if the patient was in Group T. The z-scores for the
other 2 groups are both greater than 2. We know that z-scores greater than 2 are rather
unusual.
2.114
a.
The 50th percentile is also called the median.
b.
The QL is the lower quartile. This is also the 25th percentile or the score which has 25%
of the observations less than it.
c.
The QU is the upper quartile. This is also the 75th percentile or the score which has
75% of the observations less than it.
2.116
For mound-shaped distributions, we can use the Empirical Rule. About 95% of the
observations will fall within 2 standard deviations of the mean. Thus, about 95% of the
measurements will have z-scores between -2 and 2.
2.118
We first compute z-scores for each x value.
a.
z๏ฝ
b.
z๏ฝ
c.
z๏ฝ
d.
z๏ฝ
x๏ญ๏ญ
๏ณ
x๏ญ๏ญ
๏ณ
x๏ญ๏ญ
๏ณ
x๏ญ๏ญ
๏ณ
๏ฝ
100 ๏ญ 50
๏ฝ2
25
๏ฝ
1๏ญ 4
๏ฝ ๏ญ3
1
๏ฝ
0 ๏ญ 200
๏ฝ ๏ญ2
100
๏ฝ
10 ๏ญ 5
๏ฝ 1.67
3
The above z-scores indicate that the x value in part a lies the greatest distance above the mean
and the x value of part b lies the greatest distance below the mean.
2.120
The mean score is 153. This is the arithmetic average score of U.S. twelfth graders on the
mathematics assessment test. The 25th percentile score is 111. This indicates that 25% of the
U.S. twelfth graders scored 111 or lower on the assessment test. The 75th percentile score is
177. This indicates that 75% of the U.S. twelfth graders scored 177 or lower on the
Copyright ยฉ 2017 Pearson Education, Inc.
30
Chapter 2
assessment test. The 90th percentile score is 197. This indicates that 90% of the U.S. twelfth
graders scored 197 or lower on the assessment test.
2.122
2.124
a.
From Exercise 2.35, the proportion of fup/fumic ratios that fall above 1 is 0.034. The
percentile rank of 1 is (1 ๏ญ 0.034)100% ๏ฝ 96.6 th percentile.
b.
From Exercise 2.35, the proportion of fup/fumic ratios that fall below 0.4 is 0.695. The
percentile rank of 0.4 is (0.695)100% ๏ฝ 69.5 th percentile.
a.
The z-score associated with a score of 30 is z ๏ฝ
b.
x ๏ญ x 30 ๏ญ 39
๏ฝ
๏ฝ ๏ญ1.50 . This means
s
6
that a score of 30 is 1.5 standard deviations below the mean.
x ๏ญ x 39 ๏ญ 39
๏ฝ
๏ฝ 0 . Half or 0.5 of the
The z-score associated with a score of 39 is z ๏ฝ
s
6
observations are below a score of 39.
2.126
Since the 90th percentile of the study sample in the subdivision was 0.00372 mg/L, which is
less than the USEPA level of 0.015 mg/L, the water customers in the subdivision are not at
risk of drinking water with unhealthy lead levels.
2.128
a.
If the distribution is mound-shaped and symmetric, then the Empirical Rule can be used.
Approximately 68% of the scores will fall within 1 standard deviation of the mean or
between 53% ๏ฑ 15% or between 38% and 68%. Approximately 95% of the scores will
fall within 2 standard deviations of the mean or between 53% ๏ฑ 2(15%) or between 23%
and 83%. Approximately all of the scores will fall within 3 standard deviations of the
mean or between 53% ๏ฑ 3(15%) or between 8% and 98%.
b.
If the distribution is mound-shaped and symmetric, then the Empirical Rule can be used.
Approximately 68% of the scores will fall within 1 standard deviation of the mean or
between 39% ๏ฑ 12% or between 27% and 51%. Approximately 95% of the scores will
fall within 2 standard deviations of the mean or between 39% ๏ฑ 2(12%) or between 15%
and 63%. Approximately all of the scores will fall within 3 standard deviations of the
mean or between 39% ๏ฑ 3(12%) or between 3% and 75%.
c.
Since the scores on the red exam are shifted to the left of those on the blue exam, a
score of 20% is more likely to occur on the red exam than on the blue exam.
2.130
Yes. From the graph in Exercise 2.129 c, we can see that there are 4 observations with zscores greater than 3. There is then a gap down to 2.18. Those 4 observations are quite
different from the rest of the data. After those 4 observations, the data are fairly similar. We
know that by ranking the data, we can reduce the influence of outliers. But, by doing this, we
lose valuable information.
2.132
The interquartile range is the distance between the upper and lower quartiles.
2.134
For a mound-shaped distribution, the Empirical Rule can be used. Almost all of the
observations will fall within 3 standard deviations of the mean. Thus, almost all of the
observations will have z-scores between -3 and 3.
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
2.136
31
The interquartile range is IQR ๏ฝ QU ๏ญ QL ๏ฝ 85 ๏ญ 60 ๏ฝ 25 .
The lower inner fence ๏ฝ QL ๏ญ 1.5(IQR) ๏ฝ 60 ๏ญ 1.5(25) ๏ฝ 22.5 .
