This document contains information about regression analysis using the least squares method. It includes examples of applying the least squares method to calculate regression equations from sample data sets with two variables. The regression equations allow predicting the value of the dependent variable from the independent variable. Formulas are provided for calculating the mean, standard deviation, and regression coefficients from the sample data.

1. Chapter 20
20-1
(a)
(b) f / (Nx) = f / [69(10)] = f / 690
f x2
7200
4900
19200
40500
80000
145200
86400
169000
156800
112500
51200
86700
64800
36100
0
44100
x
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
210
f
2
1
3
5
8
12
6
10
8
5
2
3
2
1
0
1
fx
120
70
240
450
800
1320
720
1300
1120
750
320
510
360
130
0
210

69
f / (Nx)
0.0029
0.0015
0.0043
0.0072
0.0116
0.0174
0.0087
0.0145
0.0116
0.0174
0.0029
0.0043
0.0029
0.0015
0
0.0015
8480 1 104 600
Chapter 20, Page 1/29

2. x
Eq. (20-9):
8480
 122.9 kcycles
69
12
1 104 600  8480 2 / 69 
Eq. (20-10): sx  
  30.3 kcycles Ans.
69  1


______________________________________________________________________________
20-2
Data represents a 7-class histogram with N = 197.
x
174
182
190
198
206
214
220

x
6
9
44
67
53
12
6
fx
1044
1638
8360
13 266
10 918
2568
1320
f x2
181 656
298 116
1 588 400
2 626 688
2 249 108
549 552
290 400
197
39 114
7 789 900
f
39 114
 198.55 kpsi Ans.
197
12
 7 783 900  39 114 2 / 197 
s
  9.55 kpsi Ans.
197  1


______________________________________________________________________________
20-3
Form a Table:
x
64
68
72
76
80
84
88
92

f
2
6
6
9
19
10
4
2
58
fx
128
408
432
684
1520
840
352
184
4548
fx2
8192
27 744
31 104
51 984
121 600
70 560
30 976
16 928
359 088
Chapter 20, Page 2/29

3. x
4548
 78.4 kpsi
58
Ans.
 359 088  45482 / 58 
sx  

58  1


12
 6.57 kpsi
Ans.
From Eq. 20-14
 1  x  78.4  
1
exp   
f  x 
Ans.
 
6.57 2
 2  6.57  


______________________________________________________________________________
2
20-4
(a)
x
5.625
5.875
6.125
6.375
6.625
6.875
7.125
7.375
7.625
7.875
8.125

f
1
0
0
3
3
6
14
15
10
2
1
55
fy
5.625
0
0
19.125
19.875
41.25
99.75
110.625
76.25
15.75
8.125
396.375
f y2
31.64063
0
0
121.9219
131.6719
283.5938
710.7188
815.8594
581.4063
124.0313
66.015 63
2866.859
y
5.625
5.875
6.125
6.375
6.625
6.875
7.125
7.375
7.625
7.875
8.125
For a normal distribution,
y  396.375 / 55  7.207,
f / (Nw)
0.072 727
0
0
0.218 182
0.218 182
0.436 364
1.018 182
1.090 909
0.727 273
0.145 455
0.072 727

f (y)
0.001 262
0.008 586
0.042 038
0.148 106
0.375 493
0.685 057
0.899 389
0.849 697
0.577 665
0.282 608
0.099 492

 2866.859  396.3752 / 55 
sy  

55  1


g(y)
0.000 295
0.004 088
0.031 194
0.140 262
0.393 667
0.725 002
0.915 128
0.822 462
0.544 251
0.273 138
0.106 720
12
 0.4358
 1  x  7.207  2 
1
f  y 
exp   
 
0.4358 2
 2  0.4358  


For a lognormal distribution,
x  ln 7.206 818  ln 1  0.060 474 2  1.9732,
g  y 
1
x  0.0604

s x  ln 1  0.060 4742  0.0604
 1  ln x  1.9732  2 
exp   
 
2
 2  0.0604  



Chapter 20, Page 3/29

4. (b) Histogram
______________________________________________________________________________
20-5
Distribution is uniform in interval 0.5000 to 0.5008 in, range numbers are a = 0.5000 in,
b = 0.5008 in.
a  b 0.5000  0.5008
(a) Eq. (20-22)

 0.5004
x 
2
2
x 
Eq. (20-23)
b  a 0.5008  0.5000

 0.000 231
2
2 3
(b) PDF, Eq. (20-20)
1250
f ( x)  
0
0.5000  x  0.5008 in
otherwise
(c) CDF, Eq. (20-21)
0

F ( x )  ( x  0.5) / 0.0008
1

x  0.5000 in
0.5000  x  0.5008 in
x  0.5008 in
If all smaller diameters are removed by inspection, a = 0.5002 in, b = 0.5008 in,
0.5002  0.5008
 0.5005 in
2
0.5008  0.5002
ˆ
x 
 0.000 173 in
2 3
1666.7 0.5002  x  0.5008 in
f ( x)  
otherwise
0
x 
Chapter 20, Page 4/29

