Question

Consider a system with one component that is subject to failure, and suppose that we have 115 copies of the component. Suppose further that the lifespan of each copy is an independent exponential random variable with mean 20 days, and that we replace the component with a new copy immediately when it fails.
(a) Approximate the probability that the system is still working after 3500 days.
(b) Now, suppose that the time to replace the component is a random variable that is uniformly distributed over (0, 0.5). Approximate the probability that the system is still working after 4125 days.

Answer 1

Answer:

Step-by-step explanation:

From the given information:

the mean $(\mu) = 115 \times 20$

= 2300

Standard deviation = $20 \times \sqrt{115}$

Standard deviation (SD) = 214.4761

TO find:

a) $P(x > 3500)= P(Z > \dfrac{3500-\mu}{214.4761})$

$P(x > 3500)= P(Z > \dfrac{3500-2300}{214.4761})$

$P(x > 3500)= P(Z > \dfrac{1200}{214.4761})$

$P(x > 3500)= P(Z >5.595)$

From the Z-table, since 5.595 is > 3.999

$P(x > 3500)=1-0.9999$

P(x > 3500) = 0.0001

b)

Here, the replacement time for the mean $(\mu) = \dfrac{0+0.5}{2}$

= 0.25

Replacement time for the Standard deviation $\sigma = \dfrac{0.5-0}{\sqrt{12}}$

$\sigma = 0.1443$

For 115 component, the mean time = (115 × 20)+(114×0.25)

= 2300 + 28.5

= 2328.5

Standard deviation = $\sqrt{(115\times 20^2) +(114\times (0.1443)^2)}$

= $\sqrt{(115\times 400) +(114\times 0.02082249}$

= $\sqrt{(46000) +2.37376386}$

= $\sqrt{(46000) +(2.37376386)}$

= $\sqrt{46002.374}$

= 214.482

Now; the required probability:

$P(x > 4125) = P(Z > \dfrac{4125- 2328.5}{214.482})$

$P(x > 4125) = P(Z > \dfrac{1796.5}{214.482})$

$P(x > 4125) = P(Z >8.376)$

$P(x > 4125) =1- P(Z$

From the Z-table, since 8.376 is > 3.999

P(x > 4125) = 1 - 0.9999

P(x > 4125) = 0.0001