Respuesta :
Answer:
a.) To observe one component fail it would not be unusual
b.) it would be unusual to have two components fail ( for a significance level of 0.05)
c.) A parallel structure will succeed with 2 identical components with a probability of 0.9711.
d.) The structures needed will be 6.
Step-by-step explanation:
It is given that failure of the components is independent of each other.
a.) The failure of each component is with a probability of 0.17.
In general we assume a significance level of 0.05 and since the probability of failure of one event is greater than the significance level.
Therefore we can say that it is not unusual to observe one component fail.
b.) The probability that two components fail is = (0.17)×(0.17) = 0.0289.
Here we can see that the probability of failure of two events is less than the significance level of 0.05.
So for a significance level of 0.05 it would be unusual to have two components fail.
For a significance level of 0.01 it would not be unusual for two components to fail.
c.) A parallel structures having 2 identical components to succeed the probability = 1 - P(2 components failing)
= 1 - 0.0289
= 0.9711
d.) It is required to find the number of components that would be need in the structure so that system succeeds with a probability greater than 0.9999.
So we can write
1 - [tex](0.17)^n[/tex] > 0.9999
⇒ [tex](0.17)^n[/tex] > 0.0001
⇒ n > [tex]\frac{\log{0.0001}}{\log{0.17}} = 5.198[/tex]
Therefore the number of structures needed will be 6.
