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7/28/2019 Elementary Stastistics
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Elementary Statistics Dr. Ghamsary Chapter 2 Page 1
1
Elementary Statistics
M. Ghamsary, Ph.D.
Chapter 02
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Elementary Statistics Dr. Ghamsary Chapter 2 Page 2
2
Descriptive Statistics
Grouped vs Ungrouped Data
Ungrouped data:have not been summarized in any wayare also calledraw data Grouped data: have been organized into a frequency distribution
RRaaww DDaattaa:: When data are collected in original form, they are calledrraaww ddaattaa.
The following are the scores on the first test of the statistics class in fall of 2004.
76 78 71 86 80
62 55 89 66 72
68 96 78 81 82
69 89 88 85 86
79 73 58 85 99
90 66 76 70 63
79 88 59 55 75
86 92 92 62 83
52 94 93 80 78
97 50 88 60 61
Table 2.1: Data fromTest#1 of fall 2007
Stem-and-Leaf: One method of displaying a set of data is with a stem-and-leaf plot.
Stem Leaf
5 0 2 5 5 8 9
6 0 1 2 2 3 6 6 8 9
7 0 1 2 3 5 6 6 8 8 8 9 9
80 0 1 2 3 5 5 6 6 6 8 8 8 9 9
9 0 2 2 3 4 6 7 9
Group Data: When the raw data is organized into a ffrreeqquueennccyy ddiissttrriibbuuttiioonn
FFrreeqquueennccyy DDiissttrriibbuuttiioonn: is the organizing of raw data in table form, using classes and
frequencies.
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Class: Number of classes in the above table is 5.Class Limits: represent the smallest and largest data values in each class. LLoowweerr CCllaassss:: the lowest number in each class. In above table 50 is the lower class limit of the
first class, 60 is the lower class limit of the 2ndclass, etc.
UUppppeerr CCllaassss::tthhee hhiigghheesstt nnuummbbeerriinn eeaacchh ccllaassss.. In above table 59 is the upper class limit of thefirst class, 69 is the upper class limit of the 2ndclass, etc.
CCllaassss WWiiddtthh:: for a class in a frequency distribution is found by subtracting the lower (orupper) class limit of one class minus the lower (or upper) class limit of the previous class. In
above table the class width is 10.
CCllaassss BBoouunnddaarriieess are used to separate the classes so that there are no gaps in the frequencydistribution.
ClassClass
Boundaries
Frequency
50-59 49.5-59.5 6
60-69 59.5-69.5 9
70-79 69.5-79.5 12
80-89 79.5-89.5 15
90-99 89.5-99.5 8
Class Tally Frequency
50-59 6
60-69 9
70-79 12
80-89 15
90-99 8
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Elementary Statistics Dr. Ghamsary Chapter 2 Page 4
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Cumulative Frequency:
Relative Frequency:
ClassFrequency Cumulative
FrequencyRelativeFrequency
50-59 6 6 6/50=0.12
60-69 9 9+6=15 9/50=0.18
70-79 12 12+15=27 12/50=0.24
80-89 15 15+27=42 15/50=0.30
90-99 8 8+42=50 8/50=0.16
nn==5500
MMoosstt PPooppuullaarr GGrraapphhss iinn SSttaattiissttiiccss
TThhee mmoosstt ccoommmmoonnllyy uusseeddggrraapphhss iinn ssttaattiissttiiccss aarree::
11.. TThhee HHiissttooggrraamm
22.. TThhee FFrreeqquueennccyy PPoollyyggoonn..
