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Unformatted text preview: Student: Grady Silnonton Colu'se: 1\Iat11119: Elelnentaiy Statistics  Spring 3010  C‘RN: 49339
Instructor: Shawn Pan'ini  16 weeks
Date: 3."'18."'10 Book: Triola: Elelnentaiy Statistics. 11c Time: 11:08 AM The data represents the actual high temperature for 14 consecutivedays. 60 60 65
80 65 ?5 Construct a dotplot of the actual high temperatures. What does the dotplot suggest about the T5 "3'0 7’5 distribution of the high temperatures? T0 65 80
55 85 A dotplot consists ofa graph in which each data value is plotted as a point (or dot) along a scale of values. Dots
representing equal values are stacked. First create a frequency distribution with a class width of 5 and the class midpoints starting at 55. The frequency for each
class midpoint is listed below. 55 60 65 T0 75 80 85
l 2 3 2 3 2 1 To create the dotplot, list the categories horizontally and place the appropriate
number of dots above each of the values. The number of dots corresponds to the _ _
frequency for each value. I 1' ' 1' 1' 1' ' 1' I What does the dotplot suggest about the distribution of the high temperatures? The range of temperatures tells us that the actual high temperatures range from
55 degrees to 85 degrees with a fairly uniform distribution of temperatures. 50 60 7O 80 90 Page 1 Student: Grady Simonton Colu'se: 1\Iat11119: Elelnentaiy Statistics  Spring 3010  C‘RN: 49339
Instructor: Shawn Pan'ini  16 weeks
Date: 3."'18."'10 Book: Triola: Elelnentaiy Statistics. 11c Time: 11:09 AM The data represents the actual high temperature for 14 consecutivedays. Use the 14 actual 66 52 76 high temperatures to construct a frequency polygon. For the horizontal axis, use the midpoint 60 84 59 values obtained from these class intervals: 5059, 6069, 7"07’9, 8089. T4 1'5 53
69 56 62
58 85 A frequency polygon uses line segments connected to points located directly above class midpoint values. The heights of
the points correspond to the class frequencies. and the line segments are extended to the right and left so that the graph
begins and ends on the horizontal axis. First create a frequency distribution with the given class intervals. Class interval 50—59 60—69 70—79 80—89 Class midpoint 44.5 54.5 64.5 ?4.5 84.5 94.5 Frequency 0 5 4 3 2 0 Use the frequencies and the class midpoints obtained from the class intervals 10} to construct a frequency polygon. —
8_ 45 55 65 75 85 95 Page 1 Student: Grady Simonton Instructor: Shawn Pan'ini  16 weeks Date: 3."'18.."'10 Book: Tl‘iola: Elenlentaiy Statistics. 11e
Time: 11:09 AM
The data represents the heights of eruptions by a geyser. 124
I29
Use the heights to construct a stemplot. What does the stemplot suggest about the distribution 1 10
of the heights? 1 19
139 A stemplot (or stemandleafplot) represents data by separating each value into two parts, the stem (all digits of the number except the last and rightmost digit) and the leaf (the last, rightmost digit of the number). Start by ﬁnding the largest and smallest value. This will allow us to know how many rows to create. Smallest Largest 90 150 The rows of digits in a stemplot are similar in nature to the bars in a histogram. One of the 9
guidelines for constructing histograms is that the number of classes should be between 5 and 10
20, and the same guideline applies to stemplots for the same reasons. Therefore, we will use 7" rows. Each row represents the tens digit. 1; 14