5. 0

F ( x)  1666.7( x  0.5002)
1

x  0.5002 in
0.5002  x  0.5008 in
x  0.5008 in
______________________________________________________________________________
20-6
Dimensions produced are due to tool dulling and wear. When parts are mixed, the
distribution is uniform. From Eqs. (20-22) and (20-23),
a   x  3s  0.6241  3  0.000 581  0.6231 in
b   x  3s  0.6241  3  0.000 581  0.6251 in
0.623
in Ans.
0.625
______________________________________________________________________________
We suspect the dimension was
20-7
F(x) = 0.555x – 33 mm.
(a) Since F(x) is linear, distribution is uniform at x = a
F(a) = 0 = 0.555(a) – 33
 a = 59.46 mm. Therefore at x = b
F(b) = 1= 0.555b – 33
 b = 61.26 mm. Therefore,
0

F ( x)  0.555 x  33
1

x  59.46 mm
59.46  x  61.26 mm
x  61.26 mm
The PDF is dF/dx, thus the range numbers are:
0.555
f ( x)  
0
59.46  x  61.26 mm
otherwise
Ans.
From the range numbers,
59.46  61.26
 60.36 mm
2
61.26  59.46
ˆ
x 
 0.520 mm
2 3
x 
Ans.
Ans.
Chapter 20, Page 5/29

6. (b)  is an uncorrelated quotient F  3600 lbf, A  0.112 in 2
CF  300 3600  0.083 33, C A  0.001 0.112  0.008 929
From Table 20-6, For 

 F 3600

 32 143 psi
 A 0.112
Ans.
  0.083332  0.0089292  
ˆ

   32 143 

1  0.0089292  


C  2694 / 32 143  0.0838 Ans.
1/2
 2694 psi
Ans.
Since F and A are lognormal, division is closed and  is lognormal too.
 = LN(32 143, 2694) psi
Ans.
______________________________________________________________________________
20-8
Cramer’s rule
a1
y
 xy

x
x
x
x

x
x
2
2
a2
2
x
0
0
0
1
1
1

3
a1 =
y
0.01
0.15
0.25
0.25
0.17
0.01
0.82
x
 x   y  x   xy  x
x
 x x    x 
x
y
 xy   y  xy   y  x
 x  x x    x 
x
2
3
3
2
3
2
2
2
Ans.
3
2
2
3
2
2
Ans.
3
x2
x3
xy
0
0.04
0.16
0.36
0.64
1.00
0
0.008
0.064
0.216
0.512
1.000
2.20
1.800
0
0.030
0.100
0.150
0.136
0.010
0.406
1.040 714
a 2 = 1.046 43
Ans.
Chapter 20, Page 6/29

7. Data
x
Regression
y
y
0.01
0
0.15 0.166 286
0.25 0.248 857
0.25 0.247 714
0.17 0.162 857
-0.01 -0.005 710
0
0.2
0.4
0.6
0.8
1.0
______________________________________________________________________________
20-9
Data
Su
0
60
64
65
82
101
119
120
130
134
145
180
S e
30
48
29.5
45
51
50
48
67
60
64
84
Regression
S e
Su2
Su S e
20.356 75
39.080 78
40.329 05
40.641 12
45.946 26
51.875 54
57.492 75
57.804 81
60.925 48
62.173 75
65.606 49
76.528 84
3 600
4 096
4 225
6 724
10 201
14 161
14 400
16 900
17 956
21 025
32 400
1 800
3 072
1 917.5
3 690
5 151
5 950
5 760
8 710
8 040
9 280
15 120
Chapter 20, Page 7/29

8. 
195
205
207
210
213
225
225
227
230
238
242
265
280
295
325
325
355
5462
78
96
87
87
75
99
87
116
105
109
106
105
96
99
114
117
122
2274.5
m = 0.312 067,
81.209 85
84.330 52
84.954 66
85.890 86
86.827 06
90.571 87
90.571 87
91.196 00
92.132 20
94.628 74
95.877 01
103.054 60
107.735 60
112.416 60
121.778 60
121.778 60
131.140 60
b = 20.356 75
38 025
42 025
42 849
44 100
45 369
50 625
50 625
51 529
52 900
56 644
58 564
70 225
78 400
87 025
105 625
105 625
126 025
1 251 868
15 210
19 680
18 009
18 270
15 975
22 275
19 575
26 332
24 150
25 942
25 652
27 825
26 880
29 205
37 050
38 025
43 310
501 855.5
Ans.
______________________________________________________________________________
20-10
    y  a0  a2 x 2 
2