33.. TThhee CCuummuullaattiivvee FFrreeqquueennccyy GGrraapphh
44.. TThhee BBaarr CChhaarrtt
55.. PPiiee CChhaarrtt
66.. PPaarreettoo CChhaarrttss
77.. DDoott PPlloott
88.. SStteemm--LLeeaaff99.. TTiimmee SSeerriieess GGrraapphh
11.. TThhee HHiissttooggrraamm
o Making decisions about a process, product, or procedure that could be improved afterexamining the variation (example: Should the school invest in a computer-based tutoring
program for low achieving students in Algebra I after examining the grade distribution? Are
more shafts being produced out of specifications that are too big rather than too small?)
o Displaying easily the variation in the process (example: Which units are causing the mostdifficulty for students? Is the variation in a process due to parts that are too long or parts that
are too short?)
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Elementary Statistics Dr. Ghamsary Chapter 2 Page 5
5
Test1
Frequency
9585756555
16
14
12
10
8
6
4
2
0
Mean 76.8
StDev 12.98
N 50
Histogram of Test1Normal
Test1
Percent
12011010090807060504030
99
95
90
80
70
60
50
40
30
20
10
5
1
Mean 76.8
StDev 12.98
N 50
AD 0.537
P-Value 0.161
Probability Plot of Test1Normal - 95% CI
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2. The frequency polygon
o Making decisions about a process, product or procedure that could be improved(example: a frequency polygon for 642 psychology test scores, shown below to the right.)
X Frequency
54.5 6
64.5 9
74.5 12
84.5 15
94.5 8
Midpoints x
f
10090807060
15.0
12.5
10.0
7.5
5.0
Scatterplot of f vs x
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Elementary Statistics Dr. Ghamsary Chapter 2 Page 7
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2.The Cumulative Frequency Graph (Ogive)
Cumulative frequency is used to determine the number of observations that lie above
(or below) a particular value.
Upper ClassBoundaries
CumulativeFrequency
59.5 6
69.5 15
79.5 27
89.5 42
99.5 50
Upper Class Boudaries
Cumula
tivef
10090807060
50
40
30
20
10
0
Scatterplot of Cumulative f vs x
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Elementary Statistics Dr. Ghamsary Chapter 2 Page 8
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4. The bar chart
Bar charts are useful for comparing classes or groups of data. A class or group can have a
single category of data or they can be broken down further into multiple categories for
greater depth of analysis.
Class Grade Frequency
50-59 F 6
60-69 D 9
70-79 C 12
80-89 B 15
90-99 A 8
ABCDF
16
14
12
10
8
6
4
2
0
Grade
Frequency
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Elementary Statistics Dr. Ghamsary Chapter 2 Page 9
9
55.. PPiiee CChhaarrtt
oo AA ppiiee cchhaarrtt iiss aa wwaayy ooffssuummmmaarriizziinngg aa sseett ooffccaatteeggoorriiccaall ddaattaa oorrddiissppllaayyiinngg tthhee ddiiffffeerreennttvvaalluueess ooffaa ggiivveenn vvaarriiaabbllee ((eexxaammppllee:: ppeerrcceennttaaggee ddiissttrriibbuuttiioonn))..
oo PPiiee cchhaarrttss uussuuaallllyy sshhooww tthhee ccoommppoonneenntt ppaarrttss ooffaa wwhhoollee.. OOfftteenn yyoouu wwiillll sseeee aa sseeggmmeenntt oofftthheeddrraawwiinngg sseeppaarraatteeddffrroomm tthhee rreesstt oofftthhee ppiiee iinn oorrddeerrttoo eemmpphhaassiizzee aann iimmppoorrttaanntt ppiieeccee ooff
iinnffoorrmmaattiioonn
A8, 16.0%
B15, 30.0%
C12, 24.0%
D9, 18.0%
F6, 12.0%
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Elementary Statistics Dr. Ghamsary Chapter 2 Page 10
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66.. PPaarreettoo CChhaarrttss
A Pareto chart is used to graphically summarize and display the relative importance of thedifferences between groups of data.
FADCB
16
14
12
10
8
6
4
2
0
Frequency
77.. DDoott pplloottA dot plot is a visual representation of the similarities between two sequences.
