Question

Assume that the life (X) of an airbag system follows a Weibull distribution with B=0.5 and the mean life (p) is 200 months. Weibull Distribution (pdf): 2 f(2) B 8 -G for r20 (a) What is the probability that an airbag system lasts at least 200 months? (b) What is the probability that an airbag system fails between 120 months and 150 months?

Answer 1

Sal? : Assume that the life (x) of an air bag system follows a Weibull distribution with B=0.5 and mean life (M) is 200 monts. The pdf of weibull distribution is B B fix) = L. (7)B-12-13)*, x>0 → mean = E(X) = 8 MT+) = 200 (given) 키 & [1 + 1 ) ::=0.5 d. B 2200 8.2! - 200 ñ í & - 100 f czej = 0.5 o de jour-le-Lajos = 0.005 The e () f(x) -0.05 e To xy,o Vz (as Probability that an airbag system lasts at least 200 monts = P(X), 200] jo f(x) dx 200

0 = P(x>200] = 3 0.05* 0541 in è de 200 but Vi er 11 u x=200 u=2 da = du dora > P {x>, 200] = S 0.05 * 0.05 é u * 20 du u=2 as é " di u=2 00 (94) - O + e-re U=V2 → P(X), 200] - e-2 = 0.2431 Probability that an airbag system lasts at beast 200 monts cp = 0.2431

an airbag (b) Probability that system fails between 120 monts and 150 monts = P L 120 < x <150) 150 =, frees de 12o BC e = 0.05 s toe dae 120 .: 2-12 x = 150 U- 10 1750 To > PL1204+ 21.50] = 0:05 Jogo é " die - u: V12 10 150 = TEM U 120 10 120 - (150 é e-vie - P{120<x<150] = 0.0406 » Probability that an air bag system fails between 120 ments and 150 monts is =0.0406