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Math Help - Standard Deviation & Mean Confusion

  1. #1
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    Standard Deviation & Mean Confusion

    Hi,

    Doing this question and working off an example in my book. The example is not explained properly, so I don't know why I am doing certain working out. Please help!

    I have a random sample of 50 weights of boxes, with normally distributed mean of 750g and standard deviation of 20g.

    I have worked out the mean of the sample...
    I have worked out the standard deviation of the sample...

    ***In the book it then goes to squareroot by 50 and multiply by the standard deviation. What is the purpose of this and what does it do?***

    Thanks
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  2. #2
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    Re: Standard Deviation & Mean Confusion

    Variance of the sample is usually denoted by \sigma^2, in your example \sigma^2=50, where \sigma denotes the standard deviation of the sample: \sigma^2= 50, hence \sigma=\sqrt{50}.

    So \sqrt{50}\cdot \sigma = \sqrt{50}\cdot\sqrt{50}=50=\sigma^2.

    Hope it helps a bit.
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  3. #3
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    Re: Standard Deviation & Mean Confusion

    Many thanks MathoMan...cleared that up perfectly!
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  4. #4
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    Re: Standard Deviation & Mean Confusion

    If you have a sample of size n from a distribution with standard deviation s then the distribution of the means of samples has standard deviation s/root n

    (note that for the population standard deviation I should really use the Greek letter sigma )
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