The upper inner fence ๏ฝ QU ๏ซ 1.5(IQR) ๏ฝ 60 ๏ซ 1.5(25) ๏ฝ 122.5 .
The lower outer fence ๏ฝ QL ๏ญ 3(IQR) ๏ฝ 60 ๏ญ 3(25) ๏ฝ ๏ญ15 .
The upper outer fence ๏ฝ QU ๏ซ 3(IQR) ๏ฝ 60 ๏ซ 3(25) ๏ฝ 160 .
With only this information, the box plot would look something like the following:
The whiskers extend to the inner fences unless no data points are that small or that large. The
upper inner fence is 122.5. However, the largest data point is 100, so the whisker stops at
100. The lower inner fence is 22.5. The smallest data point is 18, so the whisker extends to
22.5. Since 18 is between the inner and outer fences, it is designated with a *. We do not
know if there is any more than one data point below 22.5, so we cannot be sure that the box
plot is entirely correct.
2.138
To determine if the measurements are outliers, compute the z-score.
a.
b.
c.
d.
x ๏ญ x 65 ๏ญ 57
๏ฝ
๏ฝ .727
11
s
Since this z-score is less than 3 in magnitude, 65 is not an outlier.
z๏ฝ
x ๏ญ x 21 ๏ญ 57
๏ฝ
๏ฝ ๏ญ3.273
11
s
Since this z-score is more than 3 in magnitude, 21 is an outlier.
z๏ฝ
x ๏ญ x 72 ๏ญ 57
๏ฝ
๏ฝ 1.364
s
11
Since this z-score is less than 3 in magnitude, 72 is not an outlier.
z๏ฝ
x ๏ญ x 98 ๏ญ 57
๏ฝ
๏ฝ 3.727
11
s
Since this z-score is more than 3 in magnitude, 98 is an outlier.
z๏ฝ
Copyright ยฉ 2017 Pearson Education, Inc.
32
Chapter 2
2.140
a.
Using MINITAB, the box plot for data is given below.
Boxplot of Data
210
200
190
Data
180
170
160
150
140
b.
2.142
In this data set, there is one outlier. It corresponds to the value 140.
a. The z-score is z ๏ฝ
x ๏ญ x 175 ๏ญ 79
๏ฝ
๏ฝ 4.17.
s
23
b. Yes, we would consider this measurement an outlier. Any observation with a z-score that
has an absolute value greater than 3 is considered a highly suspect outlier.
x ๏ญ x 155 ๏ญ 67.755
๏ฝ
๏ฝ 3.25 . Because this
s
26.871
observation is more than 3 standard deviations from the mean, it is considered a highly
suspect outlier. It would not be considered typical of the study sample.
2.144
The z-score corresponding to 155 is: z ๏ฝ
2.146
a.
The approximate 25th percentile PASI score before treatment is 10. The approximate
median before treatment is 15. The approximate 75th percentile PASI score before
treatment is 27.5.
b.
The approximate 25th percentile PASI score after treatment is 3.5. The approximate
median after treatment is 5. The approximate 75th percentile PASI score after treatment
is 7.5.
c.
Since the 75th percentile after treatment is lower than the 25th percentile before
treatment, it appears that the ichthyotherapy is effective in treating psoriasis.
Using MINITAB, a boxplot of the data is:
Boxplot of Rockfall
17.5
15.0
Rockfall
2.148
12.5
10.0
7.5
5.0
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
33
From the boxplot, there is no indication that there are any outliers.
We will now use the z-score criterion for determining outliers. From Exercises 2.61 and 2.88,
x ๏ฝ 9.72 and s ๏ฝ 4.095 . The z-score associated with the minimum value is
x ๏ญ x 4.9 ๏ญ 9.72
z๏ฝ
๏ฝ
๏ฝ ๏ญ1.18 and the z-score associated with the maximum value is
s
4.095
x ๏ญ x 17.83 ๏ญ 9.72
z๏ฝ
๏ฝ
๏ฝ 1.98 . Neither of these indicates there are any outliers.
s
4.095
2.150
a.
Using MINITAB, the boxplots of the three groups are:
Boxplot of Honey, DM, Control
18
16
14
Data
12
10
8
6
4
2
0
Honey
2.152
DM
Control
b.
The median improvement score for the honey dosage group is larger than the median
improvement scores for the other two groups. The median improvement score for the
DM dosage group is higher than the median improvement score for the control group.
c.
Because the interquartile range for the DM dosage group is larger than the interquartile
ranges of the other 2 groups, the variability of the DM group is largest. The variability
of the honey dosage group and the control group appear to be about the same.
d.
There appears to be one outlier in the honey dosage group and one outlier in the control
group.
a.
z๏ฝ
b.
The z-score is low enough to suspect that the librarian’s claim is incorrect. Even without
any knowledge of the shape of the distribution, Chebyshev’s rule states that at least 8/9
of the measurements will fall within 3 standard deviations of the mean (and,
consequently, at most 1/9 will be above z ๏ฝ 3 or below z ๏ฝ ๏ญ3 ).
c.