15 Now ﬁll in the stemplot. Start with the first number, 124. The stem is 12 and the leafis 4.
Therefore, find the row that corresponds to 12 and insert a 4 as the first number. Use the same procedure to fill in the remaining numbers. Note that the leaves are arranged
in increasing order, not the order in which they occur in the original list. If the plot starts small then increases to some maximum, then decreases and it is also symmetric, then we refer to the 90
120
129
130
140 129
100
138
110
100 Colu'se: 1\1at11119: Elenlentaiy Statistics  Spring 3010  C‘RN: 49339 150
130
113
120
142
shape of the graph as bellshaped. If the plot starts small then increases to some maximum value then decreases but it is not symmetric, then we refer to the shape of the graph as being either right or left skewed. When the plot does not increase or decrease, we refer to the graph as being [miforrn Page 1 Student: Grady Sinmnten Course: 1\Iat11119: Elenlentaiy Statistics  Spring 3010  C‘RN: 49339 Instructor: Shawn Pan'ini  16 weeks
Date: 3."'18."'10 Book: T1‘icla: Elenlentaiy Statistics. Me Time: 11:09 AM Look at the shape of the stemplot. Since the values in the stemplot increase to a
maximum then decrease and the plot is symmetric, the plot appears to be bellshaped. Page 3 Student: Grady Silnonton
Instructor: Shawn Pan'ini
Date: 3."'18.."'10  16 weeks
Time: 11:10 AM The data represents the heights of eruptions by a geyser. Use the heights to construct an ogive. For the horizontal axis, use these class boundaries: 89.5, 99.5, 109.5, 119.5, 129.5, 139.5, 149.5, 159.5. How many eruptions were below 120 it? An ogive is a line graph that depicts cumulative frequencies. Note that the ogive uses class boundaries along the Book: Triola: Elelnentaiy Statistics. 11e 123
I27
110
119
130 90
120
129
130
140 123
100
132
110
100 Colu'se: 1\Iat11119: Elelnentaiy Statistics  Spring 2010  C‘RN: 49239 150
130
117
120
149 horizontal scale. The graph begins with the lower boundary of the ﬁrst class and ends with the upper boundary of the last class.
First create a frequency distribution with the given class intervals. Class Intervals 90—99 100—109110—119120—129130—139140—149 150—159
Frequency 1 2 4 6 4 2 1 Now ﬁnd the cumulative frequencies. Class Intervals
Class Boundaries 89.5 99.5 109.5 119.5 129.5 139.5
Cumulative Frequency 0 1 3 7’ 13 17" Use the cumulative frequencies to construct the ogive. Ogives are useful for determining the number of values below
some particular value. Look at the ogive. Find the yvalue that
corresponds to the xvalue of 1 19.5. Notice that there were 7
eruptions below 120 ft. Page 1 1 I‘\J
O
' ) _k
“P _L
'73 43 03
IIIIIIIIII C) 20:
16—" 12: 49.5
19 9099100109110119120129130139140149 150159 159.5 20 90100110120130140150160 90100110120130140150160 Student: Grady Silnonton Celu'se: 1\Iat11119: ElenIentaiy Statistics  Spring 3010  C‘RN: 49339
Instructor: Shawn Panini  16 weeks
Date: 3."'18."'10 Beck: Triola: ElenIentaiy Statistics. 11e Time: 11:10 AM The following data represent the number of people from a
town aged 2564 who have internet access in their homes. Construct a frequency polygon using the data.
First, find each class midpoint by adding the lower class limit and the upper class limit and dividing the result by 2. The midpoint of the first class is found using the aforementioned formula as seen below. 34+25
—= 29.5 The rest of the class midpoints, shown below, can be found by the same method. Midpoint To draw a frequency polygon, plot a point above each class midpoint on a horizontal axis at a height equal to the
frequency ofthe class. After the points are plotted, straight lines are drawn between consecutive points. The line segments are extended to the
left and right so that the graph begins and ends on the horizontal axis. Here is the frequency polygon that matches the data. 1000—
} soo
._, _
5 600—
:0: 400—
.l' 
— 200— 29.5 39.5 49.5 59.5
Age Page 1 Student: Grady Sinlonton Colu'se: 1\Iat11119: Elenlentaiy Statistics  Spring 3010  C‘RN: 49339
Instructor: Shawn Pan'ini  16 weeks