T e s t 1
9891847770635649
Dotp lot o f Te st1
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Elementary Statistics Dr. Ghamsary Chapter 2 Page 11
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88.. SStteemm--LLeeaaff
o The Stem-and-Leaf Plot summarizes the shape of a set of data (the distribution) andprovides extra detail regarding individual values.
o They are usually used when there are large amounts of numbers to analyze. Series ofscores on sports teams, series of temperatures or rainfall over a period of time, series of
classroom test scores are examples of when Stem and Leaf Plots could be used.
Stem Leaf
5 0 2 5 5 8 9
6 0 1 2 2 3 6 6 8 9
7 0 1 2 3 5 6 6 8 8 8 9 9
80 0 1 2 3 5 5 6 6 6 8 8 8 9 9
9 0 2 2 3 4 6 7 9
9. Time series Graph
NovOctSepAugJulJunMayAprMarFebJanDec
120
110
100
90
80
70
60
50
40
30
Month
AOL
MSFT
Variable
Time Series Plot of AOL, MSFT
Month Price of
AOL
Price of
MSFT
Jan
Feb
MarApr
May
JunJul
Aug
SepOct
Nov
Dec
65
60
5862
55
5048
55
5750
48
40
110
115
120100
95
9085
75
8060
50
40
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Type of Distributions:
There are several different kinds of distributions, but the following are the most common used in
statistics.
Symmetric , normal, or bell shape Positively skewed, Right tail, or skewed to the right side. Negatively skewed, Left tail, or skewed to the left side. Uniform Symmetric, Bell Shape, or Normal Distribution
1441261089072543618
600
500
400
300
200
100
0
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Positively skewed
0.630.540.450.360.270.180.090.00
500
400
300
200
100
0
Negatively skewed
0.990.900.810.720.630.540.450.36
500
400
300
200
100
0
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Uniform
1086420
1000
800
600
400
200
0
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1=Female 0=Male
Male Female
Sex
0
2
4
6
8
Count
Grade
F
D
C
B
A
F D C B A
Grade
0
3
6
9
12
15
Count
Sex
Male
Female
F D C B A
Grade
0
10
20
30
40
50
60
70
Percent
Sex
Male
Female
Test1 Sex Grade Test1 Sex Grade
76 1 C 76 1 C
62 1 D 59 1 F
68 1 D 92 1 A
69 1 D 93 1 A
79 0 C 88 0 B
90 0 A 86 0 B
79 1 C 66 0 D
86 1 B 81 1 B
52 0 F 85 0 B
97 1 A 85 0 B
78 1 C 70 1 C
55 1 F 55 1 F
96 1 A 62 1 D
89 1 B 80 1 B
73 0 C 60 1 D
66 0 D 80 1 B
88 1 B 72 1 C
92 0 A 82 0 B94 1 A 86 1 B
50 1 F 99 1 A
71 0 C 63 1 D
89 0 B 75 1 C
78 1 C 83 1 B
88 0 B 78 0 C
58 1 F 61 1 D
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Sex
Test1
MaleFemale
100
90
80
70
60
50
Boxplot of Test1 vs Sex
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Numerical measurements: Sttaattiissttiicc::: any value(s) or measure(s) obtained from a sample.
PPaarraammeetteerr:: any value(s) or measure(s) obtained from a specific population.Measures of central tendency: are Mean, Median, andMode,
MMeeaannis defined to be the sum of the scores in the data set divided by the total number of scores.
o Sample Mean: is denoted by x , and it is defined by:
x
x
n
i
i
n
= =
1
, or simply
x
x n=
.
o Population Mean: is denoted by , and it is defined by:
= =xN
i
i
N
1 , or simplyx
N = .
Note: The sample mean,x is an unbiased estimate of the population mean, .
Example1: Find the mean of 10, 7, 3, 12, 18.
x = + + + + =10 7 3 12 185
10 .
