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# 生產作業管理：可靠度的算式(2)

下列是一本原文的生產作業管理第四章「Reliability」(可靠度)的另外3題問題。

因為是原文，所以導致較難理解其意，

願您能給我下列3題的算式，讓我能理解！

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<1>

A system is composed of twoparts which must both operate in order for the system to perform as intended. The parts havereliabilities of .6 and .5. There is an identical backup system connected to the main system by a switch which has areliability of .9. The probability that the overall system will operate is closest to: A. .30 B. .40 C. .50 D. .60 E. .70 答案為C

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<2>

If the transformer has a meantime to wear-out of 20 years with a standard deviation of 2 years, what is the probability that it will wear outwithin 19 years? A. 0.1915 B. 0.3085 C. 0.383 D. 0.5 E. 0.6915 答案為B

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<3>

Four major components - video,audio, tuner and power supply - make up a big screen television set. All must function for the setto be accepted for shipment to customers. Reliabilities for each major component are 80%. ]What is the probabilitythat a newly made set will be accepted for shipment? A. 10.25% B. 80% C. 41% D. 20% E. Can't determine without further informationz

答案為C

### 1 Answer

- HuevosLv 67 years agoFavorite Answer
<1>

[(0.6*.9)+(.5*.9)]/2 = (0.54+.45)/2 = 0.99/2 = 0.495 = ~0.5 (C)

<2>

知道要用normal distribution來算，所以。。。先standardize

z = (x-20)/2

當x=0

z(0) = (0-20)/2 = -10

當x=19

z(19) = (19-20)/2 = -1/2

p(0<x<19) = p(-10<z<-1/2) = p(z<-1/2)-p(z<-10) = 0.30854-0.00003 = 0.3085

## 0.30854-0.00003 是從Z-table找出來的。

http://calculator.tutorvista.com/normal-distributi...

<3>

每一個component都是independent...

(0.8)*(0.8)*(0.8)*(0.8) = (0.8)^4 = 0.4096 =~ 0.41

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