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Math Help - Standard Deviation

  1. #1
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    Standard Deviation

    Hey guys, the following is from my year 11 maths B textbook...

    The mean of 50 discrete measurements is 18 and the standard deviation is 4. The mean of another 40 measurements is 15 and the standard deviation is 2. Find the mean and standard deviation for the entire 90 measurements.

    I'm really struggling, and would appreciate any working. Thanks!!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Megan View Post
    Hey guys, the following is from my year 11 maths B textbook...

    The mean of 50 discrete measurements is 18 and the standard deviation is 4. The mean of another 40 measurements is 15 and the standard deviation is 2. Find the mean and standard deviation for the entire 90 measurements.

    I'm really struggling, and would appreciate any working. Thanks!!
    For the first sample the sum is 50 \times 18 and the sum of squartes is 50 ( 4^2 + 18^2)

    Similarly for the second sample the sum is 40\times 15 and the sum of squares is 40 (2^2+15^2)

    Now compute the sum and sum of squares for the combined sample and so the mean and SD.

    (If the SD uses the 1/(n-1) weighting rather than 1/n the above will need adjusting in the obvious manner)

    CB
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  3. #3
    MHF Contributor Unknown008's Avatar
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    Let x represent the measurements from the first batch of 50 and y be that for the batch of 40.

    \frac{\Sigma(x+y)}{50+40} = \frac{\Sigma(x) + \Sigma(y)}{90}

    You can find \Sigma(x) and \Sigma(y)

    knowing that:

    \frac{\Sigma(x)}{50} = 18

    \frac{\Sigma(y)}{40} = 15

    Then;

    Sd(x+y) = \sqrt{\frac{\Sigma(x^2 + y^2)}{90} - (\frac{\Sigma(x+y)}{90})^2}

    You know that:

    Sd(x) = \sqrt{\frac{\Sigma(x^2)}{50} - (\frac{\Sigma(x)}{50})^2}

    Sd(y) = \sqrt{\frac{\Sigma(y^2)}{40} - (\frac{\Sigma(y)}{40})^2}

    And;

    \Sigma(x^2) + \Sigma(y^2) = \Sigma(x^2 + y^2)

    Can you solve this now?
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  4. #4
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    I understand how to calculate the mean.
    I'm still a little stuck on the standard deviation?

    Sorry for my lack of genius! (:
    M.
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  5. #5
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    Quote Originally Posted by Megan View Post
    I understand how to calculate the mean.
    I'm still a little stuck on the standard deviation?

    Sorry for my lack of genius! (:
    M.
    For the first set,

    the SD is given by \sqrt{\frac{\sum x^2}{50}-18^2}=4

    You can subsequently find the sum of all its squared measurements, \sum x^2

    Do the same with the other set.

    Then, pick up from where Unknown008 stopped.
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