Question 1 You work for a company that has 46 identical machines operating to pr

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Question 1 You work for a company that has 46 identical machines operating to produce Justin Bieber CDs. These machines operate as long as components in particular, call them A, B, and C. A machine operates as long as C an and B are working. Specifications from the manufactures tell you that the failure time of each part is Normally distributed as follows certain components are working, There are three d A or C A's failure time ~ N( = 9 months, 2-8 months) B's-N( 13,02 = 5) C's ~ N( = 24,02-9) Assum e that the parts failure times are independent. Your boss asks you to find how many machines can the company expect to fail in a year and the probability that a least 10 machines fail in a year.

Explanation / Answer

Q 1)

there are N=46 machines

each machine has 3 components A , B and C

now the component's failure time( in months) follows normal distributions

A~N(9,8) B~N(13,5) C~N(24,9) independently

a machine operates as long as C and A or C and B operates

now if A fails only then C and B will be operative hence the machine will be also operative.

similarly if B fails only then C and A will be operative hence the machine will be also operative.

hence the machine will fail if C fails or A and B fails together with C operating.

so P[C failing within a year]=P[failure time of C is less than 12 months]=P[C<12] where C~N(24,9)

=P[(C-9)/sqrt(9)<(12-24)/sqrt(9)]=P[Z<-3]=0.0013499

now P[A and B fails but C is operative within 12 months]

=P[A<12]*P[B<12]*P[C>12] since they are independent.

=P[(A-9)/sqrt(8)<(12-9)/sqrt(8)]*P[(B-13)/sqrt(5)<(12-13)/sqrt(5)]*[1-P[C<12]]

=P[Z<1.06066]*P[Z<-0.4472]*[1-0.0013499]

=0.855578*0.327365*0.9986501=0.2797082

where Z~N(0,1)

hence the probability of failure of a machine within a year is P[C<12]+P[A<12]*P[B<12]*P[C>12]

=0.0013499+0.2797082=0.2810581

hence the expected number of machines to fail within a year is N*0.2810581=46*0.2810581=12.9286726=13 machines (approx) [answer]

let X denotes the number of machines out of 46 machines that fail withina year.

hence X~Bin(46,0.2810581)

but since N=46 is greater than 25 the distribution of X can be approximated by a normal distribution.

now E[X]=46*0.2810581=12.9286726 and V[X]=46*0.2810581*(1-0.2810581)=9.29496

hence X~N(12.928676,9.29496)

hence the probability that at leats 10 machines will fail is

P[X>10]=1-P[X<10]=1-P[(X-12.928676)/sqrt(9.29496)<(10-12.928676)/sqrt(9.29496)]=1-P[Z<-0.9606]=1-0.168377=0.831623 [answer]

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