Example2: Find the mean of 10, 7, 3, 12, 18, 13, 17, 15, 25, 3
x =+ + + + + + + + +
= =10 7 3 12 18 13 17 15 25 30
10
150
10 15
Example3: Find the mean of scores in the test#1, 2004 in data set in this chapter.
76 62 78 61
57
06 8.x
+ + + += =
"
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Elementary Statistics Dr. Ghamsary Chapter 2 Page 18
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MMeeddiiaann::is defined to be the midpointof the data set that is arranged from smallest to largest.
Example4: Find the median of 10, 7, 3, 12, 15.
Solution: First we must sort the data set as follows: 3, 7, 10, 12, 15.
The median is 10.
Example5: Find the median of 10, 7, 3, 12, 15, 20.
Solution: After we sort we get: 3, 7, 10, 12, 15, 20.
As we observe, there are 2 middle observations. So to find the median we average these 2 values,
namely: Median=(10+12)/2 =11.
Example6: The median of scores in the test#1, 2004 in data is 78.50
Median = 78.50
MMooddee::is defined to be the value in the data set that occurs most frequently.
Example7A: Find the mode of 10, 7, 3, 12, 15, 3.
Mode is 3.
Example7B: Find the mode of 10, 7, 3, 10, 15, 3.
Modes are 3 and 10.
Example7C: Find the mode of 10, 7, 3, 10, 10, 3.
Mode is 10.
Example7D: Find the mode of 10, 7, 3, 10, 7, 3.
There is no mode, since all values occur with same frequency
Example7E: Find the mode of 10, 7, 3, 12, 15, 18.
There is no mode, since no values occur more than once.
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Example 8: Find the mean, the median, and the mode of data set:10, 17, 13, 12, 15, 18, 10, 17, 14, 16, 35, 28, 22, 17, 23, 12, 15, 28, 10, 20
Solution: First we must sort the data set
10, 10, 10, 12, 12, 13, 14, 15, 15, 16, 17, 17, 17, 18, 20, 22, 23, 28, 28, 35
o Mean: x = + + + + + + + = =10 10 10 12 28 28 3520
352
2017 6
. . . . ..
o Median: 16 172
16 5= . , since there are 2 middle observationso Mode: 10 17,
Example 9: Find the mean, the median, and the mode of data set:
25, 42, 18, 37, 25, 18, 40, 57, 64, 66, 85, 86, 92
85, 88, 92, 67, 33, 75, 85, 48, 60, 80, 60, 50
Example10: Find the mean, the median, and the mode of data set:
12.37, 13.33, 32.67, 12.37, 26.45
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Example11A: Find the mean for the following group data
Class Frequency
50-59
60-69
70-79
80-89
90-99
6
9
12
15
8
Solution: First we need to find the class marks(midpoints) and then we use the following formula
:[ ]x. f
x
n
=
,
wherex : is the midpoint or class mark, andf :is the frequency
n :is the number of data points
Class Frequency
f
Class marks
x x f.
50-59
60-69
70-79
80-89
90-99
6
9
12
15
8
54.5
64.5
74.5
84.5
94.5
327
580.5
894
1267.5
756
n f =50 x f. =3825
So the mean is[ ] 3825
76 550
x. fx
n.= = =
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Example11B: Find the mean for the following group dataClass Frequency
00-04
05-09
10-14
15-19
20-24
25-29
4
10
12
20
8
6
Weighted Average (Mean): The formula in above is also called weighted average or weightedmean. It can also be written as follows:
[ ]x ww.x=
where w is weight andx is the score.
Example12: Find the GPA of John who has the following courses with the corresponding units
and grades.
English 5 units with the grade of AMath 3 units with the grade of FSpanish 2 units with the grade of D
Solution: In this problem, x will be the value of the grades and w is the number of units,
[ ] [ ] [ ] [ ]4 0 15 3 20 0 2 222 2
13 0 0
2
15 2
.x . . .x
w
w.
