statistics xi ipa

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YPPI 1 Senior High School Of Surabaya Statistics_ 0




A. Statistics.

Statistics is the art of solving problems and answering questions by collecting and analysing

data.Statistic are used by government, businesses and sports organizations so that they can make

informed decisions when they are providing services such as health, transport and commerce

or developing new tactics. They also interested in using statistics as a means of analyzing the

effect of certain changes that may have been made, or in predicting what may happen in the

future.

There are two kinds of statistics :

a. Descriptive Statistics : procedures used to organize and present data in a convenient,

useable and communicable form.

b. Inferential Statistics : procedures employed to arrive at broader generalizations or

inferences sample data to populations.

Statistic is a number describing a sample characteristic. Results from the manipulation of

sample data according to certain specified procedures.

The next discussion we will learn about the descriptive statistics.

Populations and sample

a. Populations : a complete set of actual or potential observations

b. Sample : a subset of the population selected according to some scheme.

Data is characteristics or numbers are collected by observations.

There are two groups of data :

a. Quantitative data consists of :- Discrete data ( data cacahan)

- Continue data ( data ukuran )

b. Qualitative data consists of :

- Nominal data

- Ordinal data

B. Descriptive Statistics

Measures of central tendency are known as the mean , median and mode.

1. Mean ( rataan )

The mean is the average of all observations in a set of data

a. Ungrouped data ( data tunggal )

Kompetensi dasar : Kemampuan menyajikan dan

menafsirkan data dalam

bentuk tabel dan diagram;menghitung dan menafsirkan

ukuran pemusatan, ukuran

letak dan ukuran penyebarandata.

Mean = scoreof number

scoresall of sum




Symbolically this is written :

2. Median

The median is the middle value of any set of data arranged in numerical order (all of

scores have been placed in order of size from smallest to largest ). In the set of n

numbers, the median is located at the2

1+nth score. The median is :

a. The middle score for an odd number of scores arranged in numerical order

b. The average of the two middle scores for an even number of scores arranged innumerical order.

3. Mode ( Modus )

The mode is the score which occurs most often in a set of data

Example 1 : For the data set 6, 2 , 4, 3, 4, 5, 4, 5 find the :

a. mean b. median c. mode

Solution :

a.n

x x

i∑=

125,4

8

33

8

54543426

=

=

+++++++=

b. Arrange the data in numerical order

2 3 4 4 4 5 5 6

Use2

1+nto know the locate of the median.

5,4

2

18

=

+=

The median as the 4,5th

score : that is, between the 4th

and 5th

2 3 4 4 4 5 5 6

Me =4

2

44

=

+=

c. Mode ( Mo )

2 3 4 4 4 5 5 6

Mo = 4

n

x

x

n

i

i∑=

=1

x = mean ( rataan)

i x = nth

score (data ke – n)

∑=

n

ii

x1

= the sum of all scores

n = number of score




A five – number summary which includes the lowest values of the set, the lower

quartile , the median, the upper quartile, and the highest value of the set; that is Xmin, Q1,

Me, Q3 , Xmax.

Pay attention the diagram below :

Xmin : the lowest score

Xmax : the highest score

Q1 : the lower quartile ( the 25th

percentile)

Me : the median

Q3 : the upper quartile ( the 75th

percentile)

- Locate and calculate Q1, the median of the lower half of the data- Locate and calculate Q3, the median of the upper half of the data.

Measures of spread

1. Range ( Jangkauan )

It is defined as the difference between the highest and the lowest values in the set of data.

1. Interquartile range ( Jangkauan antar kuartil / Hamparan )

Example 2 : For the data set 4, 5, 5, 6, 9, 10, 12, 14, 15

Find : a. Range

b. Interquartile range

Solution : 4 5 5 6 9 10 12 14 15

a. R = Xmax – Xmin

= 15 – 4

= 11

b. Q3 =

2

1412 +Q1 =

2

55 +

= 13 = 5

H = Q3 – Q1

= 13 – 5

= 8

MeXmin Q1 Q3 Xmax

R = Xmax – Xmin

H = Q3 – Q1

Me Q3 Q1




C. Presenting and Interpreting Data

Graphs and frequency table are often used to represent data.

