This Power Point media for students and teachers to study about Statictics

1. SSTTAATTIISSTTIICCSS
By: Tito Nurdiyanto
tnk_adhyastha@yahoo.com

2. DDiiffffeerreenncceess bbeettwweeeenn
QQuuaalliittaattiivvee aanndd QQuuaannttaattiivvee
DDaattaa
Qualitative Data Quantitative Data
Overview:
·Deals with descriptions.
·Data can be observed but not
measured.
·Colors, textures, smells, tastes,
appearance, beauty, etc.
·Qualitative → Quality
·Examples :
1.Robust arom
2.Frothy appearance
3.Strong taste
4.Cup
Overview:
·Deals with numbers.
·Data which can be measured.
·Length, height, area, volume, weight,
speed, time, temperature, humidity,
sound levels, cost, members, ages,
etc.
·Quantitative → Quantity
·Examples :
1.10 ounces of latte
2.Serving temperature 140 F
3.Serving cup 6 inches in height
4.Cost $ 6

3. PPlleeaassee wwrriittee iinn aa ppiieeccee ooff ppaappeerr
yyoouurrss::
1. AGE
2. SUBJECT THAT YOU LIKE
3. HEIGHT
4. WEIGHT

4. LLEETTSS SSEEEE TTHHEE RREESSUULLTT
AGE
13 YEARS OLD : ………0……… PERSON
14 YEARS OLD : ………0……… PERSON
15 YEARS OLD : ………6…… PERSON
16 YEARS OLD : ………17…… PERSON
17 YEARS OLD : ………1…… PERSON

5. LLeettss wwee mmaakkee aa ddiissppllaayy ooff yyoouurr
ddaattaa!!!!!!
WWhhaatt kkiinndd ooff ddiissppllaayy tthhaatt wwee
wwiillll mmaaddee????????????????????????????????????
WWee ccaann ddiissppllaayy uussiinngg BBAARR
CCHHAARRTT aanndd PPIIEE CCHHAARRTT

6. Reading aanndd RReepprreesseennttiinngg
DDaattaa There are many ways to presenting data, here they are

7. PPiiccttooggrraamm

8. PPiiee cchhaarrtt
Mohammed asked his class about their favorite
flavour of ice cream.
Vanilla 12
Strawberry 6
Chocolate 5
Other 7
Their answers
were:

9. He decided to draw a pie cchhaarrtt ooff hhiiss rreessuullttss
Flavour Number of
pupils
Working Angle Percentag
e
Vanilla 12
Strawberry 6
Chocolate 5
Other 7
--
12 360
30
´ o
6 360
30
´ o
5 360
30
´ o
7 360
30
´ o
30
144o
72o
60o
84o
360o

10. Use the table to help you draw a pie chart. Don’t forget the
key
Vanilla
Strawberry
Chocolate
Other
144o
72o
84o
60o

11. Vertical Bar
Chart
Compound Bar
Chart
Horizontal Bar
Chart
BBaarr CChhaarrtt

12. LLiinnee DDiiaaggrraamm
It is usual to present data that you can find in newspaper,
magazine and the others media. You right, it is useful in the
fields of statistics and science, are one of the most common
tools used to present data. So, why line graph popular? Line
graph are more popular than all graph because their visual
characteristics reveal data trends clearly and these graph are
easy to create. Although they do not present specific data as
well as tables do but line graphs are able to show relationships
more clearly than tables do. Here is the example line graph.

13. 40
35
30
25
20
15
10
5
0
0 2 4 6 8 10 12
Number of chocolate bars
Number of pupils
School A
School B
How many pupils from school A
ate 7 bars a week?
How many pupils from school B
ate 7 bars a week?
How many pupils from A ate no
chocolate?
Which set of pupils ate the most
chocolate?

14. HHiissttooggrraamm
Salary of number of employees a factory, shown the
diagram below.
•In a histogram, frequency is measured by the area of the
column.
•In a vertical bar graph, frequency is measured by the
height of the bar.

15. PPoollyyggoonn
Polygon is a graph formed by joining the
midpoints of histogram column tops. These
graphs are used only when depicting data from
the continuous variables shown on a histogram.
It is useful demonstrating continuity of the
variable being studied. Here is good example
polygon. Here is the polygon of salary the
number of employees in a factory.

16. DDoott DDiiaaggrraamm
The dot diagram displays information
using dot. Each dot represent a piece
of data along a scale with frequency
represented on the other scale.

