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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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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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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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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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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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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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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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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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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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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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


.


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

Elementary Stastistics

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