Question

A Component has an exponential time-to-failure distribution with mean of 10,000 hours. a) The component has already been in operation for its mean life. What is the probability that it will fail by 15,000 hours? b) At 15,000 hours the component is still in operation. What is the probability that it will operate for another 5000 hours?​

Answer 1

Answer:

{eq}\text{ Exponential probability distribution has the next behavior: } \\[0.25cm]\\ \displaystyle f(x) \; = \; .

Answer 2

Answer:

Step-by-step explanation:

The exponential probability distribution has the following behavior:

$\\[0.25cm]\\ \displaystyle f(x) \; = \; \left\{\begin{matrix} \frac{ 1 }{ \beta } \; {\rm e}^{ -x / \beta } & 0 \; < \; x \; < \; \infty \\ 0 & x < 0 \end{matrix}\right. \\[0.25cm]\\ \text{ Where: } \\[0.25cm]\\ \displaystyle \mu \; = \; E(x) \; = \; \beta \; = \; \frac{1}{ \lambda} \\[0.25cm]\\ {/eq}$

Then, with $\mu=10,000$

We write:

$\displaystyle P( \; 10000 \leq x \leq 15000 \; ) \; = \; \int_{ 10000 }^{ 15000 } \; \frac{ 1 }{ 10000 } \; {\rm e}^{ -x/ 10000 } \; dx \\[0.25cm]$

$\displaystyle P( \; 10000 \leq x \leq 15000 \; ) \; = \; -{{\rm e}^{-x/ 10000 }} \bigg|_{ 10000 }^{ 15000 } \\[0.25cm]\\ \displaystyle P( \; 10000 \leq x \leq 15000 \; ) \; = \; -{{\rm e}^{- 15000 / 10000 }} \; - \; ( -{{\rm e}^{- 10000 / 10000 }} \; ) \\[0.25cm]$

$\displaystyle P( \; 10000 \leq x \leq 15000 \; ) \; = \; -{{\rm e}^{- 15000 / 10000 }} \; + \; {{\rm e}^{- 10000 / 10000 }} \\[0.25cm]\\ \displaystyle P( \; 10000 \leq x \leq 15000 \; ) \; = \; -0.22313016 + 0.367879441 \\[0.25cm]\\ \displaystyle P( \; 10000 \leq x \leq 15000 \; ) \; = \; 0.144749281 \\[0.25cm]\\ \displaystyle P( \; 10000 \leq x \leq 15000 \; ) \; = \; 0.1447 \\[0.25cm]\\ \displaystyle P( \; 10000 \leq x \leq 15000 \; ) \; = \; 14.47 \; \%$

We estimate that the remaining useful life of the component is close to 14%. So if this component is critical in our process, its timely replacement should be considered.