NTA UGC NET JUNE-2023 Paper-2
November 4, 2023Question 12775 – Nielit Scentist-B [02-12-2018]
November 5, 2023NTA UGC NET JUNE-2023 Paper-2
| Question 9 |
| Error number | 1 | 2 | 3 | 4 | 5 |
| Time since last failure (hour) | 6 | 4 | 8 | 5 | 6 |
The reliability of the system for one hour operation assuming an exponential model is
| e -9/29 | |
| e -7/29 | |
| e -5/29 | |
| e -3/29 |
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).
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).