+ + + += = = = =
+ +
.
Example13: A teacher is teaching 3 classes: There are 30 students in the first Class with the
average of 70 on the final exam. The second class has 40 students with the average of 60 on the final
exam. The 3rdclass has 20 students with the average of 80 on the final exam. Find the weighted
average of the three classes combined together.Solution: Let x be the average of and w be the number of students.
[ ] 70 30 60 40 80 20 2100 2400 1600 610067 8
30 40 20 90 90
( ) ( ) ( ).xx
w
w.
+ + + += = =
+ +=
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MMeeaassuurreess ooffVVaarriiaattiioonn
RRaannggee VVaarriiaannccee SSttaannddaarrdd DDeevviiaattiioonn
The RRaannggee: is defined to be the highest value minus the lowest value in the data set
The Variance: is defined by the following:
Sample:( )
2
2 1
1
n
ii
x xs
n
= =
or( )
2
2
2
1
x
x nsn
=
(short cut formula of the sample
variance).
Population:
( )2
2 1
N
i
i
x
N
=
=
, or 22
2
= x x
N
N
d i
(short cut formula of the sample
variance).
Standard deviation: is the positive square root of the variance.
Standard deviation = Variance
Sample:
( )21
1
n
i
i
x x
sn
=
=
, and
Population:
( )21
N
i
i
x
N
=
=
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Example14A: Find the range, variance, and the standard deviation of the following dataset.
3, 0, 7, 5, 15.
Solution:
o Range: Largest- Smallest = 15-0=15
o Variance: If we use the( )
2
2 1
1
n
i
i
x x
sn
=
=
, first we need to find the sample mean x .
So x = + + + + = =3 0 7 5 155
30
56 , then we substitute in the above formula and we get
s2
2 2 2 2 23 6 0 6 7 6 5 6 15 6
5 1= + + + +
b g b g b g b g b g
,
s 22 2 2 2 2
3 6 1 1 9
5 1= + + + +
b g b g b g b g b gs 2
9 36 1 1 81
5 1= + + + + ,
s 2128
432= , So the variance is s 2 32 .
x x x ( ) 2x x 3
0
7
5
15
3-6=-3
0-6=-6
7-6=1
5-6=-1
15-6=9
9
36
1
1
81
( )x x =0 ( ) 2x x =128
( )
2
2 1 1281
128 325 1 4
n
ii x xs
n
=
= = = =
o Standard deviation: As we know the standard deviation is positive square root ofvariance. standard deviation = Variance = 32 5 66.
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Elementary Statistics Dr. Ghamsary Chapter 2 Page 24
24
But if we use the short cut formula
( )22
2
1
xx
nsn
=
, first we need to find their sum, x , and their sum of squares,
2x .
3 0 5 05 1 37x + + + += = 2 2 2 2 22 3083 0 7 5 15 9 0 49 25 225x = + + + + = + + + + = then we have
( )2
2
2
1
x
x nsn
=
( )2
303
5
0
1
8 5=
=308308 1900 80 1285 32
4 4 4
= = = , which is exactly the
same as above.
----------------------------------------------------------------------------------
Example14B: Find the range, variance, and the standard deviation of the following data set.
10, 17, 13, 12, 15, 18, 10, 17, 14, 16
28, 22, 17, 23, 12, 15, 28, 10, 20, 35
Solution:
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Example15A: Find the standard deviation for the following group dataClass Frequency
50-59
60-69
70-79
80-89
90-99
6
9
12
15
8
Solution: First will modify the above formula for the variance. But first we need to find the class marks
(midpoints) and then we use the following formula
s
x x f
n
i
2
2
1=
b g .or
( )2
2
2
1
x f
n
x
ns
=
where
x : is the midpoint or class mark
f : is the frequency
n : is the number of data points
We already know the mean[ ] 3865
5
76 5
0
x. fx
n
.= = =
Classf x x f. x xi b g2 x x fi b g2 .
