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Standard deviation question

  1. Sep 7, 2012 #1
    Hello,

    If have the standard deviation of acceleration (acceleration > 0) for a car. I also have the standard deviation of deceleration (acceleration <0). With these two figures is there a way to get the a standard deviation of the acceleration (both positive and negative)?

    Thanks
     
  2. jcsd
  3. Sep 7, 2012 #2

    mfb

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    Staff: Mentor

    How do you define "standard deviation of acceleration (acceleration > 0)"?
    Do you know the "average acceleration (acceleration > 0)" (+ same for <0)?
     
  4. Sep 8, 2012 #3
    Maybe. What is your sample space? That is, what is the meaning of each element in the sets that are your samples?
     
  5. Sep 8, 2012 #4
    Hi, yes I do have the average acceleration (>0) and the average deceleration (<0).

    This is all the information that I have:

    1. The number of acceleration observations (both acceleration and deceleration combined)
    2. The SD of acceleration
    3. The SD of deceleration
    4. The average acceleration
    5. The average deceleration

    The SD is the SD of the population of the observations.

    For example:
    1. The number of acceleration observations (both acceleration and deceleration combined) =920
    2. The SD of acceleration =0.33 m/s2
    3. The SD of deceleration = 0.49 m/s2
    4. The average acceleration = 0.55 m/s2
    5. The average deceleration = -0.70 m/s2

    Thanks for the help
     
  6. Sep 8, 2012 #5

    chiro

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    Hey bradyj7.

    If the two distributions are independent, then you will add the variances together and take the square roots of the two.

    If they are dependent, you will need to include what is known as a co-variance term to get the final variance of the addition of the two random variables.

    Have you come across co-variance before?
     
  7. Sep 8, 2012 #6
    Hi chiro,

    I presume the distributions are dependent? Here is sample of the raw data and the calculated SDs and means.

    https://dl.dropbox.com/u/54057365/All/sample.JPG

    Basically I have computed the statistics in green for a bunch of journeys. After doing this I realised that I wanted to calculate the SD of the acceleration both positive and negative. This is the statistic in red.

    To save myself a lot of time and going back and doing it manually, I was wondering/hoping ithat there might be a way to get the statistic in red using the statistics in green?

    Thanks for the help
     
  8. Sep 8, 2012 #7

    mfb

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    Hmm... if you have the total acceleration time and the total deceleration time, you can simply add them as two datasets (correlations do not matter).

    Let ta be the total (positive) acceleration time steps. Let aa be the average (positive) acceleration and σa the standard deviation in this dataset.
    Define the same parameters with d for deceleration.
    Let t be the total time = number of steps: t=ta+td
    Your average acceleration (positive+negative, without superscript) is then given by
    [tex]\bar{a} = \frac{a^a t^a + a^d t^d}{t^a+t^d}[/tex]
    The standard deviation can be calculated via
    [tex]t \sigma^2 = \sum_{i,a_i>0} (a_i-\bar{a})^2 + \sum_{i,a_i<0} (a_i-\bar{a})^2[/tex]
    where the first sums run over all time steps with positive acceleration and the second runs over all with negative acceleration. N is the total number of steps. This can be simplified:
    [tex]t \sigma^2 = \sum_{i,a_i>0} (a_i-a^a + a^a - \bar{a})^2 + \sum_{i,a_i<0} (a_i-a^d + a^d-\bar{a})^2[/tex]
    [tex]t \sigma^2 =\sum_{i,a_i>0}
    \left((a_i-a^a)^2 + (a^a - \bar{a})^2 + 2(a_i-a^a)(a^a-\bar{a})\right)
    + \sum_{i,a_i<0}
    \left((a_i-a^d)^2 + (a^d - \bar{a})^2 + 2(a_i-a^d)(a^d-\bar{a})\right)[/tex]
    In both sums, the first expression is simply the standard deviation of the individual parts. [itex](a^a-\bar{a})[/itex] is constant, and the average ai is simply aa (same for d) so those terms vanish. The middle term is independent of i, so the sum just gives a factor of ta and d respectively.
    [tex]t \sigma^2 =t^a \sigma^a + t^d \sigma^d
    + t^a (a^a - \bar{a})^2)
    + t^d (a^d - \bar{a})^2)
    = t^a \sigma^a + t^d \sigma^d + t\bar{a}^2 + t^a {a^a}^2 + t^d {a^d}^2 - 2t^a a^a \bar{a} - 2t^d a^d \bar{a}[/tex]
    Using the expression for [itex]\bar{a}[/itex], this gives
    [tex]t \sigma^2 = t^a \sigma^a + t^d \sigma^d+ t^a {a^a}^2 + t^d {a^d}^2 - t\bar{a}^2[/tex]
    And that can be calculated, as all parts are known.


    If you do not now the umber of acceleration / deceleration steps: Maybe you can calculate the average acceleration based on the initial and final velocity and the total number of steps. This will allow you to calculate ta and td as well.
     
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