Date: 3."'18."'10 Book: Triola: Elenlentaiy Statistics. 11e Time: 11:10 AM A study was conducted to determine how people get jobs. The table lists data Job Sources Frequency
from 400 randomly selected subjects. Construct a Pareto chart that corresponds to Helpwanted ads (H) 270
the given data. If someone would like to get a job, what seems to be the most
effective approach? Executive search firms (E) 60
Networking (N) 30
Mass mailing (M) 40 A Pareto chart is a bar graph for qualitative data, with the bars arranged in order according to frequencies. Vertical scales
in Pareto charts can represent frequencies or relative frequencies. The tallest bar is at the left, and the smaller bars are
farther to the right. Start by arranging the frequencies in order from largest to smallest with the corresponding job sources. Job Source Frequencies
Helpwanted ads (H) 270 Largest
Executive search ﬁrms (E) 60
Mass mailing (M) 40
Networking (N) 30 Smallest Use the table above to construct a Pareto chart. The tallest bar is at the left, . . 300
and the smaller bars are tarther to the right. 250
200
150 100
50 HEMN By arranging the bars in order of frequency, the Pareto chart focuses attention on the more important categories. If
someone would like to get a job, the most effective approach seems to be helpwanted ads. Page 1 Student: Grady Si111o11ton
Instructor: Shawn Pan'ini
Date: 3."'18.."'10 A study was conducted to determine how people get jobs. The table below lists data from 400 randomly selected
subjects. Job Sources Frequency
Help—wanted ads (H) 270
Executive search ﬁrms (E) 60
Networking (N) 40
Mass mailing (M) 30
300
250
200
150
100 50 0 H E N M Colu'se: 1\Iat11119: Elelnentaiy Statistics  Spring 3010  C‘RN: 49339
 16 weeks Book: Triola: Elelnentaiy Statistics. 11e Time: 11:11AM A pie chart is a graph depicting qualitative data as slices of
a pie. Construction ofa pie chart involves slicing up the pie
into the proper proportions. The below pie chart was constructed by using the
frequencies to determine relative size and labeling each
frequency with its respective job source. I Helpwanted ads (H)
I Executive search ﬁlms (E)
El Networking (N) I] Mass mailing M Compare the pie chart found above to the Pareto chart
given on the left. Can you determine which graph is more
effective in showing the relative importance ofjob
sources? A Pareto chart is designed to emphasize a few large classes
of data, while a pie chart is useful for contrasting several
classes of data that are approximately equal. Page 1 Student: Grady Sinlonton Colu'se: 1\Iat11119: Elementary Statistics  Spring 3010  C‘RN: 49339
Instructor: Shawn Pan'ini  16 weeks
Date: 3."'18."'10 Book: T1‘iola: Elementary Statistics. 11e Time: 11:11AM In a recent year, 5400 people were killed while working. Here is a breakdown I Cmnbustibles (C)
ofcauses: combustibles (2350); substances (800); transportation (7’50); I Substances (3)
equipment (1'00); violence (650); falls (150). Use the data to construct a
Pareto chart. Compare the Pareto chart to the pie chart. Which graph is more effective in showing the relative importance of the causes of work—related
deaths? A Pareto chart is a bar graph for qualitative data, with the bars arranged in order according to frequencies. Vertical
scales in Pareto charts can represent frequencies or relative frequencies. The tallest bar is at the left, and the smaller bars
are farther to the right. Start by arranging the frequencies in order from largest to smallest with the corresponding cause of workrelated death. Cause of Cause of Workrelated death quuenmes Workrelated death quuenmes
Combustibles (C) 2350 Substances (S) 800
Transportation (T) 750 Equipment (E) 700
Violence (V) 650 Falls (F) 150
Use the chart above to construct a Pareto chart. The tallest bar is at the left, 2500
and the smaller bars are farther to the r1ght. 2000 1500
1000
500
0 A Pareto chart is designed to emphasize a few large classes of data, while a pie chart is useful for contrasting several
classes of data that are approximately equal. By arranging the bars in order of frequency, the Pareto chart focuses attention on the more important categories.
