1. A system consists of 5 identical components, all of which must operate for system success. If the failure rate and average repair time of each component is 0.01 failures/year and 40 hours respectively, calculate the system failure rate, average down time and unavailability.

2. The system in Problem 1 is reinforced by connecting a second identical system in parallel with the first. Evaluate the new system failure rate, average down time and unavailability.

Step-by-step explanation:

Given;

If the failure rate and average repair time of each component is 0.01 failures/year and 40 hours respectively,

Failure Rate for each component = ( Number of Failures / Hours ) = 0.01 / (365 x 24 ) = 1.14 x 10⁻⁶ failures per hour.

Failure Rate for system = 5 x 1.14 x 10⁻⁶ = 5.7 x 10⁻⁶ failures per hour.

Average Down Time = Repair Time / Failures in an year = 40 / 0.05 = 800 hours.

Mean time to Failure = 1 / Failure Rate = (5.7 x 10⁻⁶)-1= 175438.59 hours

Unavailability, U = Average Repair Time / ( Average Repair Time + MTTF )

= 800 / ( 800 + 175438.59 ) = 0.0045

describing distributions is one of the key skills you’ll need to earn a high score on the ap statistics exam. if you need proof of this, just flip through some past exam questions, which can be found at the collegeboard website. you’ll notice that the first free response question is almost always a question requiring you to look at a graph and describe it.

step-by-step explanation: just trying to get points sorry : (

b and c

step-by-step explanation: