Question

For a component with a log normal "time to failure" with mean time to failure=2days and standard deviation of time to failure=0.2days, find: a) R(3 days) b) h(3 days) c) R(5 days) d) h(5 days)

Answer 1

a)

Reliability function, R(t) = P[Z > (ln t - ) /]

= P[Z > (ln t - 2) /0.2]

R(3 days) = P[Z > (ln 3 - 2) /0.2] = P[Z > -4.5] = 0.9999967

b)

Hazard function, h(t) =

=

h(3 days) =

=

= 0.000003288492 / 0.599998

= 5.480838 x 10^-6

c)

R(5 days) = P[Z > (ln 5 - 2) /0.2] = P[Z > -1.95281] = 0.9745789

b)

Hazard function, h(t) =

h(5 days) =

=

= 0.02542106 / 0.9745789

= 0.02608415