The Empirical Rule states that almost none of the measurements should be above z ๏ฝ 3
or below z ๏ฝ ๏ญ 3 . Hence, the librarian’s claim is even more unlikely.
d.
When ๏ณ ๏ฝ 2 , z ๏ฝ
x๏ญ๏ญ
๏ณ
๏ฝ
4๏ญ7
๏ฝ ๏ญ3
1
x๏ญ๏ญ
๏ณ
๏ฝ
4๏ญ7
๏ฝ ๏ญ1.5
2
Copyright ยฉ 2017 Pearson Education, Inc.
34
Chapter 2
This is not an unlikely occurrence, whether or not the data are mound-shaped. Hence,
we would not have reason to doubt the librarian’s claim.
2.154
A bivariate relationship is a relationship between 2 quantitative variables.
2.156
A positive association between two variables means that as one variable increases, the other
variable tends to also increase. A negative association between two variables means that as
one variable increases, the other variable tends to decrease.
2.158
Using MINITAB, the scatterplot is as follows:
Scatterplot of Variable 2 vs Variable 1
18
16
14
Variable 2
12
10
8
6
4
2
0
0
1
2
3
4
5
Variable 1
It appears that as variable 1 increases, variable 2 also increases.
Using MINITAB, a scatter plot of the data is:
Scatterplot of SLUGPCT vs ELEVATION
0.625
0.600
0.575
SLUGPCT
2.160
0.550
0.525
0.500
0.475
0.450
0
1000
2000
3000
4000
5000
6000
ELEVATION
If one uses the one obvious outlier (Denver), then there does appear to be a trend in the data.
As the elevation increases, the slugging percentage tends to increase. However, if the outlier
is removed, then it does not look like there is a trend to the data.
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
2.162
a.
35
A scattergram of the data is:
Scatterplot of Strikes vs Age
90
80
70
Strikes
60
50
40
30
20
10
120
130
140
150
160
170
180
190
Age
b.
2.164
There appears to be a trend. As the age increases, the number of strikes tends to
decrease.
Using MINITAB, a scatterplot of the data is:
Scatterplot of Freq vs Resonance
7000
6000
Freq
5000
4000
3000
2000
1000
0
5
10
15
20
25
Resonance
There is an increasing trend and there is very little variation in the plot. This supports the
researcherโs theory.
a.
Using MINITAB, a graph of the Anthropogenic Index against the Natural Origin Index
is:
Scatterplot of F-Anthro vs F-Natural
90
80
70
60
F-Anthro
2.166
50
40
30
20
10
0
5
10
15
20
25
30
35
40
F-Natural
Copyright ยฉ 2017 Pearson Education, Inc.
36
Chapter 2
This graph does not support the theory that there is a straight-line relationship between
the Anthropogenic Index against the Natural Origin Index. There are several points that
do not lie on a straight line.
b.
After deleting the three forests with the largest anthropogenic indices, the graph of the
data is:
Scatterplot of F-Anthro vs F-Natural
60
50
F-Anthro
40
30
20
10
0
5
10
15
20
25
30
35
40
F-Natural
After deleting the 3 data points, the relationship between the Anthropogenic Index
against the Natural Origin Index is much closer to a straight line.
2.168
Using MINITAB, a scattergram of the data is:
Scatterplot of Mass vs Time
7
6
5
Mass
4
3
2
1
0
0
10
20
30
40
50
60
Time
Yes, there appears to be a negative trend in this data. As time increases, the mass tends to
decrease. There appears to be a curvilinear relationship. As time increases, mass decreases
at a decreasing rate.
2.170
One way the bar graph can mislead the viewer is that the vertical axis has been cut off.
Instead of starting at 0, the vertical axis starts at 12. Another way the bar graph can mislead
the viewer is that as the bars get taller, the widths of the bars also increase.
2.172
a.
This graph is misleading because it looks like as the days are increasing, the number of
barrels collected per day is also increasing. However, the bars are the cumulative
number of barrels collected. The cumulative value can never decrease.
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
b.
37
Using MINITAB, the graph of the daily collection of oil is:
Chart of Barrells
2500
Barrells
2000
1500
1000
500
0
May-16
May-17
May-18
May-19
May-20
May-21
May-22
May-23
Day
From this graph, it shows that there has not been a steady improvement in the suctioning
process. There was an increase for 3 days, then a leveling off for 3 days, then a
decrease.
2.174
The range can be greatly affected by extreme measures, while the standard deviation is not as
affected.
2.176
The z-score approach for detecting outliers is based on the distribution being fairly moundshaped. If the data are not mound-shaped, then the box plot would be preferred over the zscore method for detecting outliers.
2.178
One technique for distorting information on a graph is by stretching the vertical axis by
starting the vertical axis somewhere above 0.
2.180
From part a of Exercise 2.179, the 3 z-scores are ๏ญ1, 1 and 2. Since none of these z-scores are
greater than 2 in absolute value, none of them are outliers.
From part b of Exercise 2.179, the 3 z-scores are ๏ญ2, 2 and 4. There is only one z-score
greater than 2 in absolute value. The score of 80 (associated with the z-score of 4) would be
an outlier. Very few observations are as far away from the mean as 4 standard deviations.
