Answer :
Answer:
P(0≤T≤400)=0.3296
Step-by-step explanation:
we know probability density function is
[tex]f(x;\lambda)= \left\{\begin{matrix} \lambda e^{-\lambda x}& x\geq 0 \\0 & x< 0\end{matrix}\right.[/tex]
[tex]\lambda=\frac{1}{\mu}[/tex]
[tex]\lambda=\frac{1}{1000}[/tex]
probability that bulb will fail within first 400 hours
P(0≤T≤400)= [tex]\int\limits^{400}_0 {f(x)} \, dx[/tex]
P(0≤T≤400)= [tex]\int\limits^{400}_0 {\frac{e^{-\frac{t}{1000}}}{1000}} \, dx[/tex]
=[tex](1000)^{-1}[\frac{e^{\frac{-t}{1000}}}{-(1000^{-1})}]^{400}_0[/tex]
=[tex]-[e^{\frac{-400}{1000}}-e^0][/tex]
=1-0.67032 = 0.3296
so about 32.96 % bulb will fail within 400 hrs.