SSTTAATTIISSTTIICCSS 
By: Tito Nurdiyanto 
tnk_adhyastha@yahoo.com
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
PPlleeaassee wwrriittee iinn aa ppiieeccee ooff ppaappeerr 
yyoouurrss:: 
1. AGE 
2. SUBJECT THAT YOU LIKE 
3. HEIGHT 
4. WEIGHT
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
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
Reading aanndd RReepprreesseennttiinngg 
DDaattaa There are many ways to presenting data, here they are
PPiiccttooggrraamm
PPiiee cchhaarrtt 
Mohammed asked his class about their favorite 
flavour of ice cream. 
Vanilla 12 
Strawberry 6 
Chocolate 5 
Other 7 
Their answers 
were:
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
Use the table to help you draw a pie chart. Don’t forget the 
key 
Vanilla 
Strawberry 
Chocolate 
Other 
144o 
72o 
84o 
60o
Vertical Bar 
Chart 
Compound Bar 
Chart 
Horizontal Bar 
Chart 
BBaarr CChhaarrtt
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.
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?
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.
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.
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.
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
The Measure of Central 
Tendency
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
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
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.
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
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
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
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
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
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.
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
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. 
.
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
GROUPED DATA 
Grouped frequency Distribution 
Arrange Cummulative Frequency 
Distribution 
Arrange Cummulative Frequency 
Distribution 
Mode 
Mean 
Lower Quartile, Median Quartile, Upper 
Quartile 
Varians 
Standard Deviation
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
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
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
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
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?
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
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
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
( ) 
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
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
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
•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
( ) 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
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 = =
( ) 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
Thank you ^^

Statistics for math (English Version)

  • 1.
  • 2.
    DDiiffffeerreenncceess bbeettwweeeenn QQuuaalliittaattiivveeaanndd 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 iinnaa ppiieeccee ooff ppaappeerr yyoouurrss:: 1. AGE 2. SUBJECT THAT YOU LIKE 3. HEIGHT 4. WEIGHT
  • 4.
    LLEETTSS SSEEEE TTHHEERREESSUULLTT 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 mmaakkeeaa 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.
  • 8.
    PPiiee cchhaarrtt Mohammedasked his class about their favorite flavour of ice cream. Vanilla 12 Strawberry 6 Chocolate 5 Other 7 Their answers were:
  • 9.
    He decided todraw 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 tableto 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 Itis 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 ofnumber 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 isa 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 Thedot 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 ofCentral Tendency
  • 19.
    EXAMPLE (QUARTILE) Mathtest 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 modeis 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 modeof 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 nis 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 isthe 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. Aclass 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 yangmembagi 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. Afamily 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 apie 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. Ina 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 Groupedfrequency 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 Groupedfrequency 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 isnecessary 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 isvery 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 Midpoint 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 followingdata 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) FrequencyFrequency 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 2n = 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 heightare : 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.