Final answer:
(a) the reliability function for each of the five individual stations is R(t) = e^(-0.67t).
(b) the reliability function for the assembly line is R(t) = R1(t) * R2(t) * R3(t) * R4(t) * R5(t).
(c) the mean time to failure for the assembly line is 1/(λ1 + λ2 + λ3 + λ4 + λ5).
(d) the hazard function for the assembly line is h(t) = λ.
Step-by-step explanation:
(a) Station 1: The mean time to failure (MTTF) is 1.5, which means that the failure rate (λ) is the reciprocal of the MTTF, i.e., λ = 1/1.5 = 0.67.
The reliability function (R(t)) for exponential distribution is given by R(t) = e^(-λt), where t is the time.
Therefore, the reliability function for Station 1 is R(t) = e^(-0.67t).
(b) Assembly line: The assembly line consists of five independent stations.
The reliability of the assembly line is the product of the reliabilities of each station.
Therefore, the reliability function for the assembly line is R(t) = R1(t) * R2(t) * R3(t) * R4(t) * R5(t).
(c) Mean time to failure: The mean time to failure for the assembly line is the reciprocal of the sum of the failure rates of each station, i.e., MTTF = 1/(λ1 + λ2 + λ3 + λ4 + λ5).
(d) Hazard function: The hazard function (h(t)) for exponential distribution is given by h(t) = λ, where λ is the failure rate.
Therefore, the hazard function for the assembly line remains the same as the failure rate, i.e., h(t) = λ.