From part c of Exercise 2.179, the 3 z-scores are 1, 3, and 4. Two of these z-scores are
greater than 2 in absolute value. The scores associated with the two z-scores 3 and 4 (70 and
80) would be considered outliers.
From part d of Exercise 2.179, the 3 z-scores are .1, .3, and .4. Since none of these z-scores
are greater than 2 in absolute value, none of them are outliers.
2.182
๏ณ ๏ป range / 4 ๏ฝ 20 / 4 ๏ฝ 5
2.184 a.
๏ฅ x ๏ฝ 13 ๏ซ 1 ๏ซ 10 ๏ซ 3 ๏ซ 3 ๏ฝ 30
๏ฅ x ๏ฝ 13 ๏ซ 1 ๏ซ 10 ๏ซ 3 ๏ซ 3 ๏ฝ 288
2
x๏ฝ
2
2
2
2
2
30
๏ฅx ๏ฝ 5 ๏ฝ 6
Copyright ยฉ 2017 Pearson Education, Inc.
38
Chapter 2
s2 ๏ฝ
b.
๏ฅ
๏จ๏ฅ x๏ฉ
x ๏ญ
2
n ๏ญ1
2
n
s2 ๏ฝ
๏ฅ
๏ฝ
s ๏ฝ 27 ๏ฝ 5.196
2
2
4
๏จ๏ฅ x๏ฉ
x ๏ญ
2
2
n
๏ฝ
252
4 ๏ฝ 84.75 ๏ฝ 28.25
4 ๏ญ1
3
241 ๏ญ
s ๏ฝ 28.25 ๏ฝ 5.315
๏ฅ x ๏ฝ 1 ๏ซ 0 ๏ซ 1 ๏ซ 10 ๏ซ 11 ๏ซ 11 ๏ซ 15 ๏ฝ 49
๏ฅ x ๏ฝ 1 ๏ซ 0 ๏ซ 1 ๏ซ 10 ๏ซ 11 ๏ซ 11 ๏ซ 15 ๏ฝ 569
๏ฅ x ๏ฝ 49 ๏ฝ 7
x๏ฝ
2
n
s2 ๏ฝ
๏ฅ
2
2
2
2
2
2
7
๏จ ๏ฅ x๏ฉ
x ๏ญ
2
2
n ๏ญ1
n
๏ฝ
492
7 ๏ฝ 226 ๏ฝ 37.667
7 ๏ญ1
6
569 ๏ญ
s ๏ฝ 37.667 ๏ฝ 6.137
๏ฅ x ๏ฝ 3 ๏ซ 3 ๏ซ 3 ๏ซ 3 ๏ฝ 12
๏ฅ x ๏ฝ 3 ๏ซ 3 ๏ซ 3 ๏ซ 3 ๏ฝ 36
๏ฅ x ๏ฝ 12 ๏ฝ 3
x๏ฝ
2
2
n
s2 ๏ฝ
e.
2
n ๏ญ1
2
d.
n
302
5 ๏ฝ 108 ๏ฝ 27
5 ๏ญ1
4
288 ๏ญ
๏ฅ x ๏ฝ 13 ๏ซ 6 ๏ซ 6 ๏ซ 0 ๏ฝ 25
๏ฅ x ๏ฝ 13 ๏ซ 6 ๏ซ 6 ๏ซ 0 ๏ฝ 241
๏ฅ x ๏ฝ 25 ๏ฝ 6.25
x๏ฝ
2
c.
2
๏ฅ
2
2
2
4
๏จ๏ฅ x๏ฉ
x ๏ญ
2
2
n ๏ญ1
n
๏ฝ
122
4 ๏ฝ 0 ๏ฝ0
4 ๏ญ1
3
36 ๏ญ
s๏ฝ 0 ๏ฝ0
a)
x ๏ฑ 2s ๏ 6 ๏ฑ 2(5.2) ๏ 6 ๏ฑ 10.4 ๏ (๏ญ4.4, 16.4)
All or 100% of the observations are in this interval.
b)
x ๏ฑ 2s ๏ 6.25 ๏ฑ 2(5.32) ๏ 6.25 ๏ฑ 10.64 ๏ (๏ญ4.39, 16.89)
All or 100% of the observations are in this interval.
c)
x ๏ฑ 2s ๏ 7 ๏ฑ 2(6.14) ๏ 7 ๏ฑ 12.28 ๏ (๏ญ5.28, 19.28)
All or 100% of the observations are in this interval.
d)
x ๏ฑ 2s ๏ 3 ๏ฑ 2(0) ๏ 3 ๏ฑ 0 ๏ (3, 3)
All or 100% of the observations are in this interval.
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
Suppose we construct a relative frequency bar chart for this data. This will allow the
archaeologists to compare the different categories easier. First, we must compute the relative
frequencies for the categories. These are found by dividing the frequencies in each category
by the total 837. For the burnished category, the relative frequency is 133 / 837 ๏ฝ 0.159 . The
rest of the relative frequencies are found in a similar fashion and are listed in the table.
