Stabdard deviation HHHEELPP

i can do this on a online calculator, but when im doing it myself its total differnt answer

whast standard deviation for these

10,11,11,12,12,12,13,13,13,13,13,13,14,14,14,14,14 ,14,14,14,15,15,15,15,15,15,15,15,15,15,16,16,16,1 6,16,16,17,17,17,17,17,18,18,18,19,19,19,19,20,20

Ive got mode =15

mean =15.18

median =15

some one help me do the standard deviation pleaee.. know this takes time, ive been doing it last 6 hours none stop and cant do it

thanks

Re: Stabdard deviation HHHEELPP

Hey gary147.

Its a lot better if you use a computer for data that is this big in terms of observations.

If multiple packages give the same answer, then that should be evidence that you entered in the data correctly and used the right commands.

Re: Stabdard deviation HHHEELPP

manages to do this now, thanks mate

Re: Stabdard deviation HHHEELPP

The **definition** of "standard deviation" is the square root of the expected value of the square of the difference from the mean. That is, [tex]\{x_i\}[/itex] is the set of numbers, and there are n numbers in the set, the mean is, of course, $\displaystyle \mu=\frac{\sum_{i=1}^m x_n}{n}$.

The "variance", the square of the standard deviation, is $\displaystyle \sigma^2= \frac{\sum_{i=1}^n (x_i- \mu)^2}{n}$. And then, of course, the standard deviation is the square root of that, $\displaystyle \sigma= \sqrt{\frac{\sum_{i=1}^n (x_i- \mu)^2}{n}}$

You may have seen a simplified formula in you text: for any i, $\displaystyle (x_i- \mu)^2= x_i^2- 2\mu x_i+ \mu^2}$ so that $\displaystyle \sum_{i=1}^n (x_i-\mu)^2= \sum_{i=1^n} x_i^2- 2\mu x_i+ \mu^2= \sum_{i=1}^n x_i^2- 2\mu\sum_{i=1}^n x_i+ \mu^2\sum_{x=1}^n 1$$\displaystyle = \sum_{i=1}^n x_i^2- 2\mu (n\mu)+ \mu^2(n)$$\displaystyle = \sum_{i= 1}^n x_i^2- n\mu^2$ and once we have divided by n, $\displaystyle \sigma^2= \frac{\sum_{i= 1}^n x_i^2 x_i^2}{n}- \mu^2$. That is simpler to use because you can go ahead and keep a running sum of both $\displaystyle x_i$ and $\displaystyle x_i^2$ and the same time, determining $\displaystyle \mu$ later.