Question

The life (in hours) of a computer processing unit (CPU) is modeled by a Weibull distribution with parameters beta equals 3 and delta equals 1047.

Determine the mean life of the CPU (in hours)

Answer 1

The Weibull distribution has the density function
\[f(x;k,\lambda)=\begin{cases}\dfrac{k}{\lambda}\left(\dfrac{x}{\lambda}\right)^{k-1}\exp\left[-\left(\dfrac{x}{\lambda}\right)^{k}\right]&\text{for }x\ge0\\0&\text{for }x<0\end{cases}\]
(Formula taken from the Wikipedia page - I'm not sure which is \(\beta\) and \(\delta\))

The mean will be given by
\[\int_{0}^\infty xf(x)\,dx\]

Answer 2

ok . That is what I did , but the stupid wiley plus didnt accept my answer

Answer 3

So is \(\beta=k\) and \(\delta=\lambda\), or the other way around?

Answer 4

thats it

Answer 5

Alright, so the mean would be given by
\[\int_0^\infty \dfrac{\beta}{\delta}\left(\dfrac{x}{\delta}\right)^{\beta}\exp\left[-\left(\dfrac{x}{\delta}\right)^{\beta}\right]\,dx\]
I'll omit the actual values of \(\beta\) and \(\delta\) for now.

Answer 6

Actually, there's a slight typo above, should be
\[\int_0^\infty \dfrac{\beta}{\delta}x\left(\dfrac{x}{\delta}\right)^{\beta-1}\exp\left[-\left(\dfrac{x}{\delta}\right)^{\beta}\right]\,dx\]
Substitute \(u=-\left(\dfrac{x}{\delta}\right)^\beta\), then \(du=-\dfrac{\beta}{\delta}\left(\dfrac{x}{\delta}\right)^{\beta-1}\,dx\). Now,
\[\int_0^\infty \dfrac{\beta}{\delta}x\left(\dfrac{x}{\delta}\right)^{\beta-1}\exp\left[-\left(\dfrac{x}{\delta}\right)^{\beta}\right]\,dx=-\int_{0}^{-\infty} \delta\left(-u\right)^{1/\beta}\exp(u)\,du\]
Substitute again, \(u=-t\) so that \(du=-dt\), then you get
\[\delta\int_{0}^{\infty}t^{1/\beta}\exp(-t)\,dt\]
Is this what you have?

Answer 7

no not exactly

Answer 8

I think I might have messed up on the u-sub

Answer 9

When you've caught up, you'll need to know the gamma function to be able to find an exact value for the integral.
\[\Gamma(n)=\int_0^\infty t^{n-1}\exp(-t)\,dt\]
which almost perfectly matches what we have above. You just have to tweak the exponent a bit.