1. Displaying data in frequency table

Example 3 : a survey was conducted on the type of fuel used by 50 randomly selected

vehicles. The variable of “ fuel type” , is divided by four categories : unleaded,

lead replacement, LPG, and Diesel. The data has been tallied and organised in

the given frequency table.

Table.1

2. Displying data in graph

a. Bar Graph

Bar graph is divided by two :

a. Vertical column graph

b. Horizontal bar chart

Based on table.1, the data in vertical column graph and horizontal bar graph

Pic.1 Pic. 2

Exercise 1.11. Find Xmin, Q1, Me, Q3 , and Xmax for each of the following sets of data

a. 4, 3, 9, 12, 8, 17, 2, 16

b. 49.5, 13.7, 12.3, 36.5, 89.4, 27.8, 53.4, 66.8, 21.21

c. 19, 25, 72, 44, 68, 24, 51, 59, 36d. 70, 68, 71, 68, 66, 73, 65, 74, 65, 64, 78, 79, 61, 81, 60, 97, 44, 64, 83, 56

2. Find the mean, median, and mode of the following sets of data

a. 41, 52, 66, 86, 91, 65, 86, 88, 41, 62, 42, 59, 72, 99, 53, 69, 87, 93, 64, 44, 64,

42, 92, 54, 78, 86, 92, 100, 79, 47.

b. 48, 476, 91, 43, 39, 119, 33, 139, 493, 398, 547, 128, 708, 61, 25, 55, 16, 55,

30, 34,56, 51, 39, 134, 21, 26, 24.

c. 85, 52, 47, 35, 39, 62, 83, 52, 75, 95, 72, 65, 80, 78, 76, 56, 68, 85, 92, 43

3. Find the range and interquartile range of the sets of data from number 2.




b. Line graph ( diagram garis )

Diagram garis biasanya digunakan untuk menyajikan data statistik yang diperoleh

berdasarkan pengamatan dari waktu ke waktu.

Contoh 4 : Fluktuasi nilai tukar rupiah terhadap dolar AS dari tanggal 18 februari 2008

sampai dengan tanggal 22 Februari

Table 2

Penyelesaian :

c. Pie Chart

d. Stem and Leaf Diagrams ( diagram batang daun )

For data that we want to understand how it looks without losing the individual data

points, we use a stem and leaf diagram. To construct a stem and leaf diagram, we put

the first digit or more (the stem) on the left and that digit's corresponding list (leaf) on

the right. We can also have the high and low of the digit. If we want to compare two

Based on table 1 we can represent

the percentage the data in pie

chart .Example :

Unleaded = 10050

28 x % = 56%

Or we can show the angle of the

data.

Example :

Unleaded =00

6,20136050

28= x




data sets we can draw the digits in the middle, the first set of leaves on the right, and the

second set of leaves on the left. This is useful for comparing two data sets.

A comparative stem and leaf diagram is often used. The middle represents the stems,

and the left and right sides are the leaves of each of the two data sets.

Example 5 :

A computer retailer collected data on the number of computers sold during 20

consecutive Saturdays during the year. The results are as follows:

12, 14, 14, 17, 21, 24, 24, 25, 25, 26,

26, 27, 29, 31, 34, 35, 36, 39, 40, 42,

42, 45, 46, 47, 49, 49, 56, 59, 62.

We can put this data into a stem and leaf diagram as shown below. The first digit

represents the stem and the second digit represents the leaf. The stem is written on theleft hand side (once per value) and the leaf is written on the right hand side next to the

corresponding stem.

1| 2 4 4 7

2| 1 4 4 5 5 6 6 7 9

3| 1 4 5 6 9

4| 0 2 2 5 6 7 9 9

5| 6 9

6| 2

Exercise 1.2

1. The following data is temperature of Budi’s body in ten days

Nth

day 1 2 3 4 5 6 7 8 9 10

Temp (0

C) 35 36 37 36 37,5 38 37 38 38,5 37

a. Draw a line graph of data

b. Which day that is the lowest temperature

c. Which day that is the highest temperature2. The data set below is the test scores for a Maths test for 20 students.

