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Large numbers and standard deviation for a uniform distribution

  1. Jan 4, 2008 #1

    nicksauce

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    In doing a problem, I considered N (a large number, in the range 100,000-1,000,000) raindrops, falling into A (fixed at 100) segments on a roof, distributed using a random number generator I programmed. In considering the number of raindrops that fell into a given segment, the average would be [tex]\mu=\frac{N}{A}[/tex]. For a fixed N, I calculated the standard deviation [tex]\sigma[/tex]. I then plotted the standard deviation against the average, and found a nearly perfect relationship [tex]\sigma = \sqrt{\mu}[/tex]. Is this relationship correct? Can anyone tell me why it would be so, or give me a starting point to show it?
     
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  3. Jan 5, 2008 #2

    mda

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    This seems to be a case of a multinomial distribution: you might try reading the wikipedia entry.
     
  4. Jan 5, 2008 #3

    nicksauce

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    Ah thank you.

    [tex]\sigma_i^2 = Np_iq_i = \sigma^2[/tex]

    [tex]\sigma^2\simeq Np[/tex]

    [tex]\sigma^2 = N\frac{1}{A}=\mu[/tex]

    [tex]\sigma=\sqrt{{\mu}[/tex]
     
  5. Jan 5, 2008 #4

    Gib Z

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    I always thought that the Standard deviation was defined to be the square root of the variance. I know this to be true for discrete distributions, though I guess its a different case for continuous ones.
     
  6. Jan 6, 2008 #5

    HallsofIvy

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    No, for either discrete or continuous distributions, the standard deviation is the square root of the variance. What nicksauce showed above was that, in this case, the variance is equal to the mean.
     
  7. Jan 6, 2008 #6

    Gib Z

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    No I know that the Standard deviation IS the square root of the variance, but when I learned about discrete distributions the standard deviation was DEFINED to be the square root of the variance. I "guessed" it was a different matter for continuous distributions because otherwise the other people in this thread just wasted a lot of time lol.
     
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