TUGAS
MATA KULIAH STATISTIK DAN DESKRIPTIP
Adi
Yulianto
11100336
TENTANG
1.
DATA ANALISIS
2.
KOEFISIEN VARIASI
3.
JANGKAUAN ANTAR KUARTIL
21
|
45
|
40
|
47
|
31
|
27
|
30
|
41
|
46
|
26
|
28
|
22
|
44
|
37
|
27
|
29
|
45
|
45
|
38
|
26
|
26
|
49
|
40
|
44
|
30
|
27
|
24
|
49
|
45
|
28
|
21
|
25
|
48
|
23
|
38
|
23
|
26
|
47
|
26
|
49
|
28
|
27
|
46
|
30
|
50
|
35
|
39
|
50
|
26
|
46
|
Penyelesaian :
1.
Jangkauan =
data terbesar – data terkecil
=
50 – 15
=
35
2.
Banyak kelas interval = 1 + 3.3 log (n)
=
1 + 3.3 log (50)
=
1 + (3.3 . 1.69)
=
1 + 5.5
=
6.5
3.
Panjang kelas interval =
jangkauan / banyak kelas interval
= 35 / 7
= 5
Gambar 1.1 ( table frekwensi)
Nilai
|
f
|
fk
|
Xi
|
Fi * Xi
|
21-25
|
7
|
7
|
23
|
161
|
26-30
|
17
|
24
|
28
|
476
|
31-35
|
2
|
26
|
33
|
66
|
36-40
|
6
|
32
|
38
|
228
|
41-45
|
7
|
39
|
43
|
301
|
46-50
|
11
|
50
|
48
|
528
|
jumlah
|
50
|
1760
|
Ø Mean = Fi * Xi / f = 1760 / 50
= 35.2
Ø Q3 = 3.n / 4 = 3 *50 / 4 = 37.5
Dik ;
B = 41 – 0.5 = 40.5
P = 5
Fk = 32
F = 7
Penyelesaian
:
Q3 = b + (3.n/4 - fk / F ) . P
Q3 = 40.5 + (37.5 - 32 / 7 )
. 5
Q3 = 40.5 + ( 5.5 / 7 ) . 5
Q3 = 44.4
Q1
= n / 4 = 50 / 4
= 12.5
Dik :
B
= 26 – 0.5 = 25.5
P
= 5
Fk
= 7
F
= 17
Penyelesaian
;
Q1 = b + (n/4 - fk / F ) . P
Q1 = 25.5 + (12.5 - 7 / 17 ) . 5
Q1 = 25.5 + ( 5.5 / 7 ) .5
Q1 = 25.5 + 1.6
Q1 = 27.1
S2
= 1 / 50 ( 7 ( 23 – 35.2)2 +17 ( 28
– 35.2)2 + 2 ( 33 – 35.2)2 + 6 (38 – 35.2)2 +
7 ( 43 – 35.2)2
+ 11 ( 48 – 35.2)2 )
S2 =1 / 50 (
7 ( -12.2)2 + 17 ( -7.2)2
+2 ( -2.2)2 + 6 ( 2.8)2 + 7 ( 7.8)2
+ 11 (
12.8)2)
S2
= 1 / 50 (
7 ( 148.84) + 17 ( 51.84 ) +2 ( 4.84
) + 6 ( 7.84) + 7 ( 60.84 )
+ 11 ( 163.84) )
S2 = 1 / 50 ( 1041.88 + 881.28 + 9.68 + 47.04 + 425.88 +
1802.24 )
S2 = 1 / 50 (
4208 )
S2 = 84.16
Ø Simpangan
S = akar S2
= akar 84.16
S
= 9.16
Ketentuan soal
1.
KOEFISIEN VARIASI
2.
JANGKAUAN ANTAR KUARTIL
Jawab ;
1.
KV = S / rata2 X 100%
KV = 9.16 / 35.2 X 100%
KV = 26 %
2.
RAK = Q3 –
Q1
RAK =
44.4 – 27.1
RAK
= 17.3
0 komentar:
Posting Komentar