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Standard Deviation of a sample of a population's means

  1. Dec 11, 2008 #1
    Hey guys,

    Just a question which has been puzzling me for some time.

    I am told that means of samples of a population can be normally distributed with mean of [tex]\mu[/tex] and with standard deviation of [tex]\sigma[/tex]/[tex]\sqrt{}[/tex]n

    Can someone please explain to me how the standard deviation is derived or is it ismply as a result of experimentation?

    Thanks,
    Oscar
     
  2. jcsd
  3. Dec 11, 2008 #2

    ssd

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    It is derived. The given result follows if the population is normal.
     
  4. Dec 11, 2008 #3

    statdad

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    Homework Helper

    The mean is

    [tex]
    \overline X = \frac{\sum X}{n}
    [/tex]

    If all of the [tex] X [/tex] variables are independent, and identically distributed, then

    [tex]
    Var[X] = Var\left(\frac{\sum X}n\right) = \frac 1 {n^2} \sum Var(X) = \frac{n\sigma^2}{n^2} = \frac{\sigma} n
    [/tex]

    so the standard deviation is as you state.

    If the [tex] X [/tex] values are themselves normally distributed, then the mean is as well (linear combinations of normally distributed normal random variables are normal)

    If the [tex] X [/tex] values are not normally distributed, the distribution of [tex] \overline X [/tex] is approximately normal if certain conditions are satisfied.
     
  5. Dec 12, 2008 #4

    stewartcs

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    Science Advisor

    You forgot the root in the denominator.

    CS
     
  6. Dec 12, 2008 #5

    statdad

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    No, actually I forgot the square in the numerator, since I was writing out the variance. :frown:
    Still a stupid mistake on my part.

    The variance is

    [tex]
    \frac{\sigma^2} n
    [/tex]

    so the standard deviation is

    [tex]
    \frac{\sigma}{\sqrt n}
    [/tex]
     
  7. Dec 12, 2008 #6

    stewartcs

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    Science Advisor

    Sorry...it appeared as if you were answering the OP's question about the standard deviation directly which is why I assumed you had finished the derivation all the way out to the standard deviation and not just the variance.

    CS
     
  8. Dec 12, 2008 #7

    statdad

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    stewartcs; there is no need for you to apologize and I certainly did not mean to imply (in my earlier post) that there was. I am sorry if it seemed that way.
     
  9. Dec 13, 2008 #8
    Hey guys,

    Thanks very much for showing me the derivation - it has helped make things a lot clearer :)

    Thanks again,

    Oscar
     
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