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The document describes a mechatronic assembly process where defects occur randomly according to a Poisson distribution. It provides the parameter λ for the original and improved processes and asks to calculate the probability of getting exactly one defect or one or more defects. Improving the process by cutting the defect rate in half reduces the probability of getting one or more defects.

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1K views2 pages

1

The document describes a mechatronic assembly process where defects occur randomly according to a Poisson distribution. It provides the parameter λ for the original and improved processes and asks to calculate the probability of getting exactly one defect or one or more defects. Improving the process by cutting the defect rate in half reduces the probability of getting one or more defects.

Uploaded by

JunaidZafar
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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1) A mechatronic assembly is subjected to a final functional test.

Suppose that defects occur at


random in these assemblies, and that defects occur according to a Poisson distribution with
parameter = 0.06.
(a) What is the probability that an assembly will have exactly one defect?
(b) What is the probability that an assembly will have one or more defects?
(c) Suppose that you improve the process so that the occurrence rate of defects is cut in half to =
0.03. What effect does this have on the probability that an assembly will have one or more
defects?

Solution:

This is a Poisson distribution with parameter = 0.06, x ~ POI (0.06).


(a)
0.06 1
e ( 0.06)
Pr{x=1} = p (1) = 1! =

(b)
0.06 0
e ( 0.06)
Pr {x1} = 1- p {x=0} = 1-p (0) = 1- 0! =

(c)
Poisson distribution with parameter = 0.03, x ~ POI (0.03).
e0.03 (0.03)0
Pr {x1} = 1- p {x=0} = 1-p (0) = 1- 0! =

4) The output voltage of a power supply is normally distributed with mean 10 V and standard
deviation 0.04 V. If the lower and upper specifications for voltage are 9.95 V and 10.05 V,
respectively
a)
X N Normal ( = 10; variance= 0.042)
P (9.95 < X < 10.05)
9.9510 X 10.0510
P( < < )
0.04 S. D 0.04

P (-1.25 < Z < 1.25)

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