Statistics: Standard Deviation Prob.

[Mesmer]

Homework Statement

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. Homework Equations standard deviation= $$: s^2=\frac{\sum_{i=1}^n(x_i -\overline{x})}{n-1}$$ 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]

 

[mattmns]

Your formula is not correct, is that what messed you up? It should be:

$$s^2=\frac{\sum_{i=1}^n(x_i -\overline{x})^2}{n-1}$$

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

$x_i$ represents the data (for example x_1 is 42.4 and x_4 is 58.8)

And $\overline{x}$ is the mean = $$\frac{x_1+x_2+...+x_n}{n}$$

Here is an example.

My data are the following: 14.3, 17.5, 15.6

So the mean, $$\overline{x} = \frac{14.3+17.5+15.6}{3} = 15.8$$

So then

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

Hence,

$$s^2 = \frac{2.25 + 2.89 + .04}{3-1} = \frac{5.18}{2} = 2.59$$

So our standard deviation s is then the square root of that which means s = 1.60935

 

[Mesmer]

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.

 

[2ltben]

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.

 

[HallsofIvy]

By the way since
$$\Sigma_{i=1}^n (x_i- \overline{x})^2= \Sigma_{i=1}^nx^2- \overline{x}\Sigma_{i=1}^n x_i+ \overline{x}^2\Sigma_{i=1}^n 1$$
$$= \Sigma_{i=1}^n x_i^2- 2n\overline{x}+ n\overline{x}= \Sigma_{i=1}^n x_i^2- n\overlne{x}$$
(because
$\Sigma_{i=1}^n x_i= n\overline{x}$
and
$\Sigma_{i=1}^n 1= n$

$$\sigma^2= \frac{1}{n-1}\Sigma_{i=1}^n x_i^2- \frac{n}{n-1}\overline{x}$$

That's often easier to calculate.