\”Drug factory\” probability

2010-02-28 11:05 pm

A factory has 8 machines which produce drug A in a highly purified environment.
The production environments of machines are independent to each other and are
purified by a certain procedure in each hour. The effective time of the drug is
normally distributed with mean 400 seconds and standard deviation 60 seconds.
When the production environment is contaminated, the effective time of the drug
is still normally distributed but the mean effective time becomes 300 seconds and
standard deviation becomes 100 seconds. The engineer of this factory finds that the
probability of a production environment to be contaminated is 0.2.
In order to maintain the quality of the product, a drug sample is randomly selected
between two purifying procedures from each machine and its effective time is
tested. If the effective time of the drug sample is shorter than 340 seconds, the corresponding machine will be closely monitored by the technician of this factory. The
technician has to report to the engineer if there are more than 3 machines needed
to be monitored. Assume the tests are independent.

(a) What is the probability that a machine in contaminated environment is monitored?
(b) What is the probability that a machine in purified environment is monitored?
(c) What is the probability that a machine is monitored?
(d) What is the probability that the a monitored machine in really in a contaminated
production environment?
(e) What is the probability that the technician reports to the engineer?

回答 (1)

匿名

2010-03-02 11:31 pm

✔ 最佳答案

(a): Pr (a machine in contaminated environment is monitored) = Pr (contaminated environment) x Pr (machine is monitored | contaminated environment) = 0.2 x Pr [Z < (340 – 300 / 100)] = 0.2 x 0.6554 = 0.1311

(b): Pr (a machine in purified environment is monitored) = Pr (purified environment) x Pr (machine is monitored | purified environment) = 0.8 x Pr [Z < (340 – 400 / 60)] = 0.8 x 0.1587 = 0.1269

(c): Pr (a machine is monitored) = Pr (a machine in contaminated environment is monitored) + Pr (a machine in purified environment is monitored) = 0.1311 + 0.1269 = 0.258

(d): Pr (a monitored machine in really in a contaminated production environment) = Pr (contaminated environment | a machine is monitored) = Pr (a machine in contaminated environment is monitored) / Pr (a machine is monitored) = 0.1311 /0.258 = 0.5081

(e):
Pr (a machine in contaminated environment is reported) = 0.2 x 0.65543 = 0.0563
Pr (a machine in purified environment is reported) = 0.8 x 0.15873 = 0.0032
Pr (the technician reports to the engineer) = Pr (a machine in contaminated environment is reported) + Pr (a machine in purified environment is reported) = 0.0563 + 0.0032 = 0.0595

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