50-59
60-69
70-79
80-89
90-99
6
9
12
15
8
54.5
64.5
74.5
84.5
94.5
327
580.5
894
1267.5
756
(54.5-76.5)2=484
(64.5-76.5)2=144
(74.5-76.5)2=4
(84.5-76.5)2=64
(94.5-76.5)2=324
2904
1296
48
960
2592
n f =50 x.=3825
x x fi b g2 .= 7800
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After substitution insx x f
n
i2
2
1= b g .
we get2
50 1
7800159 18s .= =
, and hence the
standard deviation will be 159 1 12 68.s .=
If we use the short cut formula
( )2
2
2
1
x ff
n
x
ns
=
, we need the following table.
Classf x x f.
x f2 .
50-59
60-69
70-79
80-89
90-99
6
9
12
15
8
54.5
64.5
74.5
84.5
94.5
327
580.5
894
1267.5
756
(54.5)2.6 =17821.5
(64.5)2..9 =37442.25
(74.5)2.12 =66603
(84.5)2.15=107103.8
(94.5)2.8 =71442
n f =50 x f.=3825
x f2 .
=300412.5
( )2
2
38230
50412
0
5
504 1
.
s
= =
14630625300412 5
3004125 5 292612 55039 49
.. .
= =
7159 18
8
49
00.= and hence the standard deviation will be 159 1 12 68.s .= , which the same as
the above result.
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Example15B: Find the standard deviation for the following group dataClass Frequency
00-04
05-09
10-14
15-19
20-24
25-29
4
10
12
20
8
6
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Question 1. What will happen to the mean, median, mode, range, and standard deviation if we add
a fix number, c, to all values in the data set?
Answer. The mean, median, and mode will increase by c units, but the range, and standard
deviation will not change.
Question 2. What will happen to the mean, median, mode, range, and standard deviation if we
subtract a fix number, c, from all values in the data set?
Answer. The mean, median, and mode will decrease by c units, but the range, and standard
deviation will not change.
Question 3. What will happen to the mean, median, mode, range, and standard deviation if we
multiply a fix number, c, to all values in the data set?
Answer. The mean, median, and mode will be multiplied by c units, so does to the range, and
standard deviation.
Example 16:
X X+7 X-7 X*7
15
13
15
15
22
15+7=22
16+7=23
15+7=22
15+7=22
22+7=29
15-7=8
16-7=9
15-7=8
15-7=8
22-7=15
15*7=105
16*7=112
15*7=105
15*7=105
22*7=154
Mean
Median
Mode
Range
Sd
16
15
15
9
3.46
16+7=23
15+7=22
15+7=22
9
3.46
16-7=9
15-7=8
15-7=8
9
3.46
16*7=112
15*7=105
15*7=105
9*7=63
3.46*7=24.22
In general ifY aX b+ , then we have Mean of Y = a. [Mean of X]+b or y ax b+ Standard deviation of Y = |a| [standard deviation of X], S a Sy X
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Empirical RuleIf the distribution of a data is bell shape or normal, then
Approximately 68% of scores are one standard deviation away from the mean. They fall in theinterval x s1 , x s1 .
Approximately 95% of scores are two standard deviation away from the mean. They fall in theinterval x s2 , x s2 .
Approximately 99.7% of scores are two standard deviation away from the mean. They fall in theintervalx s3 , x s3 .
Example17. Suppose the IQ scores are normally distributed with the mean of = 100 andstandard deviation of = 15 . Then by the empirical rule Approximately 68% of scores are in the interval 100-15, to100+15 or85 to 115. Approximately 95% of scores are in the interval 100-2(15), to100+2(15) or70 to 130. Approximately 99.7% of scores are in the interval 100-3(15), to100+3(15) or55 to 145.
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CCooeeffffiicciieenntt ooffVVaarriiaattiioonn
The ccooeeffffiicciieenntt ooffvvaarriiaattiioonn is defined to be the standard deviation divided by the mean.
Coefficient of variation (CV) =s
x. Ifx is 0 or close to 0, then this measure shall not be used.
Normally this measure is used in the case we have 2 or more groups of data with different units.
Example18.
Class A Mean =129, and standard deviation= 11 CV=11/129=.085 or 8.5%
Class B Mean =150, and standard deviation= 25 CV=25/150=.167 or 16.7%
Class C Mean =60, and standard deviation= 15 CV=15/60 = .25 or 25.0%
The class C has the greatest relative variation.
MMeeaassuurreess ooffPPoossiittiioonn
Standard Scoresz
x x
s= or z x= ,
where, x or is the mean s or is the standard deviation.This value, z, measures the deviation from the mean in number of standard deviation which is also has
no unit.
Example19. Suppose John is taking 3 classes with the following scores. In which class has he
better score?
Class A English test score = 145Mean =129, and standard deviation= 11
Z=(145-129)/11 =1.45
Class B Physics test score = 190Mean =150, and standard deviation= 25
Z=(190-150)/25 = 1.60
Class C Statistics test score = 88Mean =60, and standard deviation= 15 Z=(88-60)/15=1.87
So his score in class C is higher relatively.
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PPeerrcceennttiilleessThe percentile corresponding to a given score (X) is denoted by P and it is given by the following
formula
#of scores less thanP .100total number of scores
x=
Example20. John has the score of 88 in a class of 20 students. Find the percentile rank of a his
score.
81, 65, 75, 76, 78, 62, 63, 65, 70, 90,61, 75, 76, 79, 58, 88, 82, 95, 90, 67.
Solution: In any problem of finding percentile, we must sort the data set from smallest to largest.
58, 61, 62, 63, 65, 65, 67, 70, 75, 75
76, 76, 78, 79, 81, 82,88, 90, 90, 95.
P#of scores less thanx
total number of scores.100 =
16
20= =.100 80
So johns score has 80th percentile, which means 80% of all scores are below 88.
FFiinnddiinngg tthhee SSccoorree CCoorrrreessppoonnddiinngg ttoo aa GGiivveenn PPeerrcceennttiillee
Example21. In data set of example 20, find the score corresponding 12th percentile.
Solution:
Step1: Make sure data is sorted
58, 61, 62, 63, 65, 65, 67, 70, 75, 75
76, 76, 78, 79, 81, 82, 88, 90, 90, 95
Step2: Compute the L = p% of n., where L is the location for the score.
In this example L=12%of 20=0.12(20)=2.4 or 3.
Step3: Go to the data set and pick the score at the 3rdposition which is 62.
It is usually written as P12=62
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NNoottee::
If L is not a whole number, round up to the next whole number. IfLLis a whole number, use the score as the average of Lth and((LL++11))tthh location score.Example22. In data set of example 20, find the score corresponding 40th percentile.
Step1: as before58, 61, 62, 63, 65, 65, 67, 70, 75, 7576, 76, 78, 79, 81, 82, 88, 90, 90, 95
Step2: L =40% of 20= 0.40(20)=8 which is a whole number so we are going to pick the average of 8th
and 9th scores.
Step3: 8th score is 70
9th score is 75 and their average is (70+75)/2=72.5. So P40=72.5.
DDeecciilleess:: divide the data set into 10 groups.
D1=10th percentile which the same as P10
D2=20th percentile which the same as P20
.
D9=90th percentile which the same as P90
QQuuaarrttiilleess::divide the data set into 4 groups.
Q1=First quartile or 25th percentile which the same as P25
Q2=second quartile or 50th percentile which the same as P50 . This is also median
Q3=third quartile or 75th percentile which the same as P75
IInntteerr--QQuuaarrttiillee RRaannggee ((IIQQRR))::iiss tthhee ddiiffffeerreennccee bbeettwweeeenn 33rrdd aanndd11sstt qquuaarrttiilleess aannddiitt iiss ddeennootteeddbbyy
IIQQRRaannddiitt iiss ddeeffiinneeddbbyyIIQQRR ==QQ33 QQ11.