 2  y  a0  a2 x 2   0
ao
Chapter 20, Page 8/29

9.  y  na
0
 a2  x 2  0 
 y  na
0
 a2  x 2

 2  y  a0  a2 x 2   2 x   0   xy a0  x  a2  x3
a2
Cramer’s rule
 y  x2
 xy  x3   x3  y   x 2  xy
ao 
n x3   x  x 2
n
 x2
x x
n
y
 x  xy
n
x
x x
Ans.
3
a1 
2

n xy   x  y
n x3   x x 2
3
x

Data Regression
y
y
20
40
60
80
200
19
17
13
7
56
19.2
16.8
12.8
7.2
x2
400
1600
3600
6400
12 000
x3
8 000
64 000
216 000
512 000
800 000
xy
380
680
780
560
2400
800 000(56)  12 000(2400)
 20
4(800 000)  200(12 000)
4(2400)  200(56)
 0.002
a1 
4(800 000)  200(12 000)
a0 
______________________________________________________________________________
Chapter 20, Page 9/29

10. 20-11
Data

x
0.2
0.4
0.6
0.8
1
2
5
Regression
y
7.1
10.3
12.1
13.8
16.2
25.2
84.7
y
7.931 803
9.884 918
11.838 032
13.791 147
15.744 262
25.509 836
ˆ
mk 
x2
y2
0.04
50.41
0.16 106.09
0.36 146.41
0.64 190.44
1 262.44
4 635.04
6.2 1390.83
6  90.44   5  84.7 
6  6.2    5 
2
xy
1.42
4.12
7.26
11.04
16.2
50.4
90.44
xx
-0.633 333
-0.433 333
-0.233 333
-0.033 333
0.166 666
1.166 666
0
x  x
2
0.401 111 111
0.187 777 778
0.054 444 444
0.001 111 111
0.027 777 778
1.361 111 111
2.033 333 333
 9.7656
84.7  9.7656(5)
ˆ
b  Fi 
 5.9787
6
(a)
5
x ;
6
Eq. (20-37):
y
84.7
 14.117
6
1390.83  5.9787(84.7)  9.7656(90.44)
62
 0.556
s yx 
Eq. (20-36):
1 5 6
sbˆ  0.556 
 0.3964 lbf
6 2.0333
Fi   5.9787, 0.3964  lbf Ans.
2
Chapter 20, Page 10/29

11. (b) Eq. (20-35):
0.556
 0.3899 lbf/in
2.0333
k  (9.7656, 0.3899) lbf/in Ans.
______________________________________________________________________________
sm 
ˆ
20-12 The expression  =  / l is of the form x / y. Now  = (0.0015, 0.000 092) in, unspecified
distribution; and l = (2, 000, 0.008 1) in, unspecified distribution;
C x = 0.000 092 / 0.0015 = 0.0613
C y = 0.0081 / 2.000 = 0.004 05
Table 20-6:
  0.0015 / 2.000  0.000 75
 0.06132  0.004 052 
ˆ
   0.000 75 

2
 1  0.004 05

1/2
 4.607 105   0.000 046
ˆ
We can predict  and   but not the distribution of .
______________________________________________________________________________
20-13  = E
 = (0.0005, 0.000 034), distribution unspecified; E = (29.5, 0.885) Mpsi, distribution
unspecified;
C x = 0.000 034 / 0.0005 = 0.068
C y = 0.0885 / 29.5 = 0.03
 is of the form xy
Table 20-6:
   E  0.0005(29.5)106  14 750 psi
ˆ
   14 750 0.0682  0.0302  0.0682  0.0302  


1/ 2
 1096.7 psi
C  1096.7 /14 750  0.074 35
______________________________________________________________________________
20-14

Fl
AE
where F = (14.7, 1.3) kip, A = (0.226, 0.003) in2, l = (1.5, 0.004) in, and
E = (29.5, 0.885) Mpsi, distributions unspecified.
Chapter 20, Page 11/29

12. C F = 1.3 / 14.7 = 0.0884; C A = 0.003 / 0.226 = 0.0133; C l = 0.004 / 1.5 = 0.00267;
C E =0.885 / 29.5 = 0.03

Table 20-6:
Fl
 1  1 
 Fl    
AE
 A  E 
  F l 1 A 1/ E   F l 1 A  1 E 

1
 1 

  0.003 31 in.
 14 700(1.5) 

6
 0.226   29.5 10  


Ans.
For the standard deviation, using the first-order terms in Table 20-6,
12
12
Fl
ˆ
 
 CF2  Cl2  C A2  CE2     CF2  Cl2  CA2  CE2 
AE
ˆ
   0.003 31 0.08442  0.002 67 2  0.01332  0.032 
12
 0.000 313 in
COV:
Ans.
ˆ
C    /   0.000 313 / 0.003 31  0.0945
Ans.
Force COV dominates. There is no distributional information on  .
______________________________________________________________________________
20-15 M = (15 000, 1350) lbf ⋅ in, distribution unspecified; d = (2.00, 0.005) in, distribution
unspecified.
32M
 
 d3
C M = 1350 / 15 000 = 0.09, C d = 0.005 / 2.00 = 0.0025
 is of the form x/y3, Table 20-6.
Mean:
M  15 000 lbf  in
1
1
 1 
2
2
3
 3   3 1  6Cx   3 1  6  0.0025    0.125 in *

d  d
2 

1
 1 
* Note:  3   3
d  d
Chapter 20, Page 12/29

13. 32M 32(15 000)
(0.125)

d 3

 19 099 psi Ans.
Standard Deviation:




12
2
ˆ
     CM  Cd2 / 1  Cd2 


Table 20-6:
3
3
Cd 3  3Cd  3(0.0025)  0.0075

2
ˆ
     CM   3Cd 
2
 1  3C   
2
12
d

 19 099  0.092  0.00752

 1725 psi Ans.
 1  0.0075  

2
12
COV:
C 
1725
 0.0903
19 099
Ans.
Stress COV dominates. No information of distribution of  .
______________________________________________________________________________
20-16
Fraction discarded is  +. The area under the PDF was unity. Having discarded  + 
fraction, the ordinates to the truncated PDF are multiplied by a.
a
1
1     
New PDF, g(x), is given by
 f ( x) 1      



g ( x)  
0

x1  x  x2
otherwise
A more formal proof: g(x) has the property
Chapter 20, Page 13/29

14. 1   g  x  dx  a 
x2
x2
x1
x1
f  x dx


x1
1  a   f  x  dx   f  x  dx   f  x  dx 
 

x2
0


1  a 1  F ( x1 )  1  F ( x2 ) 
1
1
1


F ( x2 )  F ( x1 ) 1      1     
______________________________________________________________________________
a
20-17 (a) d = U(0.748, 0.751)
0.751  0.748
 0.7495 in
d 
2
0.751  0.748
ˆ
 0.000 866 in
d 
2 3
1
1

 333.3 in 1
f  x 
b  a 0.751  0.748
x  0.748
 333.3( x  0.748)
F ( x) 
0.751  0.748
(b)
F(x 1 ) = F(0.748) = 0
F(x 2 ) = (0.750 – 0.748) 333.3 = 0.6667
If g(x) is truncated, PDF becomes
f ( x)
333.3

 500 in 1
F ( x2 )  F ( x1 ) 0.6667  0
a  b 0.748  0.750
x 

 0.749 in
2
2
b  a 0.750  0.748
ˆ
x 

 0.000 577 in
2 3
2 3
______________________________________________________________________________
g ( x) 
20-18 From Table A-10, 8.1% corresponds to z 1 = 1.4 and 5.5% corresponds to z 2 = +1.6.
ˆ
k1    z1
ˆ
k2    z2
From which

z2 k1  z1k2 1.6(9)  (1.4)11

z2  z1
1.6  (1.4)
 9.933
Chapter 20, Page 14/29

15. ˆ

k2  k1
11  9

 0.6667
z2  z1 1.6  (1.4)
The original density function is
 1  k  9.933  2 
1
exp   
f (k ) 
  Ans.
0.6667 2
 2  0.6667  


______________________________________________________________________________
ˆ
20-19 From Prob. 20-1,  = 122.9 kcycles and  = 30.3 kcycles.
x10   x10  122.9

ˆ

30.3
x10  122.9  30.3 z10
From Table A-10, for 10 percent failure, z 10 = 1.282
z10 
x 10 = 122.9 + 30.3(1.282)
= 84.1 kcycles Ans.
______________________________________________________________________________
20-20
f (x)
x
f
fx
f x2
f / (Nw)
60 2 120
7200 0.002899 0.000399
70 1
70
4900 0.001449 0.001206
80 3 240 19200 0.004348 0.003009
90 5 450 40500 0.007246 0.006204
100 8 800 80000 0.011594 0.010567
110 12 1320 145200 0.017391 0.014871
120 6 720 86400 0.008696 0.017292
130 10 1300 169000 0.014493 0.016612
140 8 1120 156800 0.011594 0.013185
150 5 750 112500 0.007246 0.008647
160 2 320 51200 0.002899 0.004685
170 3 510 86700 0.004348 0.002097
180 2 360 64800 0.002899 0.000776
190 1 190 36100 0.001449 0.000237
200 0
0
0
0 5.98E-05
210 1 210 44100 0.001449 1.25E-05
 69 8480
x = 122.8986
s x = 22.887 19
Chapter 20, Page 15/29

16. Eq. (20-14):
 1  x   2 
x
f ( x) 
exp   
 
ˆ
ˆ
 x 2
 2  x  


 1  x  122.8986  2 
1

exp   
 
22.88719 2
 2  22.88719  


1
x
55
55
65
65
75
75
85
85
95
95
105
105
115
115
125
125
135
135
f / (Nw)
0
0.002 899
0.002 899
0.001 449
0.001 449
0.004 348
0.004 348
0.007 246
0.007 246
0.011 594
0.011 594
0.017 391
0.017 391
0.008 696
0.008 696
0.014 493
0.014 493
0.011 594
f (x)
0.000 214
0.000 214
0.000 711
0.000 711
0.001 951
0.001 951
0.004 425
0.004 425
0.008 292
0.008 292
0.012 839
0.012 839
0.016 423
0.016 423
0.017 357
0.017 357
0.015 157
0.015 157
x
145
145
155
155
165
165
175
175
185
185
195
195
205
205
215
215
f / (Nw)
0.011 594
0.007 246
0.007 246
0.002 899
0.002 899
0.004 348
0.004 348
0.002 899
0.002 899
0.001 449
0.001 449
0
0
0.001 499
0.001 499
0
f (x)
0.010 935
0.010 935
0.006 518
0.006 518
0.002 21
0.003 21
0.001 306
0.001 306
0.000 439
0.000 439
0.000 122
0.000 122
2.8E-05
2.8E-05
5.31E-06
5.31E-06
______________________________________________________________________________
Chapter 20, Page 16/29