85 52 47 35 39 62 83 52 75 95

72 65 80 78 76 56 68 85 92 43

a. Sketch the bar diagram of the data above.

b. What is the highest mark scored for the test ?

c. What is the lowest mark scored for the test ?

3. The initial weights of students ( in kg ) in XI IPS class were :

Find a five – number summary




4.

D. Distribution Frequency Tables

1. Ungrouped data (data tunggal)

For the set data :5, 4, 6, 7, 8, 8, 6, 4, 8, 6, 4, 6, 6, 7, 5, 5, 3, 4, 6, 6, 8, 7, 8, 7, 5, 4, 9, 10, 5,

6, 7, 6, 4, 5, 7, 7, 4, 8, 7, 6 . The data has been tallied and organized in the given frequency

table.

Table 3 Score Tally Frequency

3

4

5

6

7

8

9

10

|

|||| ||

|||| |

|||| ||||

|||| |||

|||| |

|

|

1

7

6

10

8

6

1

1

2. Grouped data (data bergolong / kelompok )

Given the data of matematics exam of 32 students

39, 47, 57, 43, 59, 55, 58, 51

45, 52, 44, 54, 48, 53, 47, 48

50, 63, 43, 36, 43, 51, 40, 54

41, 49, 60, 51, 44, 34, 40, 53

The steps to organize the data in frequency table are :

a. Calculate the range of data

R = Xmax – Xmin

= 63 – 34

= 29

b. Numbers of class Interval

Use Sturges rule , numbers of class (C) = 1 + 3,3 log n, n = number of data

= 1 + 3,3 log 32

C = 5, 9665




Thus the number of class is 5 or 6

c. Length of class interval ( l )

l =C

R

=6

29= 4, 83 ( 4 or 5 )

d. Distribution frequency table of grouped data

Table 4

Each group is called a class

The size of grouping is called the class interval.

Score Tally Frequency

34 – 38 ll 2

39 – 43 llll ll 7

44 – 48 llll ll 7

49 – 53 llll lll 854 – 58 llll 5

59 – 63 lll 3

e. Based on distribution frequency table above

34, 39, 44, 49, 54, 59 may be called the lower boundary.

38, 43, 48, 53, 58, 63 may be called upper boundary

The lower limit 2nd

class interval : halfway between 38 and 39, then the lower limit of

each class interval are 33.5, 38.5, 43.5, 48.5, 53.5, 58.5.

The upper limit of each class interval are 38.5, 43.5, 48.5, 53.5, 58.5, 63.5.

3. Distribution cumulative frequency table

Frequency tables may include other columns which show relative frequency and

cumulative frequency.

Relative frequency of an event is the frequency of that event expressed as a fraction ( or

decimal equivalent ) of the total frequency.

Cumulative frequency of an event is the accumulation ( sum ) of frequencies up to and

including that event.

Table 5

Score Frequency Freq. relative (%) Freq.Cum ≤ Freq.Cum ≥

34 – 38 2 6,25 2 32

39 – 43 7 21,875 9 30

44 – 48 7 21,875 16 23

49 – 53 8 25 24 16

54 – 58 5 15,625 29 8

59 – 63 3 9,375 32 3

Relative frequency ( Fr )

Fr =dataof number the

classn Freq th

x 100%




4. Cumulative Frequency Graph (Polygon)

A cumulative frequency graph ( polygon ) is used to represent cumulative frequencies.

We plot the cumulative frequencies against the upper end points of each class interval.

Example 6 : Draw a cumulative frequency for the steel rod data

5. HistogramBased on the steel rod data example 5 above , it can be represented diagrammatically by a

histogram.

6. Frequency Polygon

A frequency polygon is a line graph which, like the histogram, gives a good visual

appreciatation the shape of the frequency distribution.

Instead of drawing the bars, the midpoint of each bar is found and is used to represent the

whole interval. These points are then joined by straight lines.

The following diagram is the polygon frequency based on the steel rod data example 5

above .




