You have the mean and standard deviation data for a population of individual unit pumps' reporting error. Each engine has either 4 or 6 unit pumps and the total reporting error for the combined performance of all pumps on the engine is being evaluated (i.e four pumps on an engine all of which are over reporting by 10%, the engine reporting error is 10%). Considering that each pump installed on the engines will be selected from the distribution data provided for an individual pump, what is the overall mean and standard deviation for the entire engine in terms of reporting error for both 4 and 6 pump engines?

Alice Chen

Alice Chen

Answered question

2022-11-11

Calculate the Standard Deviation for Multiple Combined Distributions
You have the mean and standard deviation data for a population of individual unit pumps' reporting error. Each engine has either 4 or 6 unit pumps and the total reporting error for the combined performance of all pumps on the engine is being evaluated (i.e four pumps on an engine all of which are over reporting by 10%, the engine reporting error is 10%). Considering that each pump installed on the engines will be selected from the distribution data provided for an individual pump, what is the overall mean and standard deviation for the entire engine in terms of reporting error for both 4 and 6 pump engines?

Answer & Explanation

metodikkf6z

metodikkf6z

Beginner2022-11-12Added 14 answers

Step 1
I think it's important to differentiate between the "real" distribution of error per pump f p ( E j ; θ p ), which we do not seem to have, and the "measured" statistics of this distribution, of which we seem to have the mean μ p and standard deviation σ p . From the wording, we seem to have the population statistics for these values; i.e. we can assume they are exactly correct.
Now, let's consider an engine with k pumps. Let's call the error of the engine to be the mean error of its pumps. Let E i be the error of some pump i and T be the average error of the engine. Then:
T = 1 k i E i
is a random variable, which depends on the random variable E i for i from 1 to k.
Let's compute the expected value (i.e. mean) of this variable:
E [ T ] = E [ 1 k i E i ] = 1 k i E [ E i ] = μ p
so the expected value is the same.
Let's also compute the expected value of the square which we can use below:
E [ T 2 ] = E [ ( 1 k i E i ) 2 ] = 1 k 2 i j E [ E i E j ] = 1 k 2 [ k ( V [ E i ] + E [ E i ] 2 ) + ( k 2 k ) E [ E i ] E [ E j ] ] = 1 k 2 [ k ( σ p 2 + μ p 2 ) + ( k 2 k ) μ p 2 ] = σ p 2 k + μ p 2
Now for the standard deviation. We can compute the variance as
V [ T 2 ] = E [ T 2 ] E [ T ] 2 = σ p 2 k + μ p 2 μ p 2 = σ p 2 k
Thus the standard deviation is:
stddev ( T ) = σ p k
which makes sense in the sense that the average error should have less variability as k increases (central limit theorem).
Step 2
In other words, for a k-pump engine, the expected error is given by the average error of the individual pumps, whereas the standard deviation is directly proportional to the standard deviation of the pumps, but scaled by the number of pumps involved.

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