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Math Help - How to calculate root mean std dev given 2 readings?

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    How to calculate root mean std dev given 2 readings?

    Hello

    I am at a loss as to how to work out a rsmd (for 52 weeks) given a rsmd for 20 and 32 weeks.

    The question is:

    In a certain district, the mean annual rainfall is 80cm, with standard deviation 4cm.

    Jake, a local meteorologist, kept a record of the weekly rainfall in his garden. His first data set, comprising 20 weeks of figures resulted in a mean weekly rainfall of 1.5cm. The rmsd was 0.1cm. His second set of data, over 32 weeks, resulted in a mean of 1.7cm and a rmsd of 0.09cm.

    Calculate the overall mean and the overall rmsd for the whole year.

    For the annual mean I calculate as ((20 x 1.5) + (32 x 1.7)) / 52 = 1.62cm. The answer book gives same answer so I assume this is correct way to calculate.

    My calculation for annual rmsd was:
    ((20 x 0.1) + (32 x 0.09)) / 52 = 0.094 - which is incorrect. Book answer gives 0.135. How would I calculate the annual rmsd (root mean standard deviation)?

    Angus
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    Quote Originally Posted by angypangy View Post
    Hello

    I am at a loss as to how to work out a rsmd (for 52 weeks) given a rsmd for 20 and 32 weeks.

    The question is:

    In a certain district, the mean annual rainfall is 80cm, with standard deviation 4cm.

    Jake, a local meteorologist, kept a record of the weekly rainfall in his garden. His first data set, comprising 20 weeks of figures resulted in a mean weekly rainfall of 1.5cm. The rmsd was 0.1cm. His second set of data, over 32 weeks, resulted in a mean of 1.7cm and a rmsd of 0.09cm.

    Calculate the overall mean and the overall rmsd for the whole year.

    For the annual mean I calculate as ((20 x 1.5) + (32 x 1.7)) / 52 = 1.62cm. The answer book gives same answer so I assume this is correct way to calculate.

    My calculation for annual rmsd was:
    ((20 x 0.1) + (32 x 0.09)) / 52 = 0.094 - which is incorrect. Book answer gives 0.135. How would I calculate the annual rmsd (root mean standard deviation)?
    First, I think you mean ((20 x (0.1)^2) + (32 x (0.09)^2)) / 52 at the end there. But that looks as though it's just a typo, because you correctly calculated it as 0.094. However, that is the wrong answer, and it should indeed be 0.135. The reason for this is quite subtle, and I'll try to explain it.

    The rmsd for the 20-week period is the square root of one-twentieth of the sum

    \displaystyle\sum(x_i-1.5)^2,\qquad (*)

    where the x_i represent the individual readings, and x_i-1.5 is the amount by which the reading differs from the 20-week mean 1.5. But when you calculate the rmsd for the 52-week period, you are no longer using 1.5 as the mean, but 1.62. So in calculating the contribution of these readings to the 52-week rmsd, you need to use the sum

    \displaystyle\sum(x_i-1.62)^2.\qquad (**)

    To get from (*) to (**), you have to take (*) and add the difference between (**) and (*), namely

    \displaystyle\sum\bigl((x_i-1.62)^2 - (x_i-1.5)^2\bigr).

    When you multiply out the brackets and subtract, the terms x_i^2 cancel out and you are left with -2(1.62-1.5)\sum x_i + 20(1.62^2-1.5^2). But \sum x_i is 20 times the mean of the x_i, namely 20\times 1.5. Using that, you should find that the sum (**) is equal to 0.488.

    Now do exactly the same process to the 32-week period, and you should get the sum there to be 0.464.

    Therefore the 52-week rmsd is \sqrt{\frac1{52}(0.488+0.464)} \approx 0.135.
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