Question 24.13: The life of a ball bearing is a normally distributed random ...

The life of a ball bearing is a normally distributed random variable, with a mean of 10 000 h and a standard deviation of 500 h. The manufacturer of the bearings wants to give a guarantee that 90% of the bearings will reach or exceed the rated life published in his catalogue. What should be the rated life?

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\text { Given } \quad \mu=10000 h \quad \hat{\sigma}=500 h \quad R=90 \% .

Step I Standard variable Z_{0} for 90 % reliability The probability of survival of the bearings or reliability is 90% or 0.9. As shown in Fig. 24.19, the shaded area below the normal curve from Z=Z_{0}  to Z = + ∞ should be 0.9. The area below the normal curve from Z = 0 to Z = + ∞ is 0.5.
Therefore, the area below the normal curve from Z = 0 to Z = -Z_{0} should be 0.4. From Table 24.6, the corresponding value of Z_{0} is approximately 1.28.
Therefore,

Table 24.6 Areas under normal curve from 0 to Z

9 8 7 6 5 4 3 2 1 0 Z
.0359 .0319 .0279 .0239 .0199 .0160 .0120 .0080 .0040 .0000 0.0
.0754 .0714 .0675 .0636 .0596 .0557 .0517 .0478 .0438 .0398 0.1
.1141 .1103 .1064 .1026 .0987 .0948 .0910 .0871 .0832 .0793 0.2
.1517 .1480 .1443 .1406 .1368 .1331 .1293 .1255 .1217 .1179 0.3
.1879 .1844 .1808 .1772 .1736 .1700 .1664 .1628 .1591 .1554 0.4
.2224 .2190 .2157 .2123 .2088 .2054 .2019 .1985 .1950 .1915 0.5
.2549 .2518 .2486 .2454 .2422 .2389 .2357 .2324 .2291 .2258 0.6
.2852 .2823 .2794 .2764 .2734 .2704 .2673 .2642 .2612 .2580 0.7
.3133 .3106 .3078 .3051 .3023 .2996 .2967 .2939 .2910 .2881 0.8
.3389 .3365 .3340 .3315 .3289 .3264 .3238 .3212 .3186 .3159 0.9
.3621 .3599 .3577 .3554 .3531 .3508 .3485 .3461 .3438 .3413 1.0
.3830 .3810 .3790 .3770 .3749 .3729 .3708 .3686 .3665 .3643 1.1
.4015 .3997 .3980 .3962 .3944 .3925 .3907 .3888 .3869 .3849 1.2
.4177 .4162 .4147 .4131 .4115 .4099 .4082 .4066 .4049 .4032 1.3
.4319 .4306 .4292 .4279 .4265 .4251 .4236 .4222 .4207 .4192 1.4
.4441 .4429 .4418 .4406 .4394 .4382 .4370 .4357 .4345 .4332 1.5
.4545 .4535 .4525 .4515 .4505 .4495 .4484 .4474 .4463 .4452 1.6
.4633 .4625 .4616 .4608 .4599 .4591 .4582 .4573 .4564 .4554 1.7
.4706 .4699 .4693 .4686 .4678 .4671 .4664 .4656 .4649 .4641 1.8
.4767 .4761 .4756 .4750 .4744 .4738 .4732 .4726 .4719 .4713 1.9
.4817 .4812 .4808 .4803 .4798 .4793 .4788 .4783 .4778 .4772 2.0
.4857 .4854 .4850 .4846 .4842 .4838 .4834 .4830 .4826 .4821 2.1
.4890 .4887 .4884 .4881 .4878 .4875 .4871 .4868 .4864 .4861 2.2
.4916 .4913 .4911 .4909 .4906 .4904 .4901 .4898 .4896 .4893 2.3
.4936 .4934 .4932 .4931 .4929 .4927 .4925 .4922 .4920 .4918 2.4
.4952 .4951 .4949 .4948 .4946 .4945 .4943 .4941 .4940 .4938 2.5
.4964 .4963 .4962 .4961 .4960 .4959 .4957 .4956 .4955 .4953 2.6
.4974 .4973 .4972 .4971 .4970 .4969 .4968 .4967 .4966 .4965 2.7
.4981 .4980 .4979 .4979 .4978 .4977 .4977 .4976 .4975 .4974 2.8
.4986 .4986 .4985 .4985 .4984 .4984 .4983 .4982 .4982 .4981 2.9
.4990 .4990 .4989 .4989 .4989 .4988 .4988 .4987 .4987 .4987 3.0
.4993 .4993 .4992 .4992 .4992 .4992 .4991 .4991 .4991 .4990 3.1
.4995 .4995 .4995 .4994 .4994 .4994 .4994 .4994 .4993 .4993 3.2
.4997 .4996 .4996 .4996 .4996 .4996 .4996 .4995 .4995 .4995 3.3
.4998 .4997 .4997 .4997 .4997 .4997 .4997 .4997 .4997 .4997 3.4
.4998 .4998 .4998 .4998 .4998 .4998 .4998 .4998 .4998 .4998 3.5
.4999 .4999 .4999 .4999 .4999 .4999 .4999 .4999 .4998 .4998 3.6
.4999 .4999 .4999 .4999 .4999 .4999 .4999 .4999 .4999 .4999 3.7
.4999 .4999 .4999 .4999 .4999 .4999 .4999 .4999 .4999 .4999 3.8
.5000 .5000 .5000 .5000 .5000 .5000 .5000 .5000 .5000 .5000 3.9

Z_{0}=-1.28 .

Step II Rated life of bearings

Z_{0}=\frac{X_{0}-\mu}{\hat{\sigma}} \text { or }-1.28=\frac{X_{0}-10000}{500} .

\text { or } X_{0}=9360 h .

Therefore 90% of the bearings will complete or exceed a life of 9360 h before fatigue failure.

24.19

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