Exercise 1.3

1. Given the following data :

80 66 74 74 70 71 78 74 72 67

72 73 73 72 75 74 74 74 72 72

66 75 74 73 74 72 79 71 75 75

78 69 71 70 79 80 75 76 68 68

Represent the data into :

a. Ungrouped Frequency Distribution tables

b. Grouped Frequency Distribution tables using classes : 65 – 67 , 68 – 70 , 71 – 73 ,

74 – 76, 77 – 79 , 80 – 82 .

2.

Height (m) Frequency

119 – 127 3

128 – 136 6137 – 145 10

146 – 154 11

155 – 163 5

164 – 172 3

173 – 181 2

3. Given the following data :

a. Organise the data into grouped data

b. How many class interval that happened ?

c. Mentioned the lower and upper limit of each class

d. Mentioned the midpoint

Represent the tables beside into freq

cumulative and relative tables, and then

construct :a. histogram

b. polygon

c. ogive




E. Summarising The Data

Measures of the middle of a Distribution

There are three commonly used measures for the middle of a distribution; the mean, the

mode, and the median.

a. Ungrouped data in distribution frequency table.

1. Mean

or

2. Mode (s)

3. Q1, Me, and Q3

Example 7 .

Score (xi) Frequency (f i) xi f i

4 5 20

5 2 10

6 4 24

7 3 21

8 2 16

16=∑ i f ∑ = 91ii f x

Solution :

a. 69,516

91

1

1===

∑

∑

=

=

n

i

i

n

i

ii

f

f x

x b. Mo = 4

c. n = 16 (even )

Q1 = 5,4

4

216

4

2 x x x n == ++ , x4 = 4 and x5 = 4

Q1 = 42

44

2

54 =+

=+ x x

Me = 5,8

2

116

2

1 x x xn == ++ , x8 = 6 and x9 = 6

Find :a. Mean

b. Mode

c. Q1, Me, and Q3

n

nn

f f f f

f x f x f x f x x

++++

++++=

...

...

321

332211

∑

∑

=

==

n

i

i

n

i

ii

f

f x

x

1

1

a. If n is an odd number of the data

Q1 =

4

1+n x Me =

2

1+n x Q3 =

4

)1(3 +n x

b. If n is an even number of the data

Q1 =

4

2+n x Me =

2

1+n x Q3 =

4

23 +n x




Me = 62

66

2

98 =+

=+ x x

Q3 = 5,12

4

216.3

4

23 x x x n == ++ , x12 = 7 and x13 = 7

Q3 = 72

77

2

1312 =+

=+ x x

b. Grouped data

1. Mean

i x : midpoint

The midpoint of a class interval is the average of its endpoints

Example 8:

Score Frequencyi x fi xi

34 – 38 2 36 72

39 – 43 7 41 287

44 – 48 7 46 322

49 – 53 8 51 408

54 – 58 5 56 280

59 – 63 3 61 183

∑i

f = 32

∑= 1552

ii f x

5,4832

1552

1

1===

∑

∑

=

=

n

i

i

n

i

ii

f

f x

x or

, Take any i x as s x and calculate id = i x – s x

∑

∑

=

==n

i

i

n

i

ii

f

f x

x

1

1

∑

∑

=

=+=

n

i

i

n

i

ii

s

f

f d

x x

1

1




Score Frequencyi x id = i x – s x fid i

34 – 38 2 36 -15 -30

39 – 43 7 41 -10 -70

44 – 48 7 46 -5 -35

49 – 53 8 51 = x s 0 0

54 – 58 5 56 5 2559 – 63 3 61 10 30

∑ i f = 32 ∑ −= 80ii f d

5,4832

8051

1

1=

−+=+=

∑

∑

=

=

n

i

i

n

i

ii

s

f

f d

x x

2. Mode

The class of mode is the class of highest frequency

L = Lower limit mode’s class

d1 = the difference freq of mode’s class and the class before mode’s class

d2 = the difference freq of mode’s class and the class after mode’s class

l = the length of class interval

Example 9 :