17. 20-21
x
174
182
190
198
206
214
222
1386
f
6
9
44
67
53
12
6
197
x = 198.6091
x
170
170
178
178
186
186
194
194
202
202
210
210
218
218
226
226
f / (Nw)
0
0.003807
0.003807
0.005711
0.005711
0.027919
0.027919
0.042513
0.042513
0.033629
0.033629
0.007614
0.007614
0.003807
0.003807
0
fx
1044
1638
8360
13266
10918
2568
1332
39126
f x2
181656
298116
1588400
2626668
2249108
549552
295704
7789204
f / (Nw)
0.003807
0.005711
0.027919
0.042513
0.033629
0.007614
0.003807
f (x)
0.001642
0.009485
0.027742
0.041068
0.030773
0.011671
0.002241
s x = 9.695 071
f (x)
0.000529
0.000529
0.004297
0.004297
0.017663
0.017663
0.036752
0.036752
0.038708
0.038708
0.020635
0.020635
0.005568
0.005568
0.00076
0.00076
______________________________________________________________________________
20-22
x
64
68
72
76
80
84
88
92
624
f
2
6
6
9
19
10
4
2
58
fx
128
408
432
684
1520
840
352
184
4548
f x2
8192
27744
31104
51984
121600
70560
30976
16928
359088
f / (Nw)
0.008621
0.025862
0.025862
0.038793
0.081897
0.043103
0.017241
0.008621
f (x)
0.00548
0.017299
0.037705
0.056742
0.058959
0.042298
0.020952
0.007165
Chapter 20, Page 17/29

18. x = 78.041379
x
62
62
66
66
70
70
74
74
78
78
82
82
86
86
90
90
94
94
f / (Nw)
s x = 6.572 229
f (x)
0.002684
0.002684
0.010197
0.010197
0.026749
0.026749
0.048446
0.048446
0.060581
0.060581
0.052305
0.052305
0.03118
0.03118
0.012833
0.012833
0.003647
0.003647
0
0.008621
0.008621
0.025862
0.025862
0.025862
0.025862
0.038793
0.038793
0.0381897
0.081897
0.043103
0.043103
0.017241
0.017241
0.008621
0.008621
0
______________________________________________________________________________
20-23

4 P 4(40)

 50.93 kpsi
 d 2  12
 
ˆ
 
ˆ
4 P 4(8.5)

 10.82 kpsi
 d 2  12
 
ˆ
 S  5.9 kpsi
y
For no yield, m = S y    0
z
m  m 0  m


 m
ˆ
ˆ
ˆ
m
m
m
m  S y    78.4  50.93  27.47 kpsi

2
ˆ
ˆ2
 m     S
z
y
  10.82
12
2
 5.92   12.32 kpsi
2
m
27.47

 2.230
ˆ
m
12.32
Table A-10, p f = 0.0129
R = 1 – p f = 1 – 0.0129 = 0.987 Ans.
______________________________________________________________________________
Chapter 20, Page 18/29

19. 20-24 For a lognormal distribution,
Eq. (20-18)  y  ln  x  ln 1  C x2
ˆ
Eq. (20-19)  y  ln 1  C x2 
From Prob. (20-23)
m  S y     x

 
2
2
 y  ln S y  ln 1  CS  ln   ln 1  C
y
 S 1 C2

 ln  y
2
  1  CS y








2
ˆ
 y  ln 1  CS2  ln 1  C  

y

12


2
2
 ln  1  CS y 1  C  


 S 1 C2 
 
ln  y
2
  1  CS y 



z 
ˆ

2
2
ln  1  CS y 1  C  





4 P 4(30)

 38.197 kpsi
 d 2  12 
ˆ
 
ˆ
4 P 4(5.1)

 6.494 kpsi
 d 2  12 
6.494
 0.1700
38.197
3.81
CS y 
 0.076 81
49.6
 49.6
1  0.1702 
ln 
2 
 38.197 1  0.07681 


 1.470
z
2
2
ln 1  0.076 81 1  0.170  


C 
Table A-10
p f = 0.0708
R = 1 – p f = 0.929 Ans.
______________________________________________________________________________
Chapter 20, Page 19/29

20. 20-25
x
93
95
97
99
101
103
105
107
109
111
n
19
25
38
17
12
10
5
4
4
2
136
nx2
nx
1767
164 311
2375
225 625
3686
357 542
1683
166 617
1212
122 412
1030
106 090
525
55 125
428
45 796
436
47 524
222
24 642
13 364 1 315 704
x = 13 364/136 = 98.26 kpsi
12
 1 315 704  13 364 2 / 136 
sx  

