Statistics: Standard Deviation Prob.

1. Jan 19, 2007

Mesmer

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

Last edited: Jan 19, 2007
2. Jan 19, 2007

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

Last edited: Jan 19, 2007
3. Jan 19, 2007

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.

4. Jan 20, 2007

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.

5. Jan 20, 2007

HallsofIvy

Staff Emeritus
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.