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An uninterruptible power supply (UPS) is used to protect a computer from short power outages. Suppose a particular UPS has three rechargeable batteries and the UPS will be able to protect the computer, for at least short outages, so long as at least one of the batteries can still hold a charge. Rechargeable batteries have a limited lifetime. Two of the batteries (9 volt) have exponential lifetimes with mean 4 years. The third battery (18 volt) is Weibull distributed with Beta(B)=0.75 and median lifetime of 6 years.
a) Assuming that the batteries fail independently, what is the probability that the UPS will fail (i.e. become unable to protect the computer for even short outages) within 2 years?
b) If the 18 volt battery fails within 2 years, the other two batteries become twice as likely to fail within two years, but otherwise fail independently. What now is the probability that the UPS will fail within 2 years?
How can we use some of the measurement methods we've learned this term to make better decisions in business and in life? Sometimes it's hard for students to draw the linkages between topics examined in quantitatively-oriented courses to the "real ..
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