136  1


 4.30 kpsi
Under normal hypothesis,
z0.01   x0.01  98.26 / 4.30
x0.01  98.26  4.30 z0.01
 98.26  4.30  2.3267 
 88.26  88.3 kpsi
Ans.
______________________________________________________________________________
ˆ
20-26 From Prob. 20.25,  x = 98.26 kpsi, and  x  4.30 kpsi.
ˆ
C x   x /  x  4.30 / 98.26  0.043 76
From Eqs. (20-18) and (20-19),
 y  ln  98.26  0.043 762 / 2  4.587
ˆ
 y  ln 1  0.043 762   0.043 74
For a yield strength exceeded by 99% of the population,
ˆ
ˆ
z0.01  ln x0.01   y /  y  ln x0.01   y   y z0.01


From Table A-10, for 1% failure, z 0.01 = 2.326. Thus,
Chapter 20, Page 20/29

21. ln x0.01  4.587  0.043 74  2.326   4.485
x0.01  88.7 kpsi
Ans.
The normal PDF is given by Eq. (20-14) as
 1  x  98.26  2 
1
exp   
f  x 
 
4.30 2
 2  4.30  


For the lognormal distribution, from Eq. (20-17), defining g(x),
 1  ln x  4.587  2 
1
g  x 
exp   
 
x  0.043 74 2
 2  0.043 74  


x(kpsi)
92
92
94
94
96
96
98
98
100
100
102
f / (Nw)
0.000 00
0.069 85
0.069 85
0.091 91
0.091 91
0.139 71
0.139 71
0.062 50
0.062 50
0.044 12
0.044 12
f (x)
0.032 15
0.032 15
0.056 80
0.056 80
0.080 81
0.080 81
0.092 61
0.092 61
0.085 48
0.085 48
0.063 56
g(x)
0.032 63
0.032 63
0.058 90
0.058 90
0.083 08
0.083 08
0.092 97
0.092 97
0.083 67
0.083 67
0.061 34
x(kpsi)
102
104
104
106
106
108
108
110
110
112
112
f / (Nw)
0.036 76
0.036 76
0.018 38
0.018 38
0.014 71
0.014 71
0.014 71
0.014 71
0.007 35
0.007 35
0.000 00
f (x)
0.063 56
0.038 06
0.038 06
0.018 36
0.018 36
0.007 13
0.007 13
0.002 23
0.002 23
0.000 56
0.000 56
g(x)
0.061 34
0.037 08
0.037 08
0.018 69
0.018 69
0.007 93
0.007 93
0.002 86
0.002 86
0.000 89
0.000 89
Note: rows are repeated to draw histogram
The normal and lognormal are almost the same. However, the data is quite skewed and
perhaps a Weibull distribution should be explored. For a method of establishing the
Chapter 20, Page 21/29

22. Weibull parameters see Shigley, J. E., and C. R. Mishke, Mechanical Engineering
Design, McGraw-Hill, 5th ed., 1989, Sec. 4-12.
_____________________________________________________________________________
 
20-27 x  Sf e
104
Eq. (20-28):
x 0 = 79 kpsi,  = 86.2 kpsi, b = 2.6
x  x0    x0  (1  1 b)
 79  (86.2  79)(1  1 2.6)
 79  7.2(1.38)
From Table A-34, Γ (1.38) = 0.888 54
x  79  7.2(0.888 54)  85.4 kpsi
Ans.
Eq. (20-29)
ˆ
 x    x0   1  2 b    2 1  1 b  


12
  86.2  79    1  2 2.6    2 1  1 2.6  


 7.2 0.923 76  0.888 542 


12
12
 2.64 kpsi Ans.
ˆ

2.64
Cx  x 
 0.031 Ans.
x 85.4
_____________________________________________________________________________
20-28 x = S ut
x 0 = 27.7 kpsi,  = 46.2 kpsi, b = 4.38
 x  27.7   46.2  27.7   1  1 4.38 
 27.7  18.5 (1.23)
 27.7  18.5(0.910 75)
 44.55 kpsi Ans.
ˆ
 x   46.2  27.7   1  2 4.38    2 1  1 4.38  


12
12
 18.5  (1.46)   2 (1.23) 


12
 18.5  0.8856  0.920 752 


 4.38 kpsi Ans.
4.38
Cx 
 0.098 Ans.
44.55
From the Weibull survival equation
Chapter 20, Page 22/29

23.   x  x b 
0
R  exp   
   1 p
  x0  
 


  x  x b 
R40  exp    40 0    1  p40
    x0  


  40  27.7  4.38 
 exp   
   0.846
  46.2  27.7  


p40  1  R40  1  0.846  0.154  15.4% Ans.
_____________________________________________________________________________
20-29 x = S ut ,
x 0 = 151.9 kpsi,  = 193.6 kpsi, b = 8
 x  151.9  193.6  151.9   1  1 8
 151.9  41.7 (1.125)
 151.9  41.7(0.941 76)
 191.2 kpsi Ans.
ˆ
 x  193.6  151.9   1  2 8   2 1  1 8 