Therefore, the Pareto chart is more effective in showing the relative importance of the causes of workrelated deaths. Page 1 Student: Grady Sinlonton Culu'se: 1\Iat11119: Elementary Statistics  Spring 3010  C‘RN: 49339
Instructor: Shawn Pan'ini  16 weeks
Date: 3."'18."'10 Book: T1‘iola: Elementary Statistics. 11e Time: 11:11AM Listed below are blood groups of O, A, B, and AB of randomly selected blood donors. Construct a pie chart depicting
the distribution of these blood groups. AAABOOBOOOAAAOAOOOAOAB A pie chart is a graph that depicts qualitative data as slices ofa circle, in which the size ofeach slice is proportional to
the frequency count for the category. Begin by finding the frequency of each blood type. The completed frequency table is shown below. Blood Group Frequency A 7
B 1
AB 2
O 10 Next calculate the percentage frequency of each blood group. To do so, first calculate the sum of all frequencies.
7 +1 +2 +10 = 20 The percentage frequency is equal to the category frequency divided by the sum of all frequencies, multiplied by 100%.
The table of percentage frequencies is shown below. Blood Group Percentage Frequency
A W20 X 100% = 35%
B H20X100%=5%
AB 290 X 100% = 10%
0 10120 X 100% = 50% The sizes of the slices of the circle representing the pie chart are given by the percentages calculated in the previous step.
That is, 35% of the pie chart shows blood group A, 5% of the pie chart shows blood group B, 10% of the pie chart
shows blood group AB, and 50% of the pie chart shows blood group 0. Use these results to construct the pie chart depicting the distribution of these blood groups. The pie chart is shown
below. Page 1 Student: Grady Sinlonton Colu'se: 1\Iat11119: Elenlentaiy Statistics  Spring 3010  C‘RN: 49339
Instructor: Shawn Pan'ini  16 weeks
Date: 3."'18."'10 Book: T1‘iola: Elenlentaiy Statistics. 11e Time: 11:13 AM Construct a scatter diagram. Use tar for the horizontal scale and use carbon Tar C Tar C Tar CO
monoxide (CO) for the vertical scale. Determine whether there appears to be 15 15 14 1 2 3
a relationship between cigarette tar and CO. 16 1 8 1 1 I 4 l3 A scatterplot (or scatter diagram) is a plot of paired (x, y) data with a horizontal xaxis and a vertical yaxis. The data are
paired in a way that matches each value from one data set with a corresponding value from a second data set. To
manually construct a scatterplot, construct a horizontal axis for the values of the ﬁrst variable. Then construct a vertical
axis for the values of the second variable, then plot the points. Use tar for the horizontal scale and use carbon monoxide (CO) for the vertical 20} C0
scale. E
15E
12—:
8:
43
E Tar
U__l_l_l__l_l_l_—l_l_l_rl_l_l_—l_l_l__’
o 4 s 12 16 20
Now plot the points. ACO
20‘.
' O
16—— .0
Z O
12—" z
: O .0
' O
8—: n
4—: .
0 3 . Tar
_ , 04 8121620 Determine whether there appears to be a relationship between cigarette tar and CO. The pattern of the plotted points is often helpful in determining whether there is 20} C0
some relationship between the two variables. Observe that the points of the g . .
scatterplot appear to lie along a straight line. 16‘: z .' 12—: o .. 8—: n " 4—: . 0—; . Tar 04 8121620 Thus, there appears to be a linear relationship between cigarette tar and carbon monoxide. Page 1 Student: Grady Sinlonton Colu'se: Mathll9: Elenlentaiy Statistics  Spring 2010  CRN: 49239
Instructor: Shawn Pan'ini  16 weeks
Date: 2.518.510 Book: T1'iola: Elenlentaiy Statistics. 11e Time: 11:12 AM iﬂsﬂwnoer 01' Drive1n lVlDVIB lnBaLEI'S Given below are the numbers of indoor movie theaters, listed in order by row for each year. Use the given data to construct a time—series graph. What is the trend?