Pot Category
Number Found
Computation
133
460
55
14
165
4
4
2
837
133 / 837
460 / 837
55 / 837
14 / 837
165 / 837
4 / 837
4 / 837
2 / 837
Burnished
Monochrome
Slipped
Curvilinear Decoration
Geometric Decoration
Naturalistic Decoration
Cycladic White clay
Conical cup clay
Total
Relative
Frequency
0.159
0.550
0.066
0.017
0.197
0.005
0.005
0.002
1.001
A relative frequency bar chart is:
Chart of Pot Category
.60
.50
Relative Frequency
2.186
39
.40
.30
.20
.10
0
Burnished Monochrome
Slipped
Curvilinear
Geometric
Naturalistic
Cycladic
Conical
Pot Category
Proportion within all data.
The most frequently found type of pot was the Monochrome. Of all the pots found,
55% were Monochrome. The next most frequently found type of pot was the Painted in
Geometric Decoration. Of all the pots found, 19.7% were of this type. Very few pots of
the types Painted in naturalistic decoration, Cycladic white clay, and Conical cup clay
were found.
Copyright ยฉ 2017 Pearson Education, Inc.
40
Chapter 2
2.188
a.
Using MINITAB, the stem-and-leaf display is as follows.
Character Stem-and-Leaf Display
Stem-and-leaf of Books
Leaf Unit = 1.0
1
5
6
(3)
5
4
1
1
b.
N
= 14
1 6
2 0124
2 8
3 044
3 9
4 002
4
5 3
The leaves that correspond to students who earned an โAโ grade are highlighted in the
graph above. Those students who earned Aโs tended to read the most books.
n
๏ฅx
i
c.
53 ๏ซ 42 ๏ซ 40 ๏ซ … ๏ซ 16 443
๏ฝ
๏ฝ 31.643 . This is the average
The mean is x ๏ฝ i ๏ฝ1 ๏ฝ
14
14
n
number of books read per student.
To find the median, the data must be arranged in order. In this problem, the data are
already arranged in order. There are a total of 14 observations, which is an even
number. The median is the average of the middle 2 numbers which are 30 and 34. The
34 ๏ซ 30 64
๏ฝ
๏ฝ 32 . Half of the students read more than 32 books and half
median is
2
2
read fewer.
The mode is the observation appearing the most. In this data set, there are two modes 34 and 40 because each appears 2 times in the data set. The most frequent number of
books read is either 34 or 40.
d.
Since the mean and the median are almost the same, the distribution of the data set is
approximately symmetric. This can be verified by the stem-and-leaf display in part a.
e.
For those students who earned A, the mean is
n
๏ฅx
i
x๏ฝ
i ๏ฝ1
n
๏ฝ
53 ๏ซ 42 ๏ซ 40 ๏ซ … ๏ซ 24 296
๏ฝ
๏ฝ 37 .
8
8
The variance is s 2 ๏ฝ
๏ฅ
๏จ๏ฅ x๏ฉ
x ๏ญ
2
2
n ๏ญ1
n
๏ฝ
296 2
8 ๏ฝ 530 ๏ฝ 75.7143
7
7
11, 482 ๏ญ
The standard deviation is s ๏ฝ s 2 ๏ฝ 75.7143 ๏ฝ 8.701 .
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
f.
41
For those students who earned a B or C, the mean is
n
๏ฅx
i
x๏ฝ
i ๏ฝ1
n
๏ฝ
40 ๏ซ 28 ๏ซ 22 ๏ซ … ๏ซ 16 147
๏ฝ
๏ฝ 24.5
6
6
The variance is s 2 ๏ฝ
๏ฅ
๏จ๏ฅ x๏ฉ
x ๏ญ
2
2
n ๏ญ1
n
๏ฝ
147 2
6 ๏ฝ 363.5 ๏ฝ 72.7
5
5
3,965 ๏ญ
The standard deviation is s ๏ฝ s 2 ๏ฝ 72.7 ๏ฝ 8.526 .
2.190
g.
The students who received Aโs have a more variable distribution of the number of books
read. The variance and standard deviation for this group are greater than the
corresponding values for the B-C group.
h.
The z-score for a score of 40 books is z ๏ฝ
i.
The z-score for a score of 40 books is z ๏ฝ
j.
The group of students who earned Aโs is more likely to have read 40 books. For this
group, the z-score corresponding to 40 books is 0.34. This is not unusual. For the B-C
group, the z-score corresponding to 40 books is 1.82. This is close to 2 standard
deviations from the mean. This would be fairly unusual.
x ๏ญ x 40 ๏ญ 37
๏ฝ
๏ฝ 0.345 . Thus, someone who
s
8.701
read 40 books read more than the average number of books, but that number is not very
unusual.
x ๏ญ x 40 ๏ญ 24.5
๏ฝ
๏ฝ 1.82 . Thus, someone who
s
8.526
read 40 books read many more than the average number of books. Very few students
who received a B or a C read more than 40 books.
A pie chart of the data is:
Pie Chart of Drive Star
2
4.1%
5
18.4%
3
17.3%
Category
2
3
4
5
4
60.2%
More than half of the cars received 4 star ratings (60.2%). A little less than a quarter of
the cars tested received ratings of 3 stars or less.