17. QUARTILE (singular data)
• The median of a set of data separates the data into
two equal parts.
•The first quartile is the median of the lower part of
the data.
•The second quartile is another name for the median of
the entire set of data.
• The third quartile is the median of the upper part of
the data.
ö
æ
If n is odd, the median is the value.
n 1 x
If n is even, the median is halfway the
value and the following value
ö
÷ ÷ø
æ
ç çè
x +
x
n n +1
1
2 2 2
÷ ÷ø
ç çè
+
2

18. The Measure of Central
Tendency

19. EXAMPLE (QUARTILE)
Math test score 80, 75, 90, 95,
65, 65, 80, 85, 70, 100. I will
represent the data in a box and
whisker plot. Could you help
me?
Write the data in
numerical order and find
the first quartile, the
median, the third quartile,
the smallest value and the
largest value.
•median = 80
•first quartile = 70
•third quartile = 90
•smallest value = 65
largest value = 100

20. MODE
The mode is the most frequently occurring score. In other
word, the mode is simply the number which appears most
often. Sometimes mode is also called modal value
Example
The number of points scored in a
series of football games is listed
below. Which score occurred often?
6, 3, 9, 5, 13, 15, 9, 18, 9, 8
Answer:
Ordering the scores from least to greatest,
we get: 3, 5, 6, 8, 9, 9, 9, 13, 15, 18
Hi.., now we can see easier that the score
which occurs most often is 9

21. Remember!!
The mode of a set of data is the
value in the set that occurs most
often. A set of data can be bimodal.
It is also possible to have a set of
data with no mode
On a cold winter day in January, the temperature for 9 North
American cities is recorded in Fahrenheit. What is the mode of
these temperatures?
-9, 0, -2, 3, 5, -1, 10, 4, 6
Answer:
Ordering data from least to greatest, we get
-9, -2, -1, 0, 3, 4, 5, 6, 10
Since each value occurs only once in the data set, there is
no mode for this set of data.

22. MEDIAN
If n is odd, the median is the value.
ö
æ
n 1 x
If n is even, the median is halfway the
value and the following value
Example
÷ ÷ø
ç çè
+
2
ö
÷ ÷ø
æ
ç çè
x +
x
n n +1
2 2
A marathon race was completed by 5 participants.
What is the median of these times given in hours?
1.5 hr, 4.2 hr, 3.5 hr, 6.7 hr 5.0 hr
The number of data is odd and the total
number of data is 5 so that:
Median = X = X
=
4.2 1 ( 1 )
5
2
+
n

23. MEAN
Mean is the average of a series score. It is
meant that the sum of the values in the data
set divided by the number of values. So, we
can represent mean with this formula ;
Example
x x x x ...
x
1 2 3 n
f
x
n
n
i
i
n
å=
=
+ + + +
=
1
1. Find mean for the following data : 5, 3, 4, 6, 7
Answer :
5
x = x + x + x + x +
x
1 2 3 4 5
= + + + + = 25
=
5
5 3 4 6 7
5
x
n

24. Example
2. A class contain 20 men and 30 women. If men’s average
heightis 167 and women’s average height is 158 cm, then find
the average height students in that class.
Answer:
The average height =
161,6
- x
= cm +
cm (20) (167 ) (30) (158 ) =
+
20 30
x

25. DESIL
Datum yang membagi data terurut menjadi sepersuluh bagian
Menentukan Dm
•Hitung m(n+1) /10
• Jika hasilnya bulat misal r maka D= xm
m
• Jika hasilnya Ganjil ( bukan bulat) tapi terletak
antara r dan r+1 maka Dm = Xm + (m(n+1)/10 – r )
(Xr+1 – Xr )
x
Contoh : 7, 5, 6, 5, 3, 6, 4, 8, 2, 6, 8, 7 Tentukan D1
Peny: 2, 3, 4, 5, 5, 6, 6,6 , 7,7, 8,8
D1
= 1 (12 + 1) / 10 = 1,3
D1
= X1 + ( 1,3 – 1) (X2 – X1) = 2 + (0,3)(3-2) = 2 + 0,3 = 2,3

26. Example
3. A family must drive an average 250 miles per day to complete their
Vacation on time. On the first five days, they travel 220 miles, 300
miles, 210 miles, 275 miles and 240 miles. How many miles must
they travel on the sixth day in order to finish their vacation on time?
x
The sum of the first 5 days is
1,245 miles. Let
represent the number of miles travelled on the sixth
day. We get
i å=
= 1
n
f
x
n
i
250 = 1,245 + x Þ 1,245 + x = 250´ 6
6
1,245 + x =1,500
= -
x
1,500 1,245
=
255
x

27. 1. Construct a pie chart of data in one school that the total 100
students that are consist of 50 student like rap music, 25
students like alternative music, 13 students like rock and roll, 10
students like country music and 2 students like classical.
2. The monthly incomes of 234 workers in a factory are given in the table
below. Find the mean of monthly incomes of workers!
Income (x) in dollars 600 700 800 900 1000 1100 1200
Numbers of workers
30 45 75 36 24 18 6
(f)
Exercise
3. The average height of 50 men is 165,2 cm. If Mr. Susilo is
included, then the average height becomes 165,28 cm . Find the
height of Mr. Susilo.

28. Exercise
4. In a crash test, 11 cars were tested to determine what impact
speed was required to obtain minimal bumper damage. Find the
mode of the speeds given in miles per hour below.
15, 24, 18, 22, 16, 26, 22, 20, 18, 25, 27
5. The test score of 8 eleventh grade students are listed below.
Find the median. 82, 92, 75, 94, 85, 100, 89, 78

29. ANSWER KEY
4.
5.
Since both 18 and 22 occur twice, the modes are 18 and 22 miles
per hour. This data set is bimodal
Ordering the data from least to greatest, we get:
75, 78, 82, 85, 89, 92, 94, 100
Oops! The number of data is even. So that, the value of
median is:
Median
87
85 89
2
X X
2
1
2
= + =
+
=
n
n
Number of
CD’s
Frequency
0 – 4 10
5 – 9 7
10 – 14 2
15 – 19 4
20 – 24 4
25 – 29 1
30 – 34 2
Ririn asked his class how many CD’s they
owned.
Find mode of the following data
6.
.

30. ANSWER KEY
1.
Income (x)
in dollars
Numbers of
workers (f)
x
600 30 18000
700 45 31500
800 75 60000
900 36 32400
1000 24 24000
1100 18 19800
1200 6 7200
235 192.900
i i å x · f
2.
So, the mean of monthly
incomes of workers
824.35
192900
å
x f
1 = =
235
1
·
=
å
=
=
k
i
i
k
i
i i
f
x

31. GROUPED DATA
Grouped frequency Distribution
Arrange Cummulative Frequency
Distribution
Arrange Cummulative Frequency
Distribution
Mode
Mean
Lower Quartile, Median Quartile, Upper
Quartile
Varians
Standard Deviation

32. Grouped frequency Distribution
Before learning about Grouped Frequency Distribution. Lets see
the differences Singular Frequency Distribution below
Singular Frequency Distribution Grouped Frequency Distribution
The
Age
(xi )
Talley Frequency
fi
10-19 |||| | 6
20-29 ||| 3
30-39 |||| 4
40-49 ||| 3
50-59 |||| 5
60-69 |||| | 6
70-79 ||| 3
The
Age
(xi )
Talley Frequency
fi
10 ||||
|
6
20 ||| 3
30 |||| 4
40 ||| 3
50 |||| 5
60 ||||
|
6
70 ||| 3

33. Terminology in Grouped frequency
Distribution
1. Class : 1st Class 10-19 , 2nd Class 20-29 etc
2. Number of Class = 7
3. Batas kelas ... Batas bawah kelas pertama 10
Batas atas kelas pertama 19
4. Tepi Kelas
The
Age
(xi )
Frequency
fi
10-19 6
20-29 3
30-39 4
40-49 3
50-59 5
60-69 6
70-79 3
Tepi bawah = batas bawah - 0,5
Tepi atas = batas atas + 0,5
5. Panjang Kelas = tepi atas – tepi
bawah

34. MODE
It is necessary for us to learn how to find the mean value of group
data. In order to present a large set of data more clearly, it can be
sorted into groups (or classes).
If the data are representated in classes, then the mode can be found by
the formula.
c
æ
L d ÷ ÷ø
1
d d
ö
ç çè
+
= +
1 2
Mode 1
1 L
1 d
2 d

35. Example (grouped data)
This is the way to the data in classes, find the upper, lower class
boundary and width of each class
Mathematics marks of mathematics examination are
45 56 30 67 78 49 80 35 85 87
65 90 54 78 89 43 78 34 84 55
80 93 46 38 93 57 69 35 70 76

36. First it is very easy to you to Make a
grouped tally chart for this data.
Class Talley Frequency
30-39 |||| | 6
40-49 ||| 3
50-59 |||| 4
60-69 ||| 3
70-79 |||| 5
80-89 |||| | 6
90-100 ||| 3
Between two classes there is a gap, the midpoints of the gaps define the
class boundaries
39 40
39.5
49 50
49.5
The lower boundary 30-39 is 29.5 and the upper class boundary is of the
class is 39.5. The difference between the upper class boundary and the
lower class boundary (39.5-29.5=10) is called the class width. Well, could
you determine the lower boundary and the upper boundary for each
class?

37. Example (grouped data)
Now, lets begin. The example of mode for grouped data
Find mode of each group
1.
i xf
Mass
(x)
(Kg)
Mid
Point
Frequency
40-44 42 5
45-49 47 8
50-54 52 10
55-59 57 7
49.5 2 ö ÷ø
çè
Mode = + 5
= 51.5
æ
+
2 3

38. Mean grouped data
å
= = r
f x
å
=
i
i
r
i
i i
f
X
1
1
__
.
Example (grouped data)
mathematics test taken by 100 Senior High School students:
Find the mean of the marks for a mathematics test above!
Marks 51-60 61-70 71-80 81-90 91-100
Number of
25 15 30 22 8
pupils

39. SOlution
Marks Mid point of
marks
Number of pupils
51-60 55.5 25 1387.5
61-70 65.5 15 982.5
71-80 75.5 30 2250
81-90 85.5 22 1881
91-100 95.5 8 764
Total
= 100
=7265
( xi ) i f i i x · f
( ) i x ( ) i f
i
n
i
i f .x
1 å=
n
å=
i
i f
1
72.65
7265
å
=
n
f x
= 1 = =
å
100
.
=
1 0
i
n
i
i i
f
x

40. ( )
c
1
æ -
n f
f
Q L
ö
÷ ÷ ÷ ÷
ø
ç ç ç ç
è
= +
å
1
1
1 1
4
( )
c
1
æ -
n f
f
Q L
ö
÷ ÷ ÷ ÷
ø
ç ç ç ç
è
= +
å
2
2
2 2
2
( )
c
3
æ -
n f
f
Q L
ö
÷ ÷ ÷ ÷
ø
ç ç ç ç
è
= +
å
3
3
3 3
4
= i f frequency of quartile
= 1 Q
= 2 Q
class
lower quartile
median
upper quartile
= i L the lower boundary of quartile
= 3 Q
class
n = the number of data
(å ) = i f the number of frequency
cummulative before the
quartile class
c =
i = 1,2,3
the length of quartile class
And

41. Example
The following data represent the weights (in kg) of 50
students
Weight (kg) Frequency
40-49 5
50-59 14
60-69 16
70-79 12
80=89 3
Total 50

42. Weight (kg) Frequency Frequency
Cumulative
40-49 5 5
50-59 14 19
60-69 16 35
70-79 12 47
80=89 3 50
Total 50
•First Quartile/lower quartile
1 n = 1
=
(50) 12.5
4
4
( )
c
1
æ -
n f
f
Q L
ö
÷ ÷ ÷ ÷
ø
ç ç ç ç
è
= +
å
1
1
1 1
4
, ccorrespondence with class 50-59
Solution
æ -
( )
( )
50 5
4
49.5 12.5 5
54.86
10
ö
.10 49.5 7.5
14
ö 14
çè
.10
14
1
49.5
Q
1
Q
1
1
=
+ = ÷ø
= +æ -
÷ ÷ ÷ ÷
ø
ç ç ç ç
è
= +
Q

43. •Second Quartile/Median
2 n = 2
=
(50) 25
4
4
correspondence with class 60-69
( )
c
1
æ -
n f
f
Q L
ö
÷ ÷ ÷ ÷
ø
ç ç ç ç
è
= +
å
2
2
2 2
2
æ -
( )
( )
50 19
2
59.5 25 19
63.25
10
ö
.10 59.5 6
16
ö 16
çè
.10
16
1
59.5
Q
2
Q
2
2
=
+ = ÷ø
= + æ -
÷ ÷ ÷ ÷
ø
ç ç ç ç
è
= +
Q
•Third Quartile/Median
3 n = 3
=
(50) 37.5
4
4
correspondence with class 70-79
( )
c
3
æ -
n f
f
Q L
ö
÷ ÷ ÷ ÷
ø
ç ç ç ç
è
= +
å
3
3
3 3
4
æ -
( )
50 35
4
69.5 37.5 35
71.58
ö çè
ö
.10
12
.10
12
3
69.5
Q
3
3
=
÷ø
= + æ -
÷ ÷ ÷ ÷
ø
ç ç ç ç
è
= +
Q

44. ( ) n
( )2
= = -
Variance = S2
2 1å=
s 1 2
Standard Deviation = S
1
i
i x x
n
sx
For Singular
Data
n
( ) å=
= = -
i
i x x
n
x
1

45. Example
The height are :
156, 160, 158, 166, 168, 170
Find out the mean, the variance, and the standard deviation
163
156 160 158 166 168 170
6
i x x
1
_
n
== + + + + å=
+ = i
( ) ( ) ( ) ( ) ( )
ö
÷ ÷ø
æ - + - + - + - + - + - =
ç çè
156 163)2 160 163 2 158 163 2 166 163 2 168 163 2 170 163 2
6
s 2
( ) ( ) ( )
7 2 3 2 5 2 32 52 72
s 2 = - + - + - + + +
6
= + + + + +
49 9 25 9 25 49
27,67
6
=
So, Standard Deviation is 27.67 5.26 s = s 2 = =

46. ( ) n
( )2
= = -
Variance = S2
2 1å=
s 1 2
Standard Deviation=S
1
i
i i f x x
n
sx
For Grouped
Data
n
( ) å=
i i f x x
n
= = -
i
x
1

47. Thank you ^^