I am doing some simulations for a factory as a project, in which case I have
>>machines in series each with a unique normal distribution, I want to find
>>how the resultant distribution will be, and also how I will have to tweek
>>individual distributions of these machines to get a resultant distribution
>>which I want.
>>Now suppose I have 10 machines in series each with a normal distribution
>>having mean 100 and standard devietion of 5, what will be the resultant
>>distribution of the capacity. Knowing this final resultant distribution, now
>>how will I have to change the 10 distributions, in this case all the
>>get a resultant distribution with a mean of 100 and sd of 5.
>>Hope you will help me with this, I am in a fix, not knowing how to proceed.
>>Thanks once again.
We need more information as to why you are asking this question. It
fits into our K-12 environment. See www.newton.dep.anl.gov for more
I am not entirely sure what you mean by having 10 machines in series. Does
the output from one machine go to the next? Are their electric circuits
wired together in series? I don't have a good enough idea of what your
question is to be able to come up with an answer.
What is the "resultant distribution?" I am not sure how you are defining
that. Perhaps this can help you answer your question - every random
variable that is a sum of independent normal random variables is itself a
normal random variable with a mean equal to the sums of the means of the
component r.v.'s and a variance equal to the sums of the variances of the
component r.v.'s. (The variance is the square of the standard deviation.)
Let's say that you have ten furnaces, each of which makes 10 plus or minus
1 (s.d.) tons of steel a day. What is the total output of all ten machines
in a day? The mean value will be 100 tons, and the standard deviation will
be the square root of 10 (or 3.16). As a fraction of the total, the
variability in the total output is much less that the variability in the
output of any of the individual furnaces.
Richard Barrans Jr.
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Update: June 2012