Copyright ยฉ 2017 Pearson Education, Inc.
42
Chapter 2
2.192
a.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Ratio
Variable
Ratio
N
26
Mean
3.507
StDev
0.634
Minimum
2.250
Maximum
5.060
x ๏ญ x 5.06 ๏ญ 3.507
๏ฝ
๏ฝ 2.45
s
0.634
x ๏ญ x 2.25 ๏ญ 3.507
๏ฝ
๏ฝ ๏ญ1.98
The z-score associated with the smallest ratio is z ๏ฝ
s
0.634
x ๏ญ x 3.507 ๏ญ 3.507
๏ฝ
๏ฝ0
The z-score associated with the mean ratio is z ๏ฝ
s
0.634
The z-score associated with the largest ratio is z ๏ฝ
b.
Yes, I would consider the z-score associated with the largest ratio to be unusually large.
We know if the data are approximately mound-shaped that approximately 95% of the
observations will be within 2 standard deviations of the mean. A z-score of 2.45 would
indicate that less than 2.5% of all the measurements will be larger than this value.
c.
Using MINITAB, the box plot is:
Boxplot of Ratio
5.0
4.5
Ratio
4.0
3.5
3.0
2.5
2.0
From this box plot, there are no observations marked as outliers.
a.
Using MINITAB, a histogram of the data is:
Histogram of pH
12
10
8
Percent
2.194
6
4
2
0
5.4
6.0
6.6
7.2
7.8
8.4
9.0
pH
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
43
From the graph, it looks like the proportion of wells with ph levels less than 7.0 is:
0.005 ๏ซ 0.01 ๏ซ 0.02 ๏ซ 0.015 ๏ซ 0.027 ๏ซ 0.031 ๏ซ 0.05 ๏ซ 0.07 ๏ซ 0.017 ๏ซ 0.05 ๏ฝ 0.295
b.
Using MINITAB, a histogram of the MTBE levels for those wells with detectible levels
is:
Histogram of MTBE-Level
MTBE-Detect = Detect
80
70
60
Percent
50
40
30
20
10
0
0
10
20
30
40
50
MTBE-Level
From the graph, it looks like the proportion of wells with MTBE levels greater than 5 is:
0.03 ๏ซ 0.01 ๏ซ 0.01 ๏ซ 0.01 ๏ซ 0.01 ๏ซ 0.01 ๏ซ 0.01 ๏ฝ 0.09
c.
The sample mean is:
n
๏ฅx
i
x ๏ฝ i ๏ฝ1
n
๏ฝ
7.87 ๏ซ 8.63 ๏ซ 7.11 ๏ซ ๏ ๏ ๏ ๏ซ 6.33 1,656.16
๏ฝ
๏ฝ 7.427
223
223
The variance is:
s2 ๏ฝ
๏ฅ
๏จ๏ฅ x ๏ฉ
x ๏ญ
2
i
2
n ๏ญ1
n
๏ฝ
1,656.162
148.13391
223
๏ฝ
๏ฝ 0.66727
223 – 1
222
12 , 447.9812 –
The standard deviation is: s ๏ฝ s 2 ๏ฝ 0.66727 ๏ฝ 0.8169
x ๏ฑ 2s ๏ 7.427 ๏ฑ 2(0.8169) ๏ 7.427 ๏ฑ 1.6338 ๏ (5.7932, 9.0608).
From the histogram in part a, the data look approximately mound-shaped. From the
Empirical Rule, we would expect about 95% of the wells to fall in this range. In fact,
212 of 223 or 95.1% of the wells have pH levels between 5.7932 and 9.0608.
d.
The sample mean of the wells with detectible levels of MTBE is:
n
๏ฅx
i
x๏ฝ
i ๏ฝ1
n
๏ฝ
0.23 ๏ซ 0.24 ๏ซ 0.24 ๏ซ ๏๏๏ ๏ซ 48.10 240.86
๏ฝ
๏ฝ 3.441
70
70
Copyright ยฉ 2017 Pearson Education, Inc.
44
Chapter 2
The variance is:
๏จ๏ฅ x ๏ฉ
๏ญ
x
๏ฅ
2
i
2
s2 ๏ฝ
n ๏ญ1
n
๏ฝ
240.86 2
5283.5011
70
๏ฝ
๏ฝ 76.5725
70-1
69
6112.266-
The standard deviation is: s ๏ฝ s 2 ๏ฝ 76.5725 ๏ฝ 8.7506
x ๏ฑ 2 s ๏ 3.441 ๏ฑ 2(8.7506) ๏ 3.441 ๏ฑ 17.5012 ๏ (๏ญ14.0602, 20.9422).
From the histogram in part b, the data do not look mound-shaped. From Chebyshevโs
Rule, we would expect at least ยพ or 75% of the wells to fall in this range. In fact,
67 of 70 or 95.7% of the wells have MTBE levels between -14.0602 and 20.9422.
2.196
a.
