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Cumulative standard deviations

  1. Mar 12, 2014 #1
    Hello
    A manufacturer produces blocks of metal of
    length 100 standard deviation 2,
    width 200 standard deviation 1
    depth 150 standard deviation 3

    The mean volume will be 100x200x150 but what will its sd function be? (The figures I've given are just for clarity)

    Thanks in advance, I've really no idea of how to tackle this problem.
     
  2. jcsd
  3. Mar 12, 2014 #2

    mfb

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    sd function? Standard deviation? If you are interested in more details, you need more details how the individual lengths vary.
    Simple error propagation in the product will work.
     
  4. Mar 12, 2014 #3

    mathman

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    Assuming that the length, width, and depth are statistically independent, the calculation is straightforward.
    Let V(X) (=E(X²) - E(X)²) be variance of X. Let L,W,D be the three dimensions.
    E((LWD)²) = E(L²)E(W²)E(D²), E(LWD) = E(L)E(W)E(D) (independence).
    E(L²) = V(L) + E(L)², etc.
    Combine to get V(LWD) in terms of the individual varances and means.

    In case you don't remember statistics, variance is square of standard deviation.
    If the dimensions are not statistically independent, you have a much harder problem.
     
  5. Mar 13, 2014 #4
    Many thanks

    Straightforward, straightforward, for somebody who doesn't know what variance means and who has never ever seen E(..) before? But more than willing to thrash through it.

    I'll be back on this in a few days, or should I say a few late nights.... Wish me luck!

    Many thanks again.
    Roger
     
  6. Mar 14, 2014 #5

    mathman

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    E(X) is mean value. For your purposes, variance is V(X) = E(X2) - (E(X))2
    Standard deviation is square root of variance.

    I am a little surprised that you are not familiar with these terms, given the problem you presented. What do you use for a definition of standard deviation?
     
    Last edited: Mar 14, 2014
  7. Mar 15, 2014 #6
    Hello

    Thanks but obviously I"ve googled a lot since and am now a little more familiar with these new terms. For me the SD was just a number linked to the width at half max. As such it represented the tightness of the distribution, no more, no less. That it should be the root of the variance, a term I didn't know either, was not a thing I needed to know.

    Now I have a problem that requires this knowlege, for your help I thank you all.
     
  8. Mar 15, 2014 #7
    Hello, using a more familiar language for me the result must be :

    Variance of (LWD) = [variance of L + (meanL)^2] * [variance of W + (meanW)^2] * [variance of D + (meanD)^2] - (meanL x meanW x meanD)^2

    which should give for my block of metal as example
    length 100 standard deviation 2,
    width 200 standard deviation 1
    depth 150 standard deviation 3

    variance of (LWD) = (4 +10000) * (1 +4000) * (9 + 2250) - 100*200*150 = approx 90E9

    SD = 300 000 for a volume of 3 million, about 10% whereas the SDs of each dimension <2%

    To tell you the truth, I wasn't at all expecting a wider dispersion, on the contrary. By the way, I put this question on a French forum where I live, nobody answered except one person who was also interested in a solution, I'll link him to this topic.

    Thanks again
     
  9. Mar 15, 2014 #8

    mfb

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    Where does the formula come from? It does not look right.

    I get ~2.9% as relative standard deviation. 10% is certainly too high.
     
  10. Mar 16, 2014 #9
    OK

    V(X) =E(X²) - E(X)² standard formula  derived from the defintion of variance which is E[(X-E(X))^2] 
    so E(X²) = V(X) + E(X)² re-arrangement                                                                        
    E((LWD)²) = E(L²)E(W²)E(D²) your formula
    = (V(L) + E(L)² )* (V(W) + E(W)² )* (V(D) + E(D)² ) my expansion of your formula
    V(LWD) =E((LWD)²) - E(LWD)² standard formula replacing X by LWD
    so = (V(L) + E(L)² )* (V(W) + E(W)² )* (V(D) + E(D)² ) - E(LWD)²
    ( 2² + 100² ) * ( 1² + 200²) * (3² + 150²) - (100*200*150)²
    which gives SD/vol = 2,873 %

    Don't know where I went wrong.
     
  11. Mar 16, 2014 #10

    mfb

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    Hmm interesting idea to re-write the variance.

    Good, now the results agree.
     
  12. Mar 16, 2014 #11

    mathman

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    Why do think its wrong? I got the same result. Your s.d. for the original dimensions are 2%, .5%, and 2% so the answer is certainly in the right ball park.
     
  13. Mar 16, 2014 #12
    Hello

    "Hmm interesting idea to re-write the variance." What line(s) are you refering to? I just took the lines from your posting,
    -adding a first line, for my oxn personal comprehension "derived from the defintion of variance which is E[(X-E(X))^2"] "
    -inserting "so E(X²) = V(X) + E(X)² re-arrangement" to make the subsequent substitution clearer

    by "went wrong" I was refering to my first calculation of 10%

    By the way, I didn't like the use of E( ) as a way of expressing averages. I, a newcomer, especially balked at the fact that:

    E(X)² means the the square of the avarage value of X whereas 3(5+6)² does not mean the square of 3*(5+6)
    But I quicky realised that it was the most precise way of expression, but you have to be very very careful with the placement of the brackets.

    I'm also a bit surprised at not having learnt this terminology at grammar and university in England back in the 60's
    Roger
     
  14. Mar 17, 2014 #13

    mfb

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    Post #9.

    They are different brackets - one is a function bracket, the other one a "grouping" bracket. But still, it can be confusing, yes.
     
  15. Mar 17, 2014 #14

    mathman

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    I suppose for clarity you could write (E(X))2.

    The use of E(X) for mean of X is standard notation.
     
    Last edited: Mar 17, 2014
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