Score Frequency

34 – 38 2

39 – 43 744 – 48 7

49 – 53 8 Mode

54 – 58 5

59 – 63 3

∑ i f = 32

The locate of mode is in the 4th

class interval

d1 = 8 – 7 = 1

d2 = 8 – 5 = 3l = 39 – 34 = 5

L = 49 – 0,5 = 48,5

l

21

1

d d

d LMo

++=

l

++=

21

1

d d

d LMo




= 5.31

15,48

++

= 48,5 + 1,25

= 49,75

3. Q1, Me = Q2, and Q3

iQ = ith

quartile, i = 1,2, 3Li = lower limit i

thclass quartile

n = number of the data

f k = cumulative freq before ith

class quartile

f Qi = freq ith

class quartile


Example 10 : Find Q1, Me = Q2, and Q3

Score Frequency Freq.Cum ≤

34 – 38 2 2

39 – 43 7 9 Q1

44 – 48 7 16 Me = Q2

49 – 53 8 24 Q3

54 – 58 5 29

59 – 63 3 32

∑ i f = 32

l

−+=

1

41

11

Q

k

f

f n LQ ,

4

1. n =

4

1. 32 = 8 (Q1 in the 2

ndclass intv)

= 38,5 + 57

28

−

= 38,5 + 4,28

= 42,47

l

−+=

2

42

22

Q

k

f

f n LQ ,

4

2. n =

4

1. 32 = 16 (Q2 in the 3

rdclass intv)

l

−+=

iQ

k i

ii f

f n LQ 4




= 43,5 + 57

916

−

= 43,5 + 5

= 48,5

l

−+=

2

43

33

Q

k

f

f n LQ ,

4

3. n =

4

3. 32 = 24 (Q3 in the 4

thclass intv)

= 48,5 + 58

1624

−

= 48,5 + 5= 53,5

4. Deciles

i D = ith

decile, i = 1,2,3, …,9

Li = lower limit ith

class decile



class decile

f Di = freq ith

class decile


5. Percentiles

i P = ith

Percentile, i = 1,2, 3, …, 99

Li = lower limit ith

class percentile



class percentile

f Qi = freq ith

class percentile


l

−+=

i D

k i

ii f

f n L D 10

l

−+=

i P

k i

ii f

f n L P 100




Exercise 1.4

Solve the following problems

1. Determine the mean of the following data

a. 5, 7, 9, 6, 4, 3, 2, 1

b.

Data 3 4 5 6 7 8 9

Frequency 4 5 7 8 12 3 1

2. The data set below is the test scores of Math test for 20 students

65 75 66 80 73 75 68 67 75 77

70 71 60 55 65 63 60 70 70 66

Determine the mean of the data above.

3. Find the mean of grouped data :

a.

The height(cm) Frequency

150 – 154 5

155 – 159 6

160 – 164 10

165 – 169 7

170 – 174 2

4. Determine the mean using xs = 57

The weight (kg) Frequency50 – 52 4

53 – 55 8

56 – 58 20

59 – 61 10

62 – 64 8

5. Determine the median of the following data :

a. 5, 5, 6, 4, 3, 7, 8, 9, 10, 6, 4, 3, 6, 8

b.

Data 5 6 7 8 9 10

Frequency 2 12 14 6 5 1c. d.

score Frequency

52 3

56 6

60 10

64 20

68 40

72 20

76 9

80 2

The weight

(kg)

Freq

45 – 47 2

48 – 50 6

51 – 53 8

54 – 56 15

57 – 59 10

60 – 62 7

63 – 65 2




Exercise 1.5

Solve the following problems

1. Determine the mode of the following data

a. 2, 4, 3, 6, 7, 8, 2, 6, 7, 5, 2, 1, 5

b. 8, 9, 5, 6, 8, 2, 1, 3, 4, 5

c.

Data (cm) 3,1 3,4 4,2 4,9 5,1 5,5 6,5

Frequency 4 6 12 15 7 3 2

d.

score Frequency

2 2

5 6

8 10

11 4

14 3

e.