Using MINITAB, the dot plot for the 9 measurements is:
Dotplot of Cesium
-6.0
-5.7
-5.4
-5.1
-4.8
-4.5
-4.2
Cesium
b. Using MINITAB, the stem-and-leaf display is:
Character Stem-and-Leaf Display
Stem-and-leaf of Cesium
Leaf Unit = 0.10
1
2
4
(3)
2
N
= 9
-6 0
-5 5
-5 00
-4 865
-4 11
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
c.
45
Using MINITAB, the histogram is:
Histogram of Cesium
2.0
Frequency
1.5
1.0
0.5
0.0
-6.0
-5.5
-5.0
-4.5
-4.0
Cesium
2.198
d.
The stem-and-leaf display appears to be more informative than the other graphs. Since
there are only 9 observations, the histogram and dot plot have very few observations per
category.
e.
There are 4 observations with radioactivity level of -5.00 or lower. The proportion of
measurements with a radioactivity level of -5.0 or lower is 4 / 9 ๏ฝ 0.444 .
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Ammonia
Variable
Ammonia
N
8
Mean
1.4713
StDev
0.0640
Minimum
1.3700
Q1
1.4125
Median
1.4900
Q3
1.5250
Maximum
1.5500
The stem-and-leaf display for the data is:
Stem-and-Leaf Display: Ammonia
Stem-and-leaf of Ammonia
Leaf Unit = 0.010
1
3
4
4
1
13
14
14
15
15
N
= 8
7
12
8
013
5
Since the data look fairly mound-shaped, we will use the Empirical Rule. We know that
approximately 99.7% of all observations will fall within 3 standard deviation of the mean.
For this data, the interval 3 standard deviations below the mean to 3 standard deviations
above the mean is:
x ๏ฑ 3s ๏ 1.471 ๏ฑ 3(0.064) ๏ 1.471 ๏ฑ 0.192 ๏ (1.279, 1.663)
We would be fairly confident that the ammonia level of a randomly selected day will fall
between 1.279 and 1.663 parts per million.
Copyright ยฉ 2017 Pearson Education, Inc.
46
Chapter 2
2.200
a.
From the histogram, the data do not follow the true mound-shape very well. The
intervals in the middle are much higher than they should be. In addition, there are some
extremely large velocities and some extremely small velocities. Because the data do not
follow a mound-shaped distribution, the Empirical Rule would not be appropriate.
b.
Using Chebyshev’s rule, at least 1 ๏ญ 1/ 4 ๏ฝ 1 ๏ญ 1/16 ๏ฝ 15 /16 or 93.8% of the velocities
will fall within 4 standard deviations of the mean. This interval is:
2
x ๏ฑ 4s ๏ 27,117 ๏ฑ 4(1,280) ๏ 27,117 ๏ฑ 5,120 ๏ (21,997, 32,237)
At least 93.75% of the velocities will fall between 21,997 and 32,237 km per second.
c.
If we assume that the distributions are symmetric and mound-shaped, then the Empirical Rule
will describe the data. We will compute the mean plus or minus one, two and three
standard deviations for both data sets:
Low income:
x ๏ฑ s ๏ 7.62 ๏ฑ 8.91 ๏ (๏ญ1.29, 16.53)
x ๏ฑ 2s ๏ 7.62 ๏ฑ 2(8.91) ๏ 7.62 ๏ฑ 17.82 ๏ (๏ญ10.20, 25.44)
x ๏ฑ 3s ๏ 7.62 ๏ฑ 3(8.91) ๏ 7.62 ๏ฑ 26.73 ๏ (๏ญ19.11, 34.35)
Middle Income:
x ๏ฑ s ๏ 15.55 ๏ฑ 12.24 ๏ (3.31, 27.79)
x ๏ฑ 2s ๏ 15.55 ๏ฑ 2(12.24) ๏ 15.55 ๏ฑ 24.48 ๏ (๏ญ8.93, 40.03)
x ๏ฑ 3s ๏ 15.55 ๏ฑ 3(12.24) ๏ 15.55 ๏ฑ 36.72 ๏ (๏ญ21.17, 52.27)
The histogram for the low income group is as follows:
.35
Relatie frequency
2.202
Since the data look approximately symmetric, the mean would be a good estimate for
the velocity of galaxy cluster A2142. Thus, this estimate would be 27,117 km per
second.
.30
.25
.20
.15
.10
.05
-19.11
-10.00
-1.29
7.62
Complexity
16.53
25.44
34.35
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
47
The histogram for the middle income group is as follows:
Relatie frequency
.35
.30
.25
.20
.15
.10
.05
-21.17
-8.93
3.31
15.55
Complexity
27.79
40.03
52.27
The spread of the data for the middle income group is much larger than that of the low
income group. The middle of the distribution for the middle income group is 15.55, while the
middle of the distribution for the low income group is 7.62. Thus, the middle of the
distribution for the middle income group is shifted to the right of that for the low income
group.
We might be able to compare the means for the two groups. From the data provided, it looks
like the mean score for the middle income group is greater than the mean score for the lower
income group.
(Note: From looking at the data, it is rather evident that the distributions are not moundshaped and symmetric. For the low income group, the standard deviation is larger than the
mean. Since the smallest measurement allowed is 0, this indicates that the data set is not
symmetric but skewed to the right. A similar argument could be used to indicate that the data
set of middle income scores is also skewed to the right.)
