MIBF = ∑(Start of downtime - Start of uptime)/No. of failures
MIBF = (6+4+8+5+6)/5 = 29/5
The probability or reliability that the product will work for a defined period of time without failure is given by
R(T) = exp(-T/MTBF); T = 1 hour
R(1) = e^(-1/(29/5)) = e^(-5/29) = 0.84

To calculate the reliability of the system for one hour of operation using an exponential model, you can use the formula: Reliability (R) = e^(-λt) Where: λ (lambda) is the failure rate, which can be calculated as the reciprocal of the mean time between failures (MTBF), and t is the time of interest, which in this case is 1 hour. First, let's calculate λ. The MTBF is the average time between failures, which can be calculated as the sum of the time since the last failure divided by the number of failures. MTBF = (6 + 4 + 8 + 5 + 6) / 5 = 29/5 hours Now, calculate λ: λ = 1 / MTBF = 5 / 29 per hour Now, calculate the reliability for 1 hour of operation: R = e^(-λt) = e^(-(5/29) * 1) = e^(-5/29) So, the reliability of the system for one hour of operation assuming an exponential model is approximately e^(-5/29).

Total faults detected = 159
Real faults detected among all detected faults = 159 - 75 = 84
Since probability distribution is same, total number of real faults is (100/75)*84 = 112
Undetected real faults = 112- 84 = 28