2000 20,595 22,1?4 21,90? 21,6?0 23,240 26,609
24,789 27,025 26,995 30,9?0 31,050 32,160 1m
36,448 33,333 34,490 34,688 35,361 Year A time—series graph is a graph of time—series data, which are data that have been collected at different points in time. Use time for the horlzontal was and the number of iiidoor movle theaters for Number Ufiwdom movie the vertical axis. heaters
A
30000é
32000é
28000é
24000é ZUOUO—Z'n—rrn—rrrn—rrn‘rrﬂ
0 4 8 12 16 Time Number Now plot the pomts. Number of' indoor movie theaters
A 36000é
32000é
28000é
24000§ 20000; O 4 81215
Time Number What is the trend? How does this trend compare to the trend for drivein movie theaters? Look at the timeseries graph. Notice that the trend appears to be upward. Number Dfimdoor movie Notice that the graph of the number of drivein movie theaters has a downward heaters
trend. 360005
E 32000—:
E 280005
7 240005
20000: 0 4 a 12 15
Time Page 1 Student: Grady Sinlonton Colu'se: 1\Iat11119: Elenlentaiy Statistics  Spring 3010  C‘RN: 49339
Instructor: Shawn Pan'ini  16 weeks
Date: 3."'18."'10 Book: Triola: Elenlentaiy Statistics. 11e Time: 11:13 AM The following table lists the marriage and divorce rates per 1000 people in a particular country for selected years since
1900. Construct a multiple bar graph ofthe data. Why do these data consist of marriage and divorce rates rather than
total numbers of marriages and divorces? Comment on any trends that you observe in these rates, and give explanations
for these trends. Year 1990 2000 Marriage .
Divorce 4.2 A bar graph uses bars of equal width to show frequencies of categories of qualitative data. The vertical scale represents
frequencies or relative frequencies. The horizontal scale identiﬁes the different categories of qualitative data. A multiple
bar graph has two or more sets of bars, and is used to compare two or more data sets. The bars may or may not be separated by small gaps. Construct a multiple bar graph of the data. The multiple bar graph is shown below. ’15:
210—:
S2
0 00 10 20 30 40 50 60 T0 80 90 00
I Marriage I Divorce Year Why do these data consist of marriage and divorce rates rather than total numbers of marriages and divorces?
Proportions should be used to show how a particular characteristic ofa population changes over time. Determine
whether there are any cases where providing the totals ofa particular characteristic ofa population over time differs in
interpretation from the proportions. Comment on any trends that you observe in these rates, and give explanations for these trends. Refer to the multiple bar graph to determine the trends of marriage and divorce rates over time for the particular country.
Think ofa possible explanation for these trends. Page 1 Student: Grady Silnonton
Instructor: Shawn Panini
Date: 3."'18.."'10 Colu'se: 1\Iat11119: Elenlentaiy Statistics  Spring 3010  C‘RN: 49339
 16 weeks Book: T1‘iola: Elenlentaiy Statistics. 11e Time: 11:13 AM A format for backtoback stemplots representing the pulse A stemplot represents quantitative data by separating each rate oi‘f‘emalcs and males from the given data is shown
below. Complete the backtoback stemplot. then compare
the results. Females 60 64 65 67 68 68 69 7’3 7"? 7’8
29 29 81 83 84 84 8‘? 8910112?
Males 50 54 56 63 64 66 66 66 67 68
69 69 72 72 76 80 81 81 83 89 Men Stem (tens) 5 046
988754 value into the stem (such as the lettmost digit) and the leaf
(such as the rightmost digit). The stems and leaves are
arranged in increasing order (beginning at the stern), not
the order in which they occur in the original list. To make it easier to complete the stemplot, ﬁrst arrange the
pulse rates in increasing order. Begin with the pulse rates
for women. Females
6O 64 65 6?
"1'9 68 68 69 T3 T7 T8
'19 8] 83 84 84 87 8910112?r Now order the pulse rates for men. Males
50 54 56 63 64 66 66 66 67 68
69 69 72 72 76 80 81 81 83 89 Notice that the pulse rates are now shown in increasing
order in the problem statement. The completed backtoback stemplot is shown below. Women Men Stem (tens) 5 046
988754 6 346667899
99873 '1' 226
974431 8 0] 139
9
l 10
1 1
7 12 Compare the results. The rows of digits in a stemplot are similar in nature to the
bars in a histogram. By turning the stemplot on its side, one
can see a distribution of the data. Compare the distributions
of pulse rates of men and women. Page 1 ...
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 Spring '10
 PARVINI
 Frequency distribution, Bar chart, Shawn Pan'ini

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