12
12
 41.7 (1.25)   2 (1.125) 


12
 41.7 0.906 40  0.941 762 


 5.82 kpsi Ans.
5.82
Cx 
 0.030
191.2
_____________________________________________________________________________
20-30 x = S ut ,
x 0 = 47.6 kpsi,  = 125.6 kpsi, b = 11.4
x  47.6  125.6  47.6   1  1 11.84 
 47.6  78  1.08 
 47.6  78  0.959 73  122.5 kpsi
ˆ
 x  125.6  47.6   1  2 11.84    2 1  1 11.84  


12
12
 78 (1.08)   2 (1.17) 


12
 78 0.959 73  0.936 702 


 22.4 kpsi
From Prob. 20-28,
  x  x b 
  100  47.6 11.84 
0
  1  exp   
p  1  exp   

   0.0090
    0  
  125.6  47.6  




Ans.
Chapter 20, Page 23/29

24. y = Sy,
y 0 = 64.1 kpsi,  = 81.0 kpsi, b = 3.77
y  64.1  (81.0  64.1) 1  1 3.77 
 64.1  16.9 (1.27)
 64.1  16.9(0.902 50)  79.35 kpsi
 y  (81  64.1) (1  2 3.77)   1  1 3.77  


12
 16.9  0.887 57   0.902 502   4.57 kpsi


3.77
  y y  
0
p  1  exp   
 
    y0  


3.77
  70  64.1  
 1  exp   
   0.019 Ans.
  81  64.1  


_____________________________________________________________________________
12
20-31 x = S ut = W[122.3, 134.6, 3.64] kpsi, p(x > 120) = 1 = 100% since x 0 > 120 kpsi
  133  122.3 3.64 
p(x  133)  exp   
 
  134.6  122.3  


 0.548  54.8% Ans.
_____________________________________________________________________________
20-32 Using Eqs. (20-28) and (20-29) and Table A-34,
n  n0    n0   1  1 b   36.9  133.6  36.9   1  1 2.66 
=122.85 kcycles
ˆ
 n    n0   1  2 b    2 1  1 b    34.79 kcycles


For the Weibull density function, Eq. (20-27),
  n  36.9  2.66 
exp   
 
  133.6  36.9  


For the lognormal distribution, Eqs. (20-18) and (20-19) give,
2.66
 n  36.9 
fW (n) 


133.6  36.9  133.6  36.9 
2.66 1
 y  ln 122.85   34.79 122.85 2  4.771
2
2
ˆ
 y  1   34.79 122.85   0.2778


From Eq. (20-17), the lognormal PDF is
Chapter 20, Page 24/29

25.  1  ln n  4.771 2 
1
exp   
 
0.2778n 2
 2  0.2778  


We form a table of densities f W (n) and f LN (n) and plot.
f LN  n  
n(kcycles)
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
210
220
f W (n)
9.1E-05
0.000 991
0.002 498
0.004 380
0.006 401
0.008 301
0.009 822
0.010 750
0.010 965
0.010 459
0.009 346
0.007 827
0.006 139
0.004 507
0.003 092
0.001 979
0.001 180
0.000 654
0.000 336
f LN (n)
1.82E-05
0.000 241
0.001 233
0.003 501
0.006 739
0.009 913
0.012 022
0.012 644
0.011 947
0.010 399
0.008 492
0.006 597
0.004 926
0.003 564
0.002 515
0.001 739
0.001 184
0.000 795
0.000 529
The Weibull L10 life comes from Eq. (20-26) with reliability of R = 0.90. Thus,
n0.10  36.9  133.6  36.9  ln 1 0.90  


1 2.66
 78.4 kcycles
Ans.
The lognormal L10 life comes from the definition of the z variable. That is,
Chapter 20, Page 25/29

26. ˆ
ln n0   y   y z
or
ˆ
n0  exp   y   y z 
From Table A-10, for R = 0.90, z = 1.282. Thus,
n0  exp  4.771  0.2778  1.282    82.7 kcycles


Ans.
_____________________________________________________________________________
20-33 Form a table
i
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
x
(105)L
3.05
3.55
4.05
4.55
5.05
5.55
6.05
6.55
7.05
7.55
8.05
8.55
9.05
9.55
10.05
fi
3
7
11
16
21
13
13
6
2
0
4
3
0
0
1
100
f i x⋅(105)
9.15
24.85
44.55
72.80
106.05
72.15
78.65
39.30
14.10
0
32.20
25.65
0
0
10.05
529.50
f i x2⋅(1010)
27.9075
88.2175
180.4275
331.24
535.5525
400.4325
475.8325
257.415
99.405
0
259.21
219.3075
0
0
101.0025
2975.95
g(x)⋅(105)
0.0557
0.1474
0.2514
0.3168
0.3216
0.2789
0.2151
0.1517
0.1000
0.0625
0.0375
0.0218
0.0124
0.0069
0.0038
Chapter 20, Page 26/29

