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

To approximate the probability that the system is still working after 3500 days, calculate P(X > 3500) using the exponential distribution formula. To approximate the probability that the system is still working after 4125 days with a random replacement time uniformly distributed between 0 and 0.5, consider two cases: when the component fails before the replacement time and when it fails after the replacement time. Take the weighted sum of the probabilities from the two cases to find the total probability.

Step-by-step explanation:

To approximate the probability that the system is still working after 3500 days, we need to find the probability that each copy of the component lasts longer than 3500 days. Let X denote the lifespan of a component. Since X is exponentially distributed with mean 20 days, the probability that a component lasts longer than 3500 days is given by P(X > 3500). Using the exponential distribution formula, this probability can be calculated as follows:

P(X > 3500) = e^(-3500/20) ≈ 0.00025.

To approximate the probability that the system is still working after 4125 days with a random replacement time uniformly distributed between 0 and 0.5, we need to consider two cases:

Case 1: The component fails before the replacement time (0.5 days). In this case, the component is replaced immediately, and the remaining time until 4125 days is 4125 - (failure time). The probability of this case is given by:

P(failure before replacement time) = 1 - P(X > 0.5) = 1 - e^(-0.5/20) ≈ 0.024.

Case 2: The component fails after the replacement time (0.5 days). In this case, the remaining time until 4125 days is 4125 - 0.5 = 4124.5 days. The probability of this case is simply:

P(failure after replacement time) = 1.

The total probability that the system is still working after 4125 days can be calculated by taking the weighted sum of the probabilities from the two cases:

P(system still working after 4125 days) = P(failure before replacement time) * P(remaining time > 4124.5) + P(failure after replacement time) * P(remaining time > 4124.5).

Answer 2

Step-by-step explanation:

From the given information:

the mean
`(\mu) = 115 * 20`

= 2300

Standard deviation =
`20 * √(115)`

Standard deviation (SD) = 214.4761

TO find:

a)
`P(x > 3500)= P(Z > (3500-\mu)/(214.4761))`

`P(x > 3500)= P(Z > (3500-2300)/(214.4761))`

`P(x > 3500)= P(Z > (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) = (0+0.5)/(2)`

= 0.25

Replacement time for the Standard deviation
`\sigma = (0.5-0)/(√(12))`

`\sigma = 0.1443`

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

= 2300 + 28.5

= 2328.5

Standard deviation =
`√((115* 20^2) +(114* (0.1443)^2))`

=
`√((115* 400) +(114* 0.02082249)`

=
`√((46000) +2.37376386)`

=
`√((46000) +(2.37376386))`

=
`√(46002.374)`

= 214.482

Now; the required probability:

`P(x > 4125) = P(Z > (4125- 2328.5)/(214.482))`

`P(x > 4125) = P(Z > (1796.5)/(214.482))`

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

`P(x > 4125) =1- P(Z <8.376)`

From the Z-table, since 8.376 is > 3.999

P(x > 4125) = 1 - 0.9999

P(x > 4125) = 0.0001