Example 1: Fatigue life, in hundreds of hours, for a certain type of bearing has

Question

Example 1: Fatigue life, in hundreds of hours, for a certain type of bearing has a uniform distribution between 16 and (a) What is the expected value of the fatigue life? (b) What is the standard deviation of the fatigue life? (c) What is the cumulative distribution function of the fatigue life? (d) What is the probability that a bearing of this type fails in less than 1800 hours? (e) What is the expected value and standard deviation of the total fatigue life for two randomly selected bearings? (f A random sample of 100 bearings was obtained. What is the expectation and variance of the number of bearings with fatigue life less than 1800 hours?

Explanation / Answer

a) expected value =(a+b)/2 =(16+36)/2 =26

b)std deviation =(b-a)/(!2)1/2 =5.7735

c)CDF =P(X<x) =(x-16)/20

d)P(X<18) =(18-16)/20 =0.1

e)expected value =26

std error =std deviation/(n)1/2 =4.0825

f)expected life =26

std error =std deviation/(n)1/2 =0.5774

g)hence from centrallimit theorum P(X<18) =P(Z<(18-26)/0.5774)= P(Z<13.856)=0.0000

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