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Math Help - Help with mean and standard deviation question.

  1. #1
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    Question Help with mean and standard deviation question.

    A summary of 24 observations of x gave the following information:
    ∑(x-a)=-73.2 and ∑(x-a)2=2115
    The mean of these values of x is 8.95.

    (i) Find the value of the constant a.
    (ii) Find the standard deviation of these values of x.

    HELP PLZ!
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  2. #2
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    Quote Originally Posted by Jay Jay View Post
    A summary of 24 observations of x gave the following information:
    ?(x-a)=-73.2 and ?(x-a)2=2115
    The mean of these values of x is 8.95.

    (i) Find the value of the constant a.
    (ii) Find the standard deviation of these values of x.

    HELP PLZ!
    Since no probability is given, we denote the mean of x as \mu = \Sigma x/n

    We know n is 24, and \mu = 8.95. Therefore, \Sigma x = (24)(8.95) = 214.8

    Your problem gives \Sigma (x - a) = -73.2

    That can be rewritten as \Sigma x - a = -73.2.
    a = 214.8 + 73.2 <------ \Sigma x = 214.8 from above
    a = 288

    I'll take a stab at part (ii) as well.

    From above, for your second equation, substitute our value of 288 for a, we get \Sigma (x - 288)^2 = 2115

    Expanding and simplifying, we get:
    \Sigma x^2 - 576\Sigma x + 82944 = 2115
    \Sigma x^2 - 576(214.8) + 82944 = 2115 <------ remember that \Sigma x = 214.8 from part i
    \Sigma x^2 = 42895.8

    Now, we know that Variance, denoted as \sigma ^ 2 = \Sigma (x - \mu)^2

    Plugging in our value for \mu = 8.95 and simplifying, we get:

    \sigma^2 = \Sigma x^2 - 17.9\Sigma x + \Sigma 80.1025
    \sigma^2 = 42895.8 - 17.9(214.8) + 80.1025 <---- we figured out earlier \Sigma x and \Sigma x ^2
    \sigma^2 = 42895.8 - 3844.92 + 80.1025
    \sigma^2  = 39130.98

    Now calculate standard deviation, denoted as \sigma
    \sigma = \sqrt{\sigma^2}
    \sigma = \sqrt{39130.98}
    \sigma = 197.8155
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  3. #3
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