27. x  529.5 105  100  5.295 105  cycles
Ans.
12
 2975.95 1010  529.5 105  2 100 
    

sx  


100  1




 1.319 105  cycles
Ans.
Cx  s x  1.319 5.295  0.249
 y  ln 5.295 105   0.2492 2  13.149
ˆ
 y  ln 1  0.2492   0.245
g ( x) 
1
ˆ
x y
 1  ln x  
y
exp   
 
ˆy
 2
2





2




 1  ln x  13.149  2 
1.628

exp   
 
x
0.245
 
 2


_____________________________________________________________________________
20-34 X = S u = W[70.3, 84.4, 2.01]
 x  70.3   84.4  70.3  1  1 2.01
Eq. (2-28):
 70.3  (84.4  70.3) 1.498
 82.8 kpsi
Ans.
Chapter 20, Page 27/29

28. 12
ˆ
 x  (84.4  70.3)  (1  2 2.01)   2 (1  1 2.01) 


12
Eq. (2-29):
ˆ
 x  14.1  0.997 91  0.886 17 2 


 6.502 kpsi
6.502
 0.079 Ans.
82.8
_____________________________________________________________________________
Cx 
20-35 Take the Weibull equation for the standard deviation
ˆ
 x    x0  (1  2 b)   2 (1  1 b) 


12
and the mean equation solved for x  x0
x  x0    x0   1  1 b 
and divide the first by the second,
  1  2 b    2 1  1 b  


 1  1 b 
x  x0
ˆ
x
12
 1  2 b 
4.2

 1  R  0.2763
49  33.8
 2 1  1 b 
Make a table and solve for b iteratively
b
3
4
4.1
1 + 2/b
1.67
1.5
1.49
1 + 1/b  1  2 b 
1.33
0.903 30
1.25
0.886 23
1.24
0.885 95
 1  1 b 
0.893 38
0.906 40
0.908 52
R
0.363
0.280
0.271
b  4.068 Using MathCad Ans.
x  x0
49  33.8
 33.8 
  x0 
(1  1/ b)
 1  1/ 4.068 
 49.8 kpsi Ans.
_____________________________________________________________________________
20-36 x = S y = W[34.7, 39, 2.93] kpsi
Chapter 20, Page 28/29

29. x  34.7   39  34.7   1  1 2.93  34.7  4.3  1.34 
 34.7  4.3  0.892 22   38.5 kpsi
ˆ
 x   39  34.7   1  2 2.93   2 1  1 2.93 


12
 4.3  1.68    2 1.34  


12
12
 4.3 0.905 00  0.892 222   1.42 kpsi Ans.


Cx  1.42 38.5  0.037 Ans.
_____________________________________________________________________________
20-37
x (Mrev)
1
2
3
4
5
6
7
8
9
10
11
12
Sum
78
f
fx
11
22
38
57
31
19
15
12
11
9
7
5
237
11
44
114
228
155
114
105
96
99
90
77
60
1193
f x2
11
88
342
912
775
684
735
768
891
900
847
720
7673
 x  1193 106  / 237  5.034 106  cycles


 
2
7673 1012  1193 106  / 237


ˆ
x 
 2.658 106 cycles
237  1
C x  2.658 / 5.034  0.528
From Eqs. (20-18) and (20-19),
 y  ln 5.034 106   0.5282 / 2  15.292


 
 
ˆ
 y  ln 1  0.5282   0.496
From Eq. (20-17), defining g(x),
 1  ln x  15.292  2 
1
g ( x) 
exp   
 
0.496  
x  0.496 2
 2


x (Mrev) f / (Nw)
0.5
0.000 00
g(x)(106)
0.000 11
Chapter 20, Page 29/29

30. 0.5
1.5
1.5
2.5
2.5
3.5
3.5
4.5
4.5
5.5
5.5
6.5
6.5
7.5
7.5
8.5
8.5
9.5
9.5
10.5
10.5
11.5
11.5
12.5
12.5
0.046414
0.046414
0.092827
0.092827
0.160338
0.160338
0.240506
0.240506
0.130802
0.130802
0.080 17
0.080 17
0.063 29
0.063 29
0.050 63
0.050 63
0.046 41
0.046 41
0.037 97
0.037 97
0.029 54
0.029 54
0.021 10
0.021 10
0.000 00
0.000 11
0.052 03
0.052 03
0.169 92
0.169 92
0.207 54
0.207 54
0.178 47
0.178 47
0.131 58
0.13158
0.090 11
0.090 11
0.059 53
0.059 53
0.038 69
0.038 69
0.025 01
0.025 01
0.016 18
0.016 18
0.010 51
0.010 51
0.006 87
0.006 87
z
ln x   y
ˆ
 ln x   y   y z  15.292  0.496 z
ˆ
y
L 10 life, where 10% of bearings fail, from Table A-10,
z = 1.282. Thus,
ln x = 15.292 + 0.496(1.282) = 14.66
 x = 2.33 (106) rev Ans.
Chapter 20, Page 30/29