Ten high-technology batteries are tested for 200 hours each. One failed at 50 hours; all others completed the test. FR(%) is __________ and FR(N) is __________.
A) 10%; 1/1850
B) 10%; 1/2000
C) 25%; 1850 hours
D) 90%; 1/2000
E) Indeterminate; no standard deviation is given.

Question

Quizplus · Accepted Answer

FR(%) is the Failure Rate in percentage, calculated as (number of failures / total number of units) * 100. In this case, 1 battery failed out of 10, so FR(%) = (1/10)*100 = 10%. FR(N) is the Failure Rate per unit time, calculated as (number of failures / total unit hours of operation). Since 9 batteries completed 200 hours and 1 failed at 50 hours, the total hours = (9*200) + 50 = 1850 hours. Thus, FR(N) = 1/1850.

Ten High-Technology Batteries Are Tested for 200 Hours Each