Question

Time to failure of a household refrigerator. The time to failure of a particular refrigerator type is represented by the following pdf: , which is valid within 0 ≤ t ≤ 10 yr, and f(t) = 0 elsewhere. a) Write the expression for R(t), integrate over t from t to infinity (which here is 10), and obtain the cumulative Reliability function, R(t). Then calculate the reliability for the first year, t = 1. Round your calculated value to 2 sd (significant digits). b) Assuming that 0.90 reliability is required, calculate the refrigerator Design Life corresponding to 0.90 reliability. Round your calculated value to 2 sd. c) Write the expression for and obtain the conditional failure rate (hazard rate) function, λ(t), and state whether it is increasing or decreasing as t approaches infinity for this case. d) To evaluate the reliability of this refrigerator, write the expression for and calculate the MTTF, mean time to failure to two sd. e) Obtain the probability and thereby the % units, or relative frequency of occurrence, of this refrigerator that are expected to survive its MTTF. f) The refrigerator company has a 1-month warranty program. Write the expression for and obtain the probability that the refrigerator will fail during the first month. F(1/12) = g) Write the expression for and calculate the probability that the refrigerator will survive the 1-month warranty program. Then write the expression for and calculate the (A B) B) f(t) = 0.003(10 − t) 2 4 conditional probability that the refrigerator will survive the rest of the first year given the probability that it will survive its warranty period of one month. Round your final result to 2 sd. R(1/12) = R(11/12|1/12) =

Answer 1

$f(t)=\left\{\begin{matrix} 0.003(10-t)^2 &0\leq t\leq10 \\ 0 & otherwise \end{matrix}\right.$

(a)

$R(t)=\int_{t}^{10}f(t)dt=\int_{t}^{10}0.003(10-t)^2dt=0.001(10-t)^3$

$\therefore R(1)=0.001(10-1)^3=0.729\approx 0.73$

(b)

$\texttt{Given:-}$

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!R(t)=0.90\Rightarrow0.001(10-t)^3=0.9\Rightarrow (10-t)^3=900 \Rightarrow t=10-\sqrt[3]{900}=0.3451\approx 0.35$

(c)

$\lambda(t)=\frac{f(t)}{R(t)}=\frac{0.003(10-t)^2}{0.001(10-t)^3}=\frac{3}{10-t}$

$\lambda'(t)=\frac{\mathrm{d} [\frac{3}{10-t}]}{\mathrm{d} t}=-\frac{3}{(10-t)^2}<0\;\;\;\;\;\forall\;\;0\leq t\leq10$

$\therefore\!\texttt{ t\;is\;decreasing\;as\;t\;approaches\;10}.$

(d)

$R'(t)=\frac{\mathrm{d} [0.001(10-t)^3]}{\mathrm{d} t}=-0.003(10-t)^2$

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!MTTF=-\int_{0}^{10}tR'(t)dt=0.003\int_{0}^{10}t(10-t)^2dt=0.003\int_{0}^{10}(100t-2t^2&plus;t^3)dt=0.003\times\frac{20500}{3}=20.5$