2.204
a.
For site A, there is no real pattern to the data that would indicate that the data are
skewed. For site G, most of the data are concentrated from 250 and up. There are
relatively few observations less than 250. This indicates that the data are skewed to the
left.
b.
For site A, there are 2 modes (two distance intervals with the largest number of
observations). Since there is no more than one mode, this would indicate that the data
are probably from hearths inside dwellings.
For site G, there is only one mode. This would indicate that the data are probably from
open air hearths.
2.206
The relative frequency for each cell is found by dividing the frequency by the total sample
size, n ๏ฝ 743 . The relative frequency for the digit 1 is 109 / 743 ๏ฝ 0.147 . The rest of the
relative frequencies are found in the same manner and are shown in the table.
Copyright ยฉ 2017 Pearson Education, Inc.
48
Chapter 2
First Digit
1
2
3
4
5
6
7
8
9
Total
Relative
Frequency
0.147
0.101
0.104
0.133
0.097
0.157
0.120
0.083
0.058
1.000
Frequency
109
75
77
99
72
117
89
62
43
743
Using MINITAB, the relative frequency bar chart is:
Chart of Freq
.16
Relative Frequency
.14
.12
.10
.08
.06
.04
.02
0
1
2
3
4
5
6
7
8
9
FirstDigit
Proportion within all data.
Benford’s Law indicates that certain digits are more likely to occur as the first significant
digit in a randomly selected number than other digits. The law also predicts that the number
“1” is the most likely to occur as the first digit (30% of the time). From the relative
frequency bar chart, one might be able to argue that the digits do not occur with the same
frequency (the relative frequencies appear to be slightly different). However, the histogram
does not support the claim that the digit “1” occurs as the first digit about 30% of the time. In
this sample, the number “1” only occurs 14.7% of the time, which is less than half the
expected 30% using Benford’s Law.
2.208
If the distributions of the standardized tests are approximately mound-shaped, then it would
be impossible for 90% of the school districts’ students to score above the mean. If the
distributions are mound-shaped, then the mean and median are approximately the same. By
definition, only 50% of the students would score above the median.
If the distributions are not mound-shaped, but skewed to the left, it would be possible for
more than 50% of the students to score above the mean. However, it would be almost
impossible for 90% of the students scored above the mean.
2.210
For the first professor, we would assume that most of the grade-points will fall within 3
standard deviations of the mean. This interval would be:
Copyright ยฉ 2017 Pearson Education, Inc.
Methods for Describing Sets of Data
49
x ๏ฑ 3s ๏ 3.0 ๏ฑ 3(.2) ๏ 3.0 ๏ฑ .6 ๏ (2.4, 3.6)
Thus, if you had the first professor, you would be pretty sure that your grade-point would be
between 2.4 and 3.6.
For the second professor, we would again assume that most of the grade-points will fall
within 3 standard deviations of the mean. This interval would be:
x ๏ฑ 3s ๏ 3.0 ๏ฑ 3(1) ๏ 3.0 ๏ฑ 3.0 ๏ (0.0, 6.0)
Thus, if you had the second professor, you would be pretty sure that your grade-point would
be between 0.0 and 6.0. If we assume that the highest grade-point one could receive is 4.0,
then this interval would be (0.0, 4.0). We have gained no information by using this interval,
since we know that all grade-points are between 0.0 and 4.0. However, since the standard
deviation is so large, compared to the mean, we could infer that the distribution of gradepoints in this class is not symmetric, but skewed to the left. There are many high grades, but
there are several very low grades.
By taking the first professor, you know you are almost positive that you will get a final grade
of at least 2.4, but almost no chance of getting a final grade of 4. By taking the second
professor, you know the grades are skewed to the left and that many of the students will get
high grades, but also a few will get very low grades.
2.212
The answers to this will vary. Some things that should be included in the discussion are:
From the graph, it is obvious that the amount of money spent on education has increased
tremendously over the period from 1966 to 2000 (from about $4.5 billion in 1966 to about
$22.5 billion in 2000). However, one should note that the number of students has also
increased. It might be better to reflect the amount of money spent as the amount of money
spent per student over the years from 1966 to 2000 rather than the total amount spent.
In the description of the exercise, it says that the horizontal line represents the annual average
fourth-grade childrenโs reading ability. It also indicates that the fourth-grade reading test
scores are designed to have an average of 250 with a standard deviation of 50. Thus,
regardless of whether the childrenโs reading abilities increase or decrease, the annual average
will always be 250. This line does not give any information about whether the childrenโs
reading abilities are improving or not.
In addition, if the reading scores of seventh and twelfth graders and the mathematics scores of
fourth graders improved over the same time period, one could conclude that the reading
scores of the fourth graders also improved over the same time period.
Thus, this graph does not support the governmentโs position that our children are not making
classroom improvements despite federal spending on education. This graph only portrays
that the total amount of money spent on education over the time period from 1966 to 2000
increased.
Copyright ยฉ 2017 Pearson Education, Inc.

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