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Homework Help: Statistics: Standard Deviation Prob.

  1. Jan 19, 2007 #1
    1. The problem statement, all variables and given/known data
    This is my data: 42.4, 65.7, 29.8, 58.7, 52.1, 55.8, 57.0, 68.7, 67.3, 67.3, 54.3, 54.0 I need to find the standard deviation of this list of data.

    2. Relevant equations standard deviation= [tex] : s^2=\frac{\sum_{i=1}^n(x_i -\overline{x})}{n-1}[/tex] Actually the standard deviation is the square root of this expression.

    3. I can use a statistics package like minitab or excell to solve this but how would I do it my hand? where would I start?[/b]
    Last edited: Jan 19, 2007
  2. jcsd
  3. Jan 19, 2007 #2
    Your formula is not correct, is that what messed you up? It should be:

    [tex]s^2=\frac{\sum_{i=1}^n(x_i -\overline{x})^2}{n-1}[/tex]

    If that was not the problem then is there something you don't know how to do with the formula? Do you know what [itex]x_i[/itex] and [itex]\overline{x}[/itex] are?

    [itex]x_i[/itex] represents the data (for example x_1 is 42.4 and x_4 is 58.8)

    And [itex]\overline{x}[/itex] is the mean = [tex]\frac{x_1+x_2+...+x_n}{n}[/tex]

    Here is an example.

    My data are the following: 14.3, 17.5, 15.6

    So the mean, [tex]\overline{x} = \frac{14.3+17.5+15.6}{3} = 15.8[/tex]

    So then

    [tex](x_1 -\overline{x})^2 = (14.3 - 15.8)^2 = (-1.5)^2 = 2.25[/tex]
    [tex](x_1 -\overline{x})^2 = (17.5 - 15.8)^2 = (1.7)^2 = 2.89[/tex]
    [tex](x_1 -\overline{x})^2 = (15.6 - 15.8)^2 = (-.2)^2 = .04[/tex]


    [tex]s^2 = \frac{2.25 + 2.89 + .04}{3-1} = \frac{5.18}{2} = 2.59[/tex]

    So our standard deviation s is then the square root of that which means s = 1.60935
    Last edited: Jan 19, 2007
  4. Jan 19, 2007 #3
    Thanks, I did latex my equation wrong, but you also cleared up my confusion about how the expression works. :) I see my mistake in understanding now.
  5. Jan 20, 2007 #4
    The equation is also for variance. If you want to find the actual standard deviation, you'll still need to take the square root of that.
  6. Jan 20, 2007 #5


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    Science Advisor

    By the way since
    [tex]\Sigma_{i=1}^n (x_i- \overline{x})^2= \Sigma_{i=1}^nx^2- \overline{x}\Sigma_{i=1}^n x_i+ \overline{x}^2\Sigma_{i=1}^n 1[/tex]
    [tex]= \Sigma_{i=1}^n x_i^2- 2n\overline{x}+ n\overline{x}= \Sigma_{i=1}^n x_i^2- n\overlne{x}[/tex]
    [itex]\Sigma_{i=1}^n x_i= n\overline{x}[/itex]
    [itex]\Sigma_{i=1}^n 1= n[/itex]

    [tex]\sigma^2= \frac{1}{n-1}\Sigma_{i=1}^n x_i^2- \frac{n}{n-1}\overline{x}[/tex]

    That's often easier to calculate.
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