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.

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