Question

The newest CREE led has a life expectancy of mu = 50000 hours and its life probability density function is given by: f(t) = [e^(-t/mu)]/[mu] if t greater or = 0 and f(t) = 0 if t < 0. Calculate the chance that a led will last at least tau = 100000.

Answer 1

Answer:

change that a lead is 0.13533

Explanation:

µ  = 50000

f(t) = [e^(-t/µ )]/[µ      if  t ≥ 0

f(t) = 0  if  t < 0

τ = 100000

to find out

the chance that a led will last

solution

we know function is f(t) = [e^(-τ/µ)]/[µ]    

τ = 100000

so we can say that probability (τ  ≥ 100000 ) that is

= 1 - Probability ( τ ≤ 100000 )

that is function of F so

= 1 - f ( 100000 )

that will be

= 1 - ( 1 - [e^(-τ/µ)]/[µ]   )

put all value here τ = 100000 and µ = 50000

= 1 - ( 1 - [e^(-100000/50000)]  )

= 1 - 1 - [e^(-100000/50000)]

= 0.13533

so that change that a lead is 0.13533