Extract
[...] Exercise 7 1. Cumulative total XiYi 0 ; 40 9 0,13 0,87 -7,38 40 ; 50 20 0,28 0,72 -4,3401 50 ; 60 36 0,507 0,49 -1,4315 60 ; 70 52 0,73 0,27 1,19 70 ; 80 63 0,89 0,11 3,4164 80 ; 90 68 0,96 0,04 5,175 90 ; 100 70 0,99 0,01 6,68 TOTAL 70 3,31 We know that So be F(t1) = = 0,127 We apply this formula for the t's that follow. We know that = 1 - So be R(t1) = 1 - F(t1) = 1-0,13 = 0,87 2. [...]
[...] The reliability function is defined by: = P(T > = for t ? 0. 4. a P(X345) = 1 - P(T ? 345) (0,002)(345) 0,69 = 0,5016 We have 100 x 0.5016 = 50.16; that is 50 components." Exercise S = 1-e-0.9t x e-0.95t x e-0.75t ) x ( 1-e-0.9t x e-0.95t x e-0.75t ) So let S = 1-e-0.9t x e-0.95t x e-0.75t P = ( 1 - ( 1-e-0.9t) x2) x - ( 1-e-0.95t) x2) x - ( 1-e-0.75t) x2) Exercice 4 1. [...]
[...] ? - 0.0266ln. ?=(0.0266 x 3.69) - Ln. ? = (2,098154/0,0266) - Ln. ? = 78,878 - ? = e78,878 3. We refer to the table F(t)=0.89 for 70;80 and F(t)=0.96 for 80;90. We deduce that for F(80)=0.90. Let 80 days. [...]
[...] Statistics and System Reliability Exercise We consider X to be var equal to the color of a randomly selected candy. We observe a sample n=370; with 6 classes of value Brun: B=C1; Yellow: J=C2; Red: R=C3; Orange: O=C4; Green: V=C5; Gold: D=C6 We assign the following respective values: n1=84; n2=79; n3=75; n4=49; n5=36; n6=47 We consider the following hypotheses: HO: X follows the probability distribution described by the responsible person H1: X does not follow the probability distribution described by the responsible person Let R be the var following the HO law, according to the statement we have: P1=0.3; P2=0.2; P3=0.2; P4=0.1; P5=0.1; and P6=0.1 We follow the following commands: Nb=c(84,79,75,49,36,47) Proba=c( 0.1) Chisq.test(nb,p=proba)$p.value We obtain the following result 0.018807 P is between 0.01 and 0.05, so H0 is rejected. [...]