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Example23. In data set of example 20, find the score corresponding toD2Q1Q3 IQR
Outlier: Anoouuttlliieerris an extremely high or an extremely low data value, To check for outlier we
compute Q1-1.5(IQR) andQ3+1.5(IQR), then if
The suspected score is below Q1-1.5(IQR) orThe suspected score is above Q3+1.5(IQR)
Then the score is said to be an outlier.
Example24. Is there any outlier in the following data set?
55 46 46 41
55 49 51 41
36 41 86 5352 47 44 51
51 61 51 48
Sorted Data
36 41 41 41 44 46 46 47 48 49
51 51 51 51 52 53 55 55 61 86
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Five commonly used Statistics: The five numbers in any data set that is used frequentlyare
Minimum, Q1, Q2, Q3, Maximum
Box pplloott oorrbox-and-whisker plot:: is another graphical representation of any data set. We
use the five commonly used statistics to graph the box plot. The box plot can provide answers to the
following questions
o Is a factor significant?o Does the location differ between subgroups?o Does the variation differ between subgroups?o Are there any outliers?
Example25. In data set of example 20, find the 5 common statistics.
58, 61, 62, 63, 65, 65, 67, 70, 75, 7576, 76, 78, 79, 81, 82, 88, 90, 90, 95
1. Minimum: is58
2. Q1: L= 25% of 20 =.25(20) = 5. Since this is a whole number we use the average of 5th and 6th
observation. In above ordered data set we have5th score is 65
6th score is 65
their average is also 65. SO Q1=65.
3. Q2: L= 50% of 20 =0.50(20) =10. Again since this is a whole number we use the average of 10th
and 11th observation. In above ordered data set we have
10th score is 75
11th score is 76
their average is (75+76)/2=75.5 SO Q2=75.5.
4. Q3: L= 75% of 20 =0.75(20) =15. This is a whole number we use the average of 15th and 16th
observation. In above ordered data set we have
15th score is 81
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16th score is 82
their average is (81+82)/2=81.5 SO Q3=81.5.
5. Maximum: is 95.
So the five statistics are 58, 65, 75.5, 81.5, and95.
C1
100
90
80
70
60
Boxplot of C1
Example26 In data set of example 24, find the 5 common statistics.
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Example27. In data set below use computer to find the descriptive statistics and plot allappropriate charts for all variables that was discussed so far.
Test1 Sex Grade Test1 Sex Grade
76 1 C 76 1 C
62 1 D 59 1 F
68 1 D 92 1 A69 1 D 93 1 A
79 0 C 88 0 B
90 0 A 86 0 B
79 1 C 66 0 D
86 1 B 81 1 B
52 0 F 85 0 B
97 1 A 85 0 B
78 1 C 70 1 C
55 1 F 55 1 F
96 1 A 62 1 D
89 1 B 80 1 B
73 0 C 60 1 D
66 0 D 80 1 B
88 1 B 72 1 C
92 0 A 82 0 B
94 1 A 86 1 B
50 1 F 99 1 A
71 0 C 63 1 D
89 0 B 75 1 C
78 1 C 83 1 B
88 0 B 78 0 C
58 1 F 61 1 D
Descriptive Statistics: Test1
Variable Sex N N* Mean SE Mean StDev Minimum Q1 Median Q3 MaximumTest1 Female 34 0 75.59 2.36 13.76 50.00 62.00 77.00 86.50 99.00
Male 16 0 79.38 2.77 11.10 52.00 71.50 83.50 88.00 92.00
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Sex
Test1
MaleFemale
100
90
80
70
60
50
Boxplot of Test1 vs Sex