The height(cm) Frequency

119 – 127 3

128 – 136 6

137 – 145 10

146 – 154 11

155 – 163 5164 – 172 3

173 – 181 2

2. Determine Q1, Q 2 , Q 3, and H of the following ungrouped data :

a. 2, 5, 4, 6, 3, 4, 8

b. 4, 9, 12, 6, 3, 11, 7, 2

c.

score Frequency

3 5

4 6

5 106 15

7 9

8 6

9 2

3. Determine Q1, Q 2 , Q 3, and H of the grouped data :

a. b.

Data Frequency

41 – 45 3

46 – 50 6

51 – 55 1056 – 60 12

61 – 65 5

66 – 70 4

The weight

(kg)

Frequency

50 – 52 4

53 – 55 856 – 58 20

59 – 61 10

62 – 64 8




F. The Absolut Deviation, Standard Deviation and Variance

a. Ungrouped data

1. The Absolut Deviation ( Simpangan absolute / simpangan rata-rata)

2. The Standard Deviation ( Simpangan baku )

3. Variance ( ragam )

b. Grouped data

1. The Absolut Deviation ( Simpangan absolute / simpangan rata-rata)

The absolute deviation is given by the formula

2. The Standard Deviation ( Simpangan baku )

The standard deviation measures the deviation between scores and the mean, is a

measure of the dispersal of the data. The differences between the scores and the mean

squared, anf the average of these squares is then found.The standard deviation is the

square root of this average.

, ( n < 30 ) , ( n 30≥ )( )n

x x f S

n

iii

∑= −=

1

2

( )

1

1

2

−

−= ∑=

n

x x f S

n

i

ii

SR = n

x x f n

i

ii∑=

−1

n

x x

S

n

i

i∑=

−

=1

2

2

)(

( )

n

x x

S

n

i

i∑=

−

=1

2

SR =n

x xn

i

i∑= −1




3. Variance ( ragam )

, ( n < 30 ) , ( n 30≥ )

4. Quartile deviation ( Simpangan Kuartil / Qd )

)(2

113 QQQd −=

Example 11 :

a. For the data 3, 4, 6, 8, 9, then find SR, S, S2.

Solution :

x = 6530

598643 ==++++

SR =n

x xn

i

i∑=

−1 =

5

6968666463 −+−+−+−+−

= 25

10

5

32023==

++++

n

x x

S

n

i

i∑=

−

= 1

2

2

)(

=( )

5

)69()68()66()64(63 22222−+−+−+−+−

= 2,55

26

5

94049==

++++

( )

n

x x

S

n

i

i∑=

−

=1

2

= 2S = =2,5 2,28

Example 12: Find SR, S, S2, x = 48,5

Score Frequencyi x l i x – x l ( i x – x )

2i f l i x – x l i f ( i x – x )

2

34 – 38 2 36 12,5 156,25 25 312,5

39 – 43 7 41 7,5 56,25 52,5 393,75

44 – 48 7 46 2,5 6,25 17,5 43,75

49 – 53 8 51 2,5 6,25 20 50

54 – 58 5 56 7,5 56,25 37,5 281,25

59 – 63 3 61 12,5 156,25 37,5 468,75

∑ i f = 32 ∑ − x x f ii

=190

∑ − 2)( x x f ii

= 1550

n

x x f

S

n

i

ii∑=

−

= 1

2

2

)(

1

)(1

2

2

−

−

=∑=

n

x x f

S

n

i

ii




a. c. S = 2S

= 50 = 7,07

=32

190

= 5,9375

b.

=31

1550= 50

Exercise 1.6

1. Determine SR of the following data

a. 6, 8, 11, 3, 2

b. 2, 4, 6, 2, 1

2. Determine S and S2

of the following data

a. 3, 11, 2, 8, 6

b. 4, 6, 5, 7, 3

3. Determine SR, S, and S2

of the following grouped data :

a.

Age Frequency

1 – 5 2

6 – 10 7

11 – 15 5

16 – 20 9

21 – 25 6

b.

Data Frequency

41 – 45 3

46 – 50 6

51 – 55 10

56 – 60 12

61 – 65 5

66 – 70 4

SR =n

x x f n

i

ii∑=

−1

n

x x f

S

n

i

ii∑=

−

= 1

2

2

)(

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