Question

3.3.26. Let X denote time until failure of a device and let $r(x)$ denote the hazard function of X. (a) If $r(x) = cx^b$; where $c$ and $b$ are positive constants, show that X has a Weibull distribution; i.e., $f(x) = \begin{cases} cx^b \exp \{ \frac{-cx^{b+1}}{b+1} \} & 0 < x < \infty \ 0 & \text{elsewhere.} \end{cases}$ (3.3.12)

          3.3.26. Let X denote time until failure of a device and let $r(x)$ denote the hazard
function of X.
(a) If $r(x) = cx^b$; where $c$ and $b$ are positive constants, show that X has a Weibull
distribution; i.e.,
$f(x) = \begin{cases} cx^b \exp \{ \frac{-cx^{b+1}}{b+1} \} & 0 < x < \infty \
0 & \text{elsewhere.} \end{cases}$ (3.3.12)

$3.3.26. Let X denote time until failure of a device and let r(x) denote the hazard function of X. (a) If r(x) = cx^b; where c and b are positive constants, show that X has a Weibull distribution; i.e., f(x) = cx^b {(-cx^b+1)/(b+1)} 0 < x < ∞0 elsewhere. (3.3.12)$

Added by Christina F.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Ameer Said

03/19/2022

Step 1

Step 1: The hazard function is given by r(x) = cx^b. Show more…

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3.3.26. Let X denote time until failure of a device and let r(x) denote the hazard function of X. (a) If r(x) = cx^b; where c and b are positive constants, show that X has a Weibull distribution; i.e., $$f(x) = \begin{cases} cx^b \exp\left\{-\frac{cx^{b+1}}{b+1}\right\} & 0 < x < \infty \\ 0 & \text{elsewhere}. \end{cases}$$ (3.3.12)