A shaft is to be assembled inside of a bearing. The mean inside diameter of the

Question

A shaft is to be assembled inside of a bearing. The mean inside diameter of the bearing is 3.25 with a variance of .0010. The mean outside diameter of the shaft is 3.195 with a variance of .0013. a) When two parts are assembled at random, what is the probability that the parts will interfere? b) When two parts are assembled at random, what percentage of the parts will have a clearance of at least .02? c) Suppose that the clearance when the parts are assembled is supposed to be at least .055. What should be the mean outside diameter of the shaft, if we want 99 % of the assemblies to meet the .055 clearance?

Explanation / Answer

let A is dia of bearing and B for shaft

hence std error of differnce =(s12+s22 )1/2 =(0.0010+0.0013)1/2 =0.048

also mean differnce of A and B =A-B =3.25-3.195 =0.055

a) P(A-B<0) =P(Z<(0-0.055)/0.048)=P(Z<-1.1468)=0.1257

b)P(A-B>0.02)=1-P(A-B<0.02)=1-P(Z<(0.02-0.055)/0.048)=1-P(Z<-0.7298)=1-0.2328=0.7672 ~76.72%

c)P(A-B>0.055)>0.99

P(A-B<0.055)<0.01

(0.055-(A-B))/0.048)<-2.326

A-B>0.1666

B<3.25-0.1666

B<3.0834

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