MEAN, MEDIAN, MODE
FREQUENCY DISTRIBUTION TABLE
A table containing array data have been grouped by
class or specific categories .
How to make
it?
2
SEE This Data
50
32
33
73
47
45
79
40
70
38
72
61
76
65
56
54
58
54
76
85
52
74
40
68
42
75
43
48
44
82
54
62
77
45
67
66
65
66
46
35
56
85
57
58
59
55
56
47
61
69
25
63
64
28
52
54
23
33
48
36
3
FIRST STEPS ARE
MAKE A FREQUENCY
DISTRIBUTION TABLE
4
STEP 01 :
SET NUMBER OF CLASSES (Jumlah Kelas JK)
JK = 1 + 3,322 log n
H.A. Sturges (1926)
JK = number of classes
n = the number of observations (data)
5
Form the data, number of classes are
JK = 1 + 3,322 log 60
= 1 + 3,322 (1,778)
= 1 + 5,907
= 6,907 ≈ 7 class
6
STEP 02 :
Set class interval ( Interval Kelas IK)
IK =
(Xt – Xr)
JK
Dimana :
IK
= class inteval
Xt
= highest data
Xr
= lower data
Xt – Xr = Range
7
From the data obtained Xt = 85 and Xr =
23, then IK = ....?
IK =
IK =
(Xt – Xr)
JK
(85 – 23)
7
= 8,8 ≈ 8
8
The frequency table is :
Class interval
Tally
Frequency ( f )
23 – 31
Ill
3
32 – 40
llll llll
8
41 – 49
IIII IIII II
10
50 – 58
IIII IIII IIII
II
14
59 – 67
IIII IIII III
11
68 – 76
IIII IIII
8
77 – 85
IIII II
6
total
60
9
WHAT YOU NEED TO
KNOW FROM THE
FREQUENCY
DISTRIBUTION TABLE
10
 (Class Limit)
• Lower Class Limit
lowest possible values within a class interval.
Example
In the class interval : 20-29; 30-39; 40-49
the lower class limit are 20, 30, dan 40.
• Upper Class Limit
highest possible values within a class interval
Example
In the class interval 20-29; 30-39; 40-49
The highest class limit 29, 39, dan 49.
11
Class Boundaries
• Lower Class Boundary
The real lower class boundary LCB – 0,5
Ex :
Class interval 20-29; 30-39; 40-49, the LCB 19,5; 29,5;
dan 39,5.
• Upper Class Boundary
The real upper class boundary UCL+ 0,5
Ex :
Class interval 20-29; 30-39; 40-49, the UCB 29,5;
39,5; dan 49,5.
12
 Mid-point or class mark
Mid-point ith = (LCL + UCL) : 2
Dimana
MP I
LCL
UCL
:
= Mid point class ith (1,2,3,4,…..i)
= Lowe class limit
= Upper class limit
13
Mid point for class interval on Table 5.1 :
Class interval
f
Mid point
23 – 31
32 – 40
41 – 49
3
8
10
27
36
45
50 – 58
59 – 67
68 – 76
14
11
8
……
……
……
77 – 85
6
60
……
total
14
 Cummulative Frequency
• Cf : frequencies results from the merger of the
class frequency with the class freqency before
• Cf can be calculated based on :
≤ (equal to or less than)
≥ (equal to or more than)
15
Comulative frequency for Table 5.1 :
Class interval
f
Mid
point
Cf
(≤)
Cf
(≥)
0
23 – 31
3
27
3
32 – 40
8
36
11
41 – 49
10
45
21
50 – 58
14
54
dst
59 – 67
11
63
25
68 – 76
8
72
14
77 – 85
6
81
6
total
60
dst
60 16
presented in
graphical is
certainly interesting
17
Polygon:…use frequency and mid point
frequency
polygon
20
15
10
5
0
27
36
45
54
63
72
81
Mid point
18
Histogram:…use freq and lower class boundary
frequency
HISTOGRAM
20
15
10
5
0
LCB
22,5 31,5 40,5 49,5 58,5 67,5 76,5 85,5
19
Ogive:…use cummulaive freq dan
lower class boundary
Cf
60
Less than
OGIVE
20
5
0
more than
22,5 31,5 40,5 49,5 58,5 67,5 76,5
Lower class
boundary
20
Exercise :
Make a complete frequency distribution tables and graphs of
polygons, histograms, and the ogive of the data distribution
follows
86 63 44 75 74 54 84 78 58 77
61 72 52 73 64 78 66 56 68 76
71 71 87 48 83 94 83 68 86 96
21
Apa yang dimaksud
UKURAN PEMUSATAN ?
• Ukuran nilai pusat yaitu nilai
yang mewakili suatu deretan/
rangkaian/gugusan data
• Ukuran Pemusatan mencakup :
MEAN, MEDIAN,dan MODUS
22
MEAN, MEDIAN, MODUS
Data Tidak Dikelompokkan
MEAN (Me) ---- rata-rata hitung
•Diperoleh dengan menjumlahkan seluruh nilai data
(x1+ x2 +…+ xi) dibagi dengan banyaknya data (n).
• Rata-rata hitung yang diambil dari data sampel
dilambangkan dengan x bar =
x
n
x = x1+x2+x3…xi
n
atau
Σ xi
x =
i=1
n
Contoh 6.1 : mean data tidak dikelompokkan
Mata Kuliah
P.Statistik
Nilai
10
Azas-azas
Manajemen
Perilaku Organisasi
MSDM
8
PPSDM
Matematika
Olah Raga
9
6
7
Jumlah
Mean
7
7
54
(54 : 7) = 7,7
MEDIAN (Md)
• Nilai yang ada di tengah-tengah rangkaian data,
setelah diurutkan dari data dengan nilai terkecil
sampai terbesar.
• Letak Md data tidak dikelompokkan dicari
dengan :
LMd = (n + 1) : 2
n adalah banyaknya data
Contoh 6.2 : Median data tidak dikelompokkan
LMd = (7 + 1) : 2
= 4 (median terletak pada urutan data ke 4)
n=7
Nilai Md
Nilai
10
8
7
7
9
6
7
Nilai setelah diurutkan
6
7
7
7
8
9
10
Urutan
Pertama
Kedua
Ketiga
Keempat
..
..
..
Bagaimana menentukan Md
jika banyaknya data adalah genap ?
Nilai
10
8
n=8
Nilai setelah
diurutkan
6
7
7
9
9
7
7
7
8
9
6
7
9
10
LMd = (8 + 1) : 2
= 4,5
Median terletak pada data
urutan ke 4,5 atau antara
urutan ke 4 dan 5.
Berapa Nilainya ?
Md = (7 + 8) : 2 = 7,5