Question

Ten high-technology batteries are tested for 200 hours each. One failed at 20 hours; another failed at 140 hours; all others completed the test. FR(%) is __________ and MTBF is __________.

Answer 1

Failure rate = 20%

MTBF = 880 hours

Step-by-step explanation:

given data

batteries = 10

tested = 200 hours

one failed = 20 hours

another fail at = 140 hours

solution

we know that Mean Time between Failures is express as = (Total up time) ÷ (number of breakdowns) ....................1

so here Total up time will be

Total up time = 200 × 10

Total up time = 2000

and here

Number of breakdown = 1 at 20 hour and another at 140 hour = 2

so it will be = (Total up time) ÷ (number of breakdowns) .......2

=
`(2000)/(2)` = 1000

so here gap between occurrences is

gap between occurrences= 140 - 20

gap between occurrences = 120 hour

and

MTBF will be

MTBF = 1000 - 120

MTBF = 880 hours

and

Failure rate (FR) will be

Failure rate (FR) = 1 ÷ MTBF ................3

Failure rate (FR) = R÷T ......................4

as here R is the number of failures and T is total time

so Failure rate (FR) = 20%