The pdf of timetofailure for a household appliance is ft 20

The pdf of time-to-failure for a household appliance is f(t) = 200/(t + 10)^3 where t is in years. The appliance is being marketed with a 1-year warranty. a) Find the reliability function, the failure rate and MTTF for the appliance b) Find the probability that the appliance will survive the warranty period. c) What is the design life of the appliance if it is expected to operate with a minimum reliability of 0.95?

Solution

Use I[a, b] for the integral from a to b and E[a, b] for the evaluation from a to b

a) R(t) = 1 - I[0, t] f(t) dt = 1 - I[0, t] 200/(10+t)³ dt = 1 + 100/(10+t)² E[0, t] =

1 + 100/(10+t)² - 100/(10+0)² = 100/(10+t)²

lambda(t) = f(t)/R(t) = 200/(10+t)³ /(100/(10+t)²)   = 2/(10+t)

MTTF = I[0, inf] t f(t) dt = I[0, t] t 200/(10+t)³ dt = I[0, t] (t +10-10) 200/(10+t)³ dt =

-200/(10+t) + 1000/(10+t)²E[0, inf] =

0 - (-20 + 10) =10

b) The warranty period is 1 year

R(1) =  100/(10+t)² = 100/(10+1)² = 100/121 = 0.826446280991736

c) R(t) = .95

100/(10+t)² = .95

100/.95 = (10+t)²

t = sqrt(100/.95) - 10 = 100/sqrt(95) - 10

t = 0.259783520851542