The chief design engineer for Colonial Warning Systems wants to know the reliability of a two-lamp device intended to warn the operator of potential problems with a commercial steam kettle.One (either)lamp is to be lit if steam pressure drops below the desired level,and two (both)are to be lit if temperature drops below a desired level.His supplier estimates that a lamp has a failure rate of .2 per hour of operation,and a mean wear-out time of seven years with a standard deviation of eight months.If the temperature drops below the desired level,what is the probability the system will perform reliably for one hour?
A).04
B).40
C).60
D).64
E).96

Question

Quizplus · Accepted Answer

We can use the formula for reliability of a system composed of independent components:
R = R1 * R2 * ... * Rn
where Ri is the reliability of component i.
For the two-lamp device, we can consider the two lamps as independent components. The probability that one lamp will operate for one hour without failing is 0.8, and the probability that two lamps will operate for one hour without failing is (0.8)^2 = 0.64. Therefore, the reliability of the two-lamp device for one hour is 0.8 * 0.64 = 0.512 or 51.2%. However, we also need to consider the wear-out time of the lamps. Since the mean wear-out time is seven years (or 84 months), the probability that a lamp will fail due to wear-out in one hour is 1/((84/12)*60) = 0.000198. Therefore, the overall probability that the system will perform reliably for one hour is 0.512 * (1 - 0.000198) = 0.5118, which is closest to 0.64 (D) in the answer choices.

The Chief Design Engineer for Colonial Warning Systems Wants to Know