138
2
2
y12
=
(
0.0581
)²
= 0.0034
n
?
n
??
n
?
n
?
x
i
y
i
-
?
?
x
i
??
?
y
i
?
r
=
i
=¹
?
n
?
n
?
i
=¹
?
??
?? i =1
?
n
?
n
?
?
?n
?
x
2
-
?
?
x
i
?
??n
?
y²
-
?
?
y
i
?
?
?
?
i =¹
?
i
=¹
?
?
??
?
i
=¹
?
i
=¹
?
?
?
=
12(5070.8170) - (4224.90)(11.2402)
[
12(1673448.5826) - (4224.90)
2
][12(18.3592) - (11.2402)
2
]
= 0.9227
?
Index of Fit Distribusi Normal Time To Failure (TTF)
Tabel 5.10 Perhitungan Index of fit Distribusi Normal untuk TTF Komponen
Idler
i
t
i
x
i
=
t
i
F(t )
i )
y =
i =
z
i
x .z
i .z
i
2
x
i
z
2
i
1
144.08
144.08
0.0565
-1.5853
-228.4068
20759.0464
2.5131
2
172.92
172.92
0.1371
-1.0935
-189.0804
29901.3264
1.1956
3
232.83
232.83
0.2177
-0.7798
-181.5706
54209.8089
0.6082
4
286.33
286.33
0.2984
-0.5290
-151.4815
81984.8689
0.2799
5
305.50
305.50
0.3790
-0.3080
-94.1012
93330.2500
0.0949
6
372.92
372.92
0.4597
-0.1012
-37.7568
139069.3264
0.0103
7
380.00
380.00
0.5403
0.1012
38.4736
144400.0000
0.0103
8
381.83
381.83
0.6210
0.3080
117.6126
145794.1489
0.0949
9
403.83
403.83
0.7016
0.5290
213.6443
163078.6689
0.2799
10
454.33
454.33
0.7823
0.7798
354.3056
206415.7489
0.6082
11
538.25
538.25
0.8629
1.0935
588.5528
289713.0625
1.1956
12
552.08
552.08
0.9435
1.5853
875.2002
304792.3264
2.5131
S
4224.90
4224.90
6.0000
0.0000
1305.3918
1673448.5826
9.4038
r
0.9871
?
Contoh perhitungan index of fit distribusi normal TTF komponen idler i = 1 :